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1 SARVAJNYA 840942174 ELECTRIC FIELD A charged particle does not interact directly with another charged particle kept at a distance. They interact through an invisible communicating medium, called electric field. It is present every where in the space surrounding the charge. The concept of electric field present around the charge, is analogous to the gravitational field produced by a mass. Electric charge produces an electric field and electric field exerts a force on electric charges kept in it. Strength of Electric Field (Electric Intensity) The strength of an electric field at a point is described in terms of force experienced by a charge kept at that point. Electric intensity or electric field strength at any point is defined as the force acting on an unit positive charge(UPC) kept at that point. It is donated by E. It is a vector quantity. The direction of electric field at a place is same as the direction of the force experienced by a test charge (an unit positive charge) kept at that point. If ‘F’ is the force experienced by a charge ‘q’ kept in the electric field then the strength of electric field ‘E’ at that point is written as, E = F/q or F = q E An electric field in the space is said to be uniform, if a charge kept in it experiences same force at all points & the field is said to be non-uniform, if the charge kept in it experiences different force at different points. In S.I., the electric field strength is measured with unit NC-1 Expression for the electric field at a point due to an isolated charge : +Q X q Field due to + ve charge Y d (a) Q a X Y Field due to a – ve charge Let a positive charge of magnitude Q is kept at X. Let Y is the point at a distance d from the charge, where intensity of electric field is to be determined. An arbitrary charge say q is placed at Y. The force acting on this q can be calculated using Coulomb’s law as follows: 1 Qq F= 2 4 0 d 1 Qq 2 F 4 0 d But from the definitions of electric field strength As E = E q q i.e., 1 Q E= 4 0 d 2 2 SARVAJNYA 840942174 The direction of E is taken from X to Y for a positive charge[ i.e. away form the charge] & for the negative charge field is taken along Y to X [ i.e. towards the charge]. The electric field due to an isolated charge varies with distance ‘d’ as follows: d2 d2 An electric field at a point due to large number of charges can be calculated by taking vector sum of individual electric fields. En Ei E2 If E1 , E2 , E3 ,...... are fields at a point due to charges q1, q2, q3 Then net field at that point is E E1 E2 + ……. This is E3 also called as super position principle of electric fields. ELECTRIC DIPOLE: Electric dipole is an arrangement of two equal and opposite point charges separated by a small distance. Many molecules exist as an electric dipoles. For example the hydrogen chloride molecule, in which hydrogen part shows positive charge and chloride the negative charge, this constitute an electric dipole. The line passing through the charges forming electric dipole is called as a axial line and the line passing through the mid point of dipole axis and perpendicular to the axial line is called as equatorial line of dipole. 2d O The strength of electric dipole is measured in the terms of a quantity called Electric Dipole moment, which is denoted by ‘P’. The magnitude of electric dipole moment of an electric dipole is expressed as product of magnitude of the charge possessed by the particles forming a dipole and the separation between them. If ‘q’ is the magnitude of charge and ‘2d’ is the separations between the, electric dipole moment of dipole ‘P’ is written as, P = q . 2d Dipole moment is a vector quantity. Its direction is taken from negative charge to positive charge. In S.I. it is measured in Coulomb meter ( C-m). The molecules like H-Cl, show permanent dipolement. These molecules are called as Polar molecules. Their dipole moment arises due to asymmetric distribution of positive and negative charges in the molecules. In these molecules, the center of all positive charges do not coincide with the center of all negative charges. However, there are other molecules like H2, N2 etc in which the centers of positive and negative charges coincide. This is due to the symmetric distribution of charges +ve & –ve in the molecules. Such molecules do not exhibit dipolement. They are called as Non-polar molecules. 3 SARVAJNYA 840942174 The molecules that are non polar basically can be forced to become a polar by subjecting them to an electric field. The force exerted by applied electric field disturbs the coincidence of centers of positive and negative charges of the molecule. “The molecule which exhibits dipole property under the action of external field is called as induced dipole”. Electric Filed due to an electric Dipole: Field on axial point of a dipole: Consider an electric dipole consisting of charges +q & -q which are d distance apart. A line passing through the charges forming a dipole is called as its axial line. And a line that acts as a perpendicular bisector of axial line, which passes through center of a dipole is called as its equatorial line. The intensity of electric field at a point on the axial line of an electric dipole is given by, E For a short dipole, x >> d = axial Eaxial = Where ‘P’ represents dipolement and where the field is expected. 1 40 2 Px (x d 2 )2 (1) 2 1 2P 3 4 0 x ‘x’ represent the distance from the center of dipole to the point Electric field on the equatorial point of a dipole: Electric field at a point on equatorial line of a dipole, which is E For short dipole x >> d. equit = Eequit = 1 40 ‘x’ m apart from center of dipole is given by, P (x d ) 2 2 (2) 3 2 1 P . 4 0 x 3 From equations (1) & ( 2) it can be shown that E axial = 2 E equit 4 SARVAJNYA 840942174 Also in both cases, E 1/x3 Electric field due to an electric dipole at any point: Consider an electric dipole made of charges +q and q separated by a distance 2d. Let P be a point at a distance x from the centre of the dipole. Let be the angle made by the axis of the dipole and the line joining the point and the midpoint of the dipole. The magnitude of electric field at P is given by E= 1 P 4 0 x 3 3 cos 2 1 Behavior of an electric dipole kept in External uniform Electric Field: Consider an electric dipole consisting of charges +q & -q which are distance 2d apart. Let the dipole is kept in uniform electric field of strength E as shown in fig. Axis of dipole makes an angle with the field. (dipole moment vector making angle with the field) Each of charge forming the dipole, experience the force of ‘qE’ amount. These forces are oppositely directed. Although the net force on dipole is zero, the dipole experiences the torque due to non zero perpendicular distance between these forces. This torque tends the dipole to rotate and align itself parallel to the applied field. The torque actions on dipole can be calculated as follows: From definition : Torque = Force x perpendicular distance between the force = q E . 2d sin = q .2d . E sin = PE sin …………(1) ( P = q.2d) i.e. = P.E Definitions of Dipole moment: From above expression for the torque, = PE sin Dipole is kept at right angles to the uniform field of unit strength then , for = 90 & E = 1 NC-1 = p x1 x sin 900 = p OR = P Thus dipole moment of an electric dipole is defined as numerical value of, torque experienced by it when it is kept at right angles to an uniform electric field of unit strength Polarization of Dielectric Medium : A dielectric material is one which has negligible number of free charges. They are non-conducting electrically. Dielectric substances are composed of either polar atoms or non-polar atoms. In case of a polar dielectric material, although the individual atoms or molecules exhibit finite non zero dipole moment, because of random orientation of all dipoles, the net dipole moment of entire material turns to be zero. 5 SARVAJNYA 840942174 It is called as an un polarized dielectric material, see fig Un polarized dielectric polarized dielectric When such un polarized dielectric is kept in external electric field, the dipoles of dielectric material experience torque. This torque compels them to align parallel to the directions of field as in fig off course alignment is not 100% perfect, because the thermal agitations in the material tends to disalign some dipoles. This is called as polarization of a dielectric. The polarized dielectric exhibits finite dipole moment. As a result of polarization, there is net negative charge on one side of the dielectric and equal net positive charge on its other side. This establishes small induced electric field ‘Ep’ inside the dielectric (due to polarization) whose direction is opposite to the external field ‘E0’. Thus resultant field in the space, where dielectric is kept, is less than ‘E0’, and it is written as, Ed= E0 – Ep. The quantity dielectric constant (K) for a medium describes, how many fold the electric field in the space is reduced due to introduction of dielectric. IAs, E0 field in the space in the absence of dielectric Ep induced electric field in a dipole due to polarisation. Then the dielectric constant k is written as, K = E0/Ed= E0 / (E0-Ep) Note: If a conductor is kept is electric field, the free electrons in the conductor experience the force and move opposite to the external field. The movement of electrons from one side of the conductor to the other side, establish reverse electric field in conductor. This reverse field grows till it turns equal and opposite to external field. Thus, net field in the space where conductor is kept is zero. Therefore, Ep=E0 OR Ed=E0E0=0. Thus, for a good conductor of electricity, its dielectric constant is, K = E0/0 = In case of a perfect insulator, Ep=0, OR Ed=E0-0= E0. therefore, its dielectric constant is, K = E0/ E0 = 1 Polarizability: For a single molecule induced electric dipole moment ( p ) is proportional to the applied electric field (E) i.e., p 0 E p = 0 E Or where is a constant and is called atomic or molecule polarizability and is given by 6 SARVAJNYA 840942174 = P 0 E SI Unit of = Unit of p Cm 2 1 2 m3 Unit of 0 unit of E C N m NC 1 Thus unit of is m3 (i.e., the unit of volume). Dielectric Strength of a medium : ‘The electric dipole moment acquired per unit volume of a dielectric is called as polarization’. It is denoted by ‘p’ The polarization ‘p’ of a dielectric can be increased, with the strength of external field. This continues till the external field takes a critical value. Beyond this value, the electrons from the molecules or atoms of dielectric material, get themselves detached, because of strong force exerted by field. This stage of dielectric is called as breakdown. As result of breakdown, dielectric loses its insulation property and starts conducting. The maximum strength of external field that can be applied to a dielectric without causing a breakdown is called as dielectric strength of the medium. Esafe = Vsafe d Where Vsafe is the maximum potential that can be applied to a dielectric of thickness d without its electric breakdown. * Dielectric strength of air medium is 3x106vm-1 Electric Lines of Force : Electric lines of Force are imaginary, hypothetical invisible lines, which are used to visualize the electric field due to a charge or group of charges. The concept of lines of force was introduced by Faraday. They exist in 3 dimensional space around the charge. Electric line of force in an electric field is an imaginary curved line such that the tangent drawn to it at a given point gives, the direction of electric field at that point. Properties of Lines of Force: 1. They originate from the surface of an isolated positive point charge and are directed radially outward. The lines of force, terminate on an isolated negative point charge and are directed radially inwards. 2. The tangent drawn to the electric line of force at any point gives the directions of electric field at that point. Line of force represents a path on which a test positive charge moves or tends to move. 3. The no. of lines of forces per unit area perpendicular to the direction Electric field, is equal to the intensity of electric field at that point. 7 SARVAJNYA 840942174 4. Electric lines of force, do not intersect with each other. That is, they do not meet at a common point. If it happens so, then it leads to the following contradiction. If two lines of forces intersect at a point, then at the point of intersection, one can draw two tangents, one tangent to each line of force. It means that, a test charge kept at this point will move simultaneously along two directions which is impossible! 5. We know that, electric field inside good conductor is zero. Therefore the lines of force do not exist inside it. However, in case of an insulator or a dielectric, the electric field does exist in it. Hence the lines of force also can exist in it. 6. Electric lines of force are perpendicular to the surface of a charged body. 7. Electric lines of force exert, lateral pressure on one another. This explains repulsion between like charges. 8. Lines of force shows longitudinal contractions. This explains attraction between unlike charges. Lateral Pressure Longitudinal Contractions. Electric Flux: Electric Flux is the measure of the no. of electric lines of force, penetrating some surface. The no. of lines of force penetrating a surface, in turn is directly proportional to the charge within that surface. Electric Flux of electric field through a surface is defined as the no. of lines of force passing through it. If the lines of force are passing normally then it called as Normal flux. It is denoted by . Consider small plane area ‘ds’ in the electric field, which is normal to the direction of electric field of strength E. The Normal flux through ds is given as d= E. ds ……….(1) If is angle between normal to the surface and Field E, then flux through the surface ds is, d= E . ds Cos ………….(2) Electric Flux is a scalar quantity. If the directions of lines of force, are directed out of the given surface then the flux is taken positive and if lines of force are directed into the given surface flux is taken negative. Area vector: Area is a scalar quantity. But in certain situations are of a surface has to be represented by a vector so as to specify it completely. The area vector is represented by a vector along the normal to the area. The length of the vector is proportional to the magnitude of the area. Its direction indicates the orientation of the area. Gauss Theorem : Gauss Theorem relates the Electric Flux of electric field and the charge. It states that total normal Electric Flux through any closed surface is equal to enclosed by it. (1 / ) times the net charge 8 SARVAJNYA 840942174 = 1 xQ Where Q = q i = total charged enclosed. Explanations of Gauss Theorem : Consider a closed surface of any shape S as shown in the adjacent fig, which contains the charges q1, q2…..qn. Let surface S is divided into large number of elementary areas ds1, ds2, ds3 ……..etc. Let E1, E2….. etc are the strengths of electric fields on these surfaces respectively. The individual normal flux through these elementary areas can be given as, d1 = E1ds1 cos1 d2 = E2ds2 cos2 .. .. .. and so on. Then the total normal Electric Flux through the entire surface is given by total = d1 + d2 +…. = dn1 According to Gauss theorem this total normal flux ie. total is equal to ( 1/ enclosed. i.e., total = 0) times net charge 1 . Q , Where Q = q1 +q2 +….. E ds Cos = OR 1 .Q Applications of Gauss theorems 1. Electric intensity near a charged conductor: Consider a charged conductor of any shape. It is because of its irregular shape, the density of charges at various places, on its surface, is different Let p is the point close to the surface at which intensity of electric field E is to be calculated. Let a cylindrical Gaussian surface is constructed, which consists of point p on its circular top surface, of area ds. part of cylinder lies inside the charged conductor and the other part outside. As ‘p’ is close to the charged conductor, the electric field at point p is perpendicular to the surface of conductor just below it. As there is no electric field inside the conductor the flux of electric field through the part of cylinder inside the charged conductor is zero. The electric flux through the cylindrical region of Gaussian surface is also zero, as there is no any component of electric field perpendicular to it. Thus, net normal flux through entire cylindrical Gaussian surface is , = E . ds … (1) Using Gauss theorem = [ 1/] . ( charge enclosed say Q ) …. (2) if ‘’ is surface density of the charge on the conductor at a place just below point P then, charge enclosed by Gaussian surface will be, Q = . ds Substituting in equation (1) = 1 . ( . ds) E= 9 SARVAJNYA 840942174 comparing equation (1) and (3) E . ds = 1 0 ( . ds) , Or Thus electric intensity at a point near charged conductor is directly proportional to the charge density at that point. 1. Electric Intensity near a charged spherical conductor : Consider a spherical conductor of radius R which is given with a charge Q. the charge gets uniformly distributed over the surface of the sphere because of its symmetrical shape. Let P is the point at distance d from the sphere where electric intensity is to be determined. Case1: Point is outside the conducting sphere. Let us construct a spherical Gaussian surface of radius d, that encloses the charged sphere and contains the point P. Because of symmetrical shape of charged sphere, the electric field at all points on Gaussian surface is same in magnitude and is radially outwards from the centre O. The electric flux through small area ds around point p is given as d= Edscos as the field every where is perpendicular, = 00, d = Eds … (1) The total normal electric flux through the spherical Gaussian surface is equal to, = E.∑ds or = E . ∑ds but for spherical Gaussian surface of radius d ∑ ds = 4d2 = E . (4d2) …. (2) As we know, total charge enclosed by Gaussian surface is equal to the charge given to the sphere i.e. Q. then the above total normal flux can be also obtained by Gauss theorem as, = (1/ 0 ). Q …. (3) Comparing (2) and (3) E . 4d2 = 1 .Q 0 or E = 1 4 0 Q d2 This expression is same as expression for the field at a point due to a point charge Q kept at O. Thus it can be concluded that, for a point outside the charged sphere, it appears that as if the total charge on sphere is concentrated at its centre. Case 2 : Point is outside & close to surface of the conducting sphere If P is close to surface of charged spherical conductor , then , the distance of point P can be taken equal to the radius of the sphere( this is due to the reason that the radius of the Gaussian surface drawn in this case, will be same as the radius of the sphere). Thus substituting d = R , expression for the electric field at a point close and outside the charged spherical conductor is written as, E = Case 3: 1 40 Q R2 Point is inside the conducting sphere Let point P be inside the sphere which is at a distance d its center, as shown in the fig. A Gaussian surface is constructed consists of point P. if E is the electric field intensity at P, then field from which at 10 SARVAJNYA 840942174 every point on Gaussian surface is, = E . 4d2 From the Gauss theorem, = 1 0 (ch arg e enclosed ) As in this case Gaussian surface does not enclose the charge, therefore Q = 0 =0 E . 4d2 = 0 OR E=0 “Thus electric field at every point inside the charges conductor is zero “ The adjacent figure shows variation of electric field at a point near a charged spherical conductor: 3. Electric field near a charged plate Consider thin plate with surface area A. If charge Q is given to it, then the surface density of Q charge on the plate is given as, = . As the total charge given to it is equally distributed on its A both surfaces the charge density on each surface is equal to, . 2 Electric intensity at a point close to such plate is given as, E 2 0 Electric potential: In the topic of gravitation, it is have learnt that, a mass kept in gravitational field experiences force. If the test mass is moved in the gravitational field, the work is done. This work done is stored in it, in the form of gravitational potential energy. ( even in the absence of test mass, every point in gravitational field, has capacity to exert force on a mass kept in it, thus every point in gravitational field is said to have gravitation potential ) In analogy to gravitation, it is observed that, every point in electric field has a capacity to exert force on a test charge kept in it. Thus a charge kept in electric field acquires an energy called electric potential energy. If a test charge is moved from one place to another place in the electric field, against the direction of field, then there is a work done. This work done itself is stored in the charge, in the form of its potential energy. The potential energy stored in a charge at infinite distance from the source of electric field is taken zero, since it does not experience force. This place of zero potential acts as a reference level for measurement of potential energy of a charge. The electric potential at a point in the electric field, is defined as amount of work done in bringing an unit positive charge, from infinity to the concerned point in the electric field. If the work is done against the field, then the Potential at that point is taken positive and if work is done by the field itself then potential is taken negative. Electric potential at a point in the electric field is also defined as, the potential energy stored in an unit charge kept at that point If w is the work done in bringing a charge Q from infinity to the point in the electric field the from definitions he electric potential at that point is, v = w/Q 11 SARVAJNYA 840942174 In S.I. it is measured with the unit volt. Electric potential at a point is said to be one volt, if one joule of work is done in bringing an Unit Positive Test Charge from infinity to that point against the field. Electric Potential at a point due to an isolated point charge. Let Q be a positive point charge placed at O. Its electric field is outwardly directed and extends towards infinity as shown in the fig 1. P is the point where electric potential is desired. X and Y are two points in the electric field which are separated by a small distance dx such that an unit positive charge experiences almost same force at X and Y. Let OX=x Let an unit positive charge is kept at X. The force F that acts on this UPC is, F= 1 40 Q X2 …(1) In order to displace it against this force, an external force of same magnitude but in opposite direction is to be applied on the UPC ( i.e. – F ). This applied force is such that there is no acceleration of UPC. The work done in displacing UPC from X to Y is given as, dw = - F . dx ..(2) Negative sign in expression also represent that, the UPC is displayed against the electric field. Using (1) dw= - 1 40 Q . dx X2 …(3) Electric potential at Point P is the total work done in bringing UPC from to point P. To find this total work done, equations (3) can be integrated between the limits x = and x =d p W= dW = dW = dW = dW = xd x 1 40 Q 40 xd x Q dX X2 1 dX X2 1 1 4 0 X Q 1 1 40 d Q 1 x dW = 4 0 d p But from the definition of electric potential , dW = V, therefore, V= 1 4 0 Q d Relation between electric filed and electric potential : Let A and B are the two points in electric field, produced by point charge +Q. the distance between A& B points (dx) is so small that UPC kept at A and B experiences same force (F) i.e. Electric field E from A to B is same. From definition, the potential difference between A and B points ( dv) is equal to work done in moving UPC from A to B. 12 SARVAJNYA 840942174 Or dv = dw = F . dx dv = - E . dx = - E . dx E = - ( dv/dx) The quantity (dv/dx) is called as potential gradient. It is defined as rate of fall of potential with the distance. Thus electric intensity at a point in an electric field is equal to the negative potential gradient at that point. Differences between electric field and electric potential : Electric field 1. Electric field at a point is defined as force experienced by UPC kept at that point. 2. Electric field is a vector quantity. 3. It is measured in NC-1 4. Electric field at point due to a point charge is E= 1/40 x Q/d2 5. Electric field inside charged spherical conductor is zero. Electric Potential 1. Electric potential at point is defined as amount of work done in bringing UPC from infinity to that point. 2. Electric potential is a scalar quantity 3. It is measured in volt or JC-1 4. Electric potential at a point due to point charge is V = ¼0 x Q/d 5. Electric potential inside charged conductor is constant. Electric potential at a point due to large no. of charges. Electric potential is a scalar quantity. Therefore the total electric potential at a point due to group of charges can be obtained by adding the electric potentials due to individual charges. Let V1 is the potential at a point P due to charge Q1 at a distance r1, V2 due to charge Q2 at distance R2 and so on at P is, V = V1+V2+…..Vn i.e. V q1 1 q2 1 qn . ....... 40 r1 40 r2 40 rn 1 V 1 40 i n qi i 1 i r Electric potential at a point due to an electric dipole: Electric potential at a point, at a distance of ‘x’ from the center and on the axis of electric dipole is given by the following formula, V= 1 40 P (x d 2 ) 2 Electric potential at any point on its equatorial line is zero. It is due to the reason that every point on equatorial line of electric dipole is equidistant from the opposite point charges forming the dipole. Hence equatorial line of dipole is also called as zero potential line. Equipotential surface: Any surface which has the same electrostatic potential at every point is known as equipotential surface. An equipotential surface may e the surface of a body or simply a surface in space near a charged body. Equipotential surfaces cane drawn properties of an equipotential surface: 1. No work is done in moving a test charge from one point to another in an equipotential surface. 2. Electric field is always perpendicular to the equipotential surface. 13 SARVAJNYA 840942174 3. 4. The equipotential surfaces help in determination of strong electric fields from weak electric fields. Equipotential surfaces will never intersect. Potential energy of system of charge: Group of charges forms a charge system. As the charges in a group interact each other, the charge system posses potential energy. If charges in the charge system, maintain infinite separation from each other, the PE of system is zero. To remove a charge from the charge system or to add an external charge to the system, work is necessary to be done. The potential energy of system of charges is defined as the amount of work done in bringing all the charges from infinity to the present positions in the system . Explanation : let q1 and q2 are two charges forming a system are separated by a distance d as shown in the fig. Let V is the potential at the point where Q2 is placed. Let q1 and q2 are two charges are kept at ‘0’ and the other at . As they do not interact potential energy of either charges is zero. Let charge q2 is brought from to a point at a distance d from charge q1. The work done in this case can be shown as W 1 q1 q 2 40 d This work done is stored as PE of system of two charges. If both of the charges are of same type PE of system is positive and if both charges are of opposite type PE of system negative. If number of charge are more than two, then total PE of system of charges is given as PE 1 40 qi q j d ij The summations is taken over all possible pair of the number of charges forming the charge system. **** 14 SARVAJNYA 840942174 QUESTIONS’ BANK ELECTRIC CHARGES: Questions carrying one mark each: 1. How many electrons make one coulomb of electric charge? [Model QP., Oct. 2002] 2. The repulsive force between two electrons is F. How will the force change if the electrons are replaced by protons? [Model QP.] 3. What is the practical unit of electric charge? [Oct. 83, 86, April 85] 4. Define unit charge. [April 84, 88] 5. Define surface density of charge. [Oct. 84, 2001] 6. If the shape of an object is non-uniform, where will the surface charge density be maximum? [April 89,90] 7. In textile mills, the atmosphere is rendered humid. Why? [Oct. 91] 8. How many electrons are to e added/ removed to charge a body to +1 coulomb? (Charge of an electron = 1.6x1019 C) [April 92] 9. In electrostatics, gravitational forces are not taken into account. Why? [Oct. 92] 10. Force between two point charges is 0.5 N. If the distance between them is doubled what will e the force between them? [April 95] 11. Mention one of the methods of charging a body. [April 99] 12. How many electrons do constitute a charge of 1 C? [April 2000,2003] 13. How does the surface density of charge depend on the curvature of the surface? [April 2000] 14. How many electrons make 1 nanocoulomb of electric charge? [April 2001] 15. Two spherical droplets have equal surface density of charge. If they are combined to form a single droplet, what happens to the surface density of charge? [April 91] 16. What is the effect of dielectric on a force between two charged spheres separated by a distance? [Oct. 98] 17. What happens to the magnitude of the force between two charges if glass plate is interposed between them? [April 2007] Questions carrying two marks each: 1. State and explain Coulomb’s law in electrostatics. [MQP Apr. 88, 05, July 2006] 2. Distinguish between conductor and insulator. [April 85] 3. What is the effect of dielectric on a force between two charged spheres separated by a distance? [Oct. 98] 4. Write the expression for Coulomb force between two point charges separated by a distance in vector form and explain the terms. [Apr. 2006, M.Q.P] 5. What is an electric dipole? Define dipole moment. [July 2006] Questions carrying five marks each: 1. Two small spheres each of mass 10 mgm are suspended from a point by threads 0.5 m long. They are equally charged & repel each other to a distance of 0.28 m. If g=10 m/s 2, what is the charge on each? [Apr. 92] 2. Two identical pith balls each weighing 30 mg are hung from the same point of a rigid support by two in extensible threads each of 0.10 m. When equal amount of identical charges are given to the pith balls they get separated by 0.10 m. Calculate the charge on each of them. (G=10 m/s 2) [April 95] 3. Three positive charges 3 nC, 4 nC and 5 nC are placed at the vertices A, B and C respectively of an equilateral triangle ABC of side 0.2 m. Find only the magnitude of the force on the largest charge. [April. 2002] 4. Two point charges of +3nC and 3nC are placed at corners A and B of an equilateral triangle ABC of side 0.3 m in air. If a charge +2nC is placed at the point C, what is the force acting on it? [Oct. 2002] 5. Two small metal spheres are charged so that they repel each other with a force of 2x10 5 N. The charge on one sphere is twice that on the other. When they are moved 0.1 m farther apart, the force reduces to 5x106N. What are the charges and what is the initial separation between them? [April 97] 10 10 10 6. ABCD is a square of side 1 m. Point charges of +2x10 C, 4x10 C and +8x10 C are placed at corners A, B, C respectively. Calculate the work done in transferring a charge of 10 C from D to the point of intersection of diagonals. [July 2006] 7. A metal sphere of radius 0.01 m is charged to a potential of 45 kV and kept at the corner B of a rightangled triangle ABC (AB=0.3 m, BC=0.4 m and ABC=90). Calculate the work required to shift a charge of 2x106 C from corner C to corner A of the triangle. [April 2007] 15 SARVAJNYA 840942174 ELECTROSTATIC FIELD: Questions carrying one mark each: 1. A charged spherical conductor is equivalent to a point charge at the centre with respect to a point outside the conductor, explain. [Model Q.P.] 2. What is the potential of the earth? [Oct. 88, April 95] 3. Two point charge Q and 4Q are separated by a distance of 12 cm. Find the distance of the neutral point from Q. [Oct. 89] 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. o Find the magnitude of electric field due to a point charge of 1.6x1010 C at a distance of 1 A from it. [Oct. 89] If 4x104 joules of work is done in moving a charge of 5 microcoulomb from a point A to another point B, then find the potential difference between A and B. [April 90] A potential difference of 300 mV is applied between two plates separated by a distance of 1 mm. What is the electric intensity in the region between the plates. [Oct. 90] What is the magnitude of electric intensity inside a charged hollow sphere? [Oct. 90, April 97] If a dielectric is introduced between two charges what happens to the electric field between them?[April 91] If the electric field at a distance r from a charge q is E, what is the electric potential at that point. [April 91] Find the electric intensity due to a point charge of 10 mC at a distance of 3 cm. [April 93] Three charges +3C, +2C and 5C are placed at the vertices of an equilateral triangle of side ‘d’ m. What is the potential at the circumcentre of the triangle? [April 93] A hollow spherical conductor of radius 0.1 m is charged. At which point is the electric intensity due to this conductor maximum? [Oct. 95] At which point is the electric intensity due to a charged sphere is zero? [April 96] State Gauss’ theorem. [Oct. 97, April 02, 2006] What is an electric line of force in an electric field? [April 98] Is electric potential a scalar or a vector? [Oct. 2002] A spherical surface surrounds a point charge Q. What happens to the total flux throught he surface when the surface is changed to a cube? [Oct. 2003] Mention the S.I. unit of electric intensity. [Oct. 2000] Can two electric lines of forces intersect? [April 2004] The potential at any point inside a hollow charged spherical conductor of radius 0.2 m is 1 V. What is the potential on the surface of the conductor? [Oct. 04] What is an electric dipole? Mention the S.I unit of electric field. [Apr. 2005] What is an equipotential surface? [July 2006] Questions carrying two marks each: 1. Electric intensity inside a charged conductor is zero this follows from Gauss’ theorem. Explain.[Model Q.P.] 2. Calculate the potential due to a charge of 100 microcoulomb at a distance of 9 metres. [Oct. 84] 3. Obtain an expression for the electric intensity at a point due to a point charge. [Oct. 93] 4. State and explain Gauss’ theorem in electrostatics. [April 96, 2007] 5. Explain why the work done in moving a charge on equipotnetial surface is zero. [April 99] 6. State Gauss theorem and mention one of its uses. [Model QP] 7. What is electric dipole moment? What is its unit. 8. Show that a dipole is placed in a electric field experiences a torque. 9. Give the expression for electric potential due to a system of three point charge. 10. Obtain the expression for capacity of a spherical conductor. [Apr. 2006] Questions carrying four/ five marks each: 1. State Gauss’ Theorem. Obtain an expression for the electric field at any point near the surface of a charged conductor of irregular shape by applying Gauss’ theorem. [Model. Q.P., Oct. 97, Apr. 2003] 2. Define electric intensity and electric potential. Arrive at the relation E = dV/dx. [Model Q.P.] OR Define electric intensity and electric potential. Deduce the relation connecting them. [Oct. 88, 89, 91, Apr. 04] 3. State Gauss’ theorem in electrostatic. Hence derive the formula for the electric field at a point outside a uniformly charged Spherical conductor. [April 83, Oct. 86] OR Obtain an expression for electric intensity at a point outside a charged spherical conductor by using Auss’ theorem. [April 83, Oct. 99, 2001] 4. Derive an expression for the electric potential at a point due to a point charge. [Oct. 83, 87, 88, 2000, 04, April. 97, 98, 2000, 02, July 2006] 16 840942174 5. 6. SARVAJNYA Distinguish between electric intensity and electric potential at a point in an electric field. Obtain the relation between them. [April 87, 2002] Write the expression for electric field at any point due to an electric dipole using this expression deduce the expression for the electric intensity at any point on (i) On the axis of the dipole (ii) the perpendicular bisector of the axis of the dipole. CET QUESTIONS 1. 2. 3. 4. 5. 6. 7. A body a) b) c) d) A body a) b) c) is said to be charged if. it attracts light objects it repels light objects it attracts heavy objects it repels heavy objects is said to be positively charged if. it has no electrons at all it has only protons it has more number of electrons than protons d) it has less number of electrons than protons The sure test to know whether a body is charged or not is. a) repulsion b) attraction c) both 1 and 2 d) neither 1 nor 2 A body is said to e negatively charged if it has. a) only electrons b) only protons c) less number of electrons than protons d) less number of protons than electrons Charges are. a) quantized b) not quantized c) always positive integral multiples of the least unit of charge d) always negative multiples of the least unit of charge The smallest unit of charge is equal to. a) 1.6x1019C b) 1.6x1019C c) 1C d) 1 C To charge a body to +1C. a) 8. 9. 6.25 x 1018 electrons have to be added to the body b) 6.25 x 1018 electrons must be removed from the body c) one electron must be added to the body d) one electron must be removed from the body To charge a body to 1 C. a) 6.25 x 1012 electrons must be removed from the body b) 106 electrons must be removed from the body c) 106 electrons must be added to the body d) 6.25 x 1012 electrons must be added to the body. When a glass rod is rubbed against silk. a) glass rod loses electrons and becomes negatively charged b) silk loses electrons and becomes positively charged c) glass rod loses electrons and becomes positively charged d) silk loses electrons and becomes negatively charged 10. When an ebonite rod is rubbed against fur. a) ebonite rod loses electrons and becomes negatively charged b) fur loses electrons and becomes negatively charged c) ebonite rod gains electrons and becomes positvely charged d) ebonite rod gains electrons and becomes vely charged 11. When a positively charged body is made to touch an uncharged conductor. a) the uncharged conductor also becomes positvely charged b) the uncharged conductor becomes negatively charged c) neither 1 nor 2 12. When a body is charged by friction. a) +ve charges are newly created b) ve charges are newly created c) charges are conserved. They are simply transferred from one body to the other d) a body cannot be charged by friction 13. A charged body induces. a) an equal amount of the same kind of charge b) an equal amount of the opposite kind of charge c) no charge d) charge depending on the temperature of the body 14. When a conductor is charged. a) all the charges are localised in the central region b) all the charges are distributed on the surface of the conductor c) more charges are localised at the centre than on the surface d) none of the above 15. When an insulator is charged. a) the charge remains localised b) the charges are distributed over the surface c) either 1 of 2 depending on the shape of the insulator d) an insulator cannot be charged 16. The distribution of charges on a conductor is such that. a) it is always uniform irrespective of the shape of the body b) more charges reside at points where the curvature is less c) more charges reside at points where the curvature is more none of the above 17. The surface density of charge is defined as the ratio of the total charge of the body to its. a) length b) surface area c) volume d) cross-sectional area 17 SARVAJNYA 840942174 18. Repulsion exists between. a) a positvely charged body and another positvely charged only b) a positively charged body and an uncharged body c) a negatively charged body and an uncharged body d) two uncharged bodies 19. The force ‘F’ between two charged bodies separated by a distance ‘d’ is such that. 1 a) F b) F d d 1 c) F d2 d) F d2 20. The force between two charges. a) depends on the medium between them b) does not depend on the medium between them c) depends on the medium when it is a liquid only d) depends on the medium when it is a gas only 21. The force between two charges in different media are different because. a) different media have different densities b) different media have different viscosities c) different media have different permittivites d) different media have different permeabilities 22. Relative permittivity of a medium r is related to the permittivity of the medium and the permittivity of vacuum as follows: a) r = 0 b) r = . 0 c) T = d) 0 = . r 0 23. The force between two charges is. a) high if the medium in between the two charges has a high permittivity b) high if the medium in between the two charges has a low permittivity c) independent of the permittivity of the medium d) none of the above 24. The permittivity of free space is. a) 8.85 x 1012 F.m1 b) 8.85 x 1018 F.m1 c) 8.85 x 1018 F.m1 d) 8.85 x 1012 F.m1 25. The relative permittivity of water is 80. This means that. a) the force between two charges in water is 80 times less than that in vacuum at the saem distance b) the force between two charges in water is 80 times less than that in vacuum at the same distance c) relative permittivity has nothing to do with force between two charges 26. n identical mercury droplets charged to the same potential V coalesce to form a single bigger drop. The potential of the new drop will be. (CET 2000) a) n2/3 V b) nV2 c) nV d) V/n 27. Two pith balls carrying identical charges are suspended from a point in air. Afterwards they are 28. 29. 30. 31. 32. 33. suspended inside water. The distance between the two pith balls is now. a) increased b) reduced c) unchanged d) doubled When two charges +1C each are separated by a distance of 1 m in air, they. a) attract each other with a force of 1 N b) repel each other with a force of 1 N c) attract each other with a force of 9x109 N d) repel each other by a force of 9x109 N Charge on a body can be detected using. (CET 1983) a) an electroscope b) an electrometer c) a voltmeter d) an ammeter When a glass plate is introduced between two bodies, the force between them. (CET 1984) a) increases b) decreases c) remains the same d) becomes zero Two positive charges are placed with a fixed separation. A slab of dielectric medium is introduced between them. As a result, the repulsion between the charges. (CET 1985) a) changes to attraction b) remains constant c) decreases d) increases The intensity of an electric field at a point is defined as. a) the force experienced by a charge of +1C placed at that point b) the force experienced by a charge of 1 electron placed at the point c) the force experienced by a proton placed at that point d) the force experienced by a neutron placed at that point The intensity of an electric field at a point at distance ‘d’ from a point charge q is given by E= a) q2 4 0 d b) c) q 4 0 d d) 4 0 q d q 4 0d2 34. The unit of electric intensity is. a) Nm b) NC c) N.C1 d) N.m1 35. An electric field is said to be uniform if. a) the intensity at all pints in the field is the same b) the direction at all points in the field is the same c) neither 1 nor 2 d) both 1 and 2 36. A charge q when placed in an electric field of intensity E experiences a force F given by. a) F = qE2 b) F = q2E q c) F = d) F = qE E 37. A proton and an -particle are subjected to the same electric field. If a1 and a2 denote their respective accelerations, then a1/a2 is equal to. a) 1:1 b) 1:2 c) 2:1 d) 1:4 38. If is the surface density of a charged plane sheet, the electric intensity at a point t a distance ‘r’ very close to it is. (CET 1984, 1985) 18 SARVAJNYA 840942174 39. 40. 41. 42. 43. 44. 45. 46. 47. 4q a) b) r r a 1 c) d) 4 r 4 qr The electric potential at a distance ‘r’ from a point charge q is given by V = (CET 1984) qr 4q a) b) 4 r q 1 c) d) 4 r 4 qr The unit of electric potential is. a) the same b) V/2 c) 1 Volt d) 1 Farad The electric potential at a point distant r from a charge ‘q’ is V. When the charge is replaced by a charge 4q, the potential at the point will be. (CET 1984) a) the same b) V/2 c) 2V d) 4V Charges present on the clouds are due to. (CET 1984) a) motion of water drops b) earth’s magnetic field c) lighting d) motion of the clouds A deuterium nucleus and a helium nucleus are placed in the same electric field. The acceleration of helium is. (CET 1985) a) greater than that of deuterium b) less than that of deuterium c) equal to that of deuterium d) zero Inside a hollow charged spherical conductor, the electric field is found to be. (CET 1985) a) zero b) a function of the area of the sphere c) proportional to the distance from the centre d) a function of the charge density of the sphere Consider the electric potential due to a point charge at a distance. The potential due to eight times this charge at four times the above distance is. (CET 1985) a) half the original potential b) twice the original potential c) the same as the original potential d) four times the original potential The accumulation of charge on clouds, which produces lighting, is caused by. (CET 1985) a) ran drops changing into electrons b) the electric field of the earth c) ionization of the sun d) electrification due to motion of water molecules State which of the following statements is true. (CET 1985) a) the number of times electric lines of force cross depends on the charge distribution b) no two lines of force intersect each other c) two lines of force intersect each other at lest once d) lines of force in a dielectric medium can intersect each other 48. A closed surface contains negative charges. Electric flux across it is. a) zero b) directed inwards c) directed along the normal to the surface outwards d) parallel to the surface 49. Four charges of magnitude 1 nC, 4nC, 3 nC and 5 nC are placed at the corners A, B, C and D of a square ABCD of side 6 m. The potential at the centre of the square (in volts) is. a) 7.5 b) 15 2 2 c) 7.5 2 d) 1.5 2 50. Two point charges +2C and +6C repel each other with a force of 12 N. When a charge q is added to each of the charges, they attract each other with a force of 4 N. Value of q is (in C). a) +4 b) 4 c) +1 d) 1 51. A slab of dielectric is introduced between two equal negative charges with a fixed separation. As a result. (CET 1987) a) the force between the charges decreases b) the slab gets heated up c) the two charges attract each other d) an electric current passes through the slab 52. A hollow spherical conductor carries negative charge. A positive charge is placed at the centre of the sphere. Then this positive charge will. (CET 1987) a) oscillate between the opposite points of the conductor b) stick to the conductor c) move in circle d) stay at the centre 53. Point charges +50 C, +100C and 75 C are placed on the circumference of a circle of radius 0.5 m to form an equilateral triangle ABC. The potential at the centre of the circle is (Cet 1988) a) 150 C/m b) 450 C/m c) 150 C/m d) none of these 54. The mass of a proton is bout 2000 times the mass of an electron. An electron and a proton are injected into a uniform electric field at right angles to the direction of the field with the same initial kinetic energy. Then (CET 1988) a) the electron trajectory will be less curved than that of the proton. b) The proton trajectory will be less curved than that of the electron c) Both the trajectories will be equally curved d) Both the trajectories will be straight 55. Two fixed point charges +4q and q units are separated by a distance ‘a’. The point where the resultant field intensity is zero is. 2 a 3 3a c) 2 a) b) a 2 d) none of these 56. To move a unit +charge from one point to another point on an equipotential surface (CET 1990) a) work is done b) work is done on the charge c) no work is done d) work done is a constant 57. Potential at any point inside a charged hollow sphere. 19 SARVAJNYA 840942174 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. a) increases with distance b) is a constant c) no work is done d) work done is a constant Two point charges +2C and +6C repel each other with a force of 12 N.If a charge of 2C is given to each of these chargse, what will be the force now? (CET 1991) a) zero b) 8 N (attraction) c) 8 N (repulsion) d) none of these 10 Coulombs of charge are situated at each of the vertices of a 1 cm cube. Then the electric field at the centre of the cube is. (CET 1991) a) 5 N/C b) 10 N/C c) 40 N/C d) zero P and Q are two points lying on the perpendicular bisector of the line AB. Work done in taking a charge of 5 nC from P to Q. (CET 1995) a) depends only on the charge shifted b) is zero c) depends on the distance PQ d) depends on the distance AB A conducting sphere of rdius 10 cm is charged with 10 C. Another uncharged sphere of radius 20 cm is allowed to touch it for enough time. After the two are separated, the surface density of charge on the two spheres will be in the ratio (CET 1995) a) 2:1 b) 1:1 c) 4:1 d) 3:1 n identical mercury droplets charged to the same potential V coalesce to form a single bigger drop. The potential of the new drop will be (CET 1995) a) nV b) V/n c) n2/3V d) nV2 The electric field intensity due to a hollow spherical conductor is maximum (CET 1996) a) outside the sphere b) on the surface of the sphere c) at any point inside the sphere d) only at the centre of the sphere A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10V. The potential at the centre of the sphere is. a) Zero b) 10 V c) the same as that at a point 5 cm away from the surface d) the same as that at a point 25 cm away from the surface Two equal negative charges q are fixed at points (O, q) and (O, a) on the y-axis. A positive charge Q is released from rest at the point (2a, O) on the Xaxis. The charge Q will a) execute simple harmonic motion about the origin b) move to the origin and remain at rest there c) move to infinity d) execute oscillatory but not simple harmonic motion Three point charges 4q, a and Q are placed in a straight line of length 1 at points ½ and zero. The net force on the charge q is zero. The value of Q is. a) q b) 2q c) q/2 d) 4q Two point charges 12 C and 8 C respectively are placed 10 cm apart in air. The work done to bring them 4 cm closer is. a) zero b) 3.8 J c) 4.8 J d) 5.8 J 68. The work done in carrying a charge q once round a circle of radius r with a charge Q at the centre is. a) qQ/ 40r b) qQ/ 40r c) qQ/ 40 (1/2r) d) zero 69. A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q= a) Q/2 b) Q/2 c) + Q/2 d) + Q/4 70. If dV is the potential difference between two points separated by a distance, then electric intensity E is given by. dV dx dx c) E = dV a) E = dx dV dV d) E = dx b) E = 71. The electric potential in a region along along the Xaxis varies with x according to the relation V (x) = 5+4x2. Then a) p.d. between the points x=1 and x=3 is 32V b) force exerted by a charge of 1 C at x=1 m is 8 N c) the force exerted by the above charge is along the +ve X-axis. d) a uniform electric field exists in this region along the x-axis 72. Inside a uniform charged spherical conductor, the electric a) field is zero every where b) potential is zero everywhere c) potential is same as at any point outside d) field has the same magnitude everywhere but it is not zero 73. The distance of closet approach between two protons in vacuum is 1014 metre and the force between them is 2.31N. The charge of a proton is. a) 1.6x1017C b) 1.6x1018C c) 1.6x1019C d) 2.31x1014C 74. A charge A of 25 C is placed on the line between two charges B of 5 C and C of 30C. The charge is 0.05 m from B and 0.1 m from C. What is the force on A? a) 1.125 N b) 11.25 N c) 112.5 N d) 1125 N 75. What is the strength of the electric field such that an electron placed in the field would experience an electrical force equal to its weight? [Charge of the electron = 1.6 x 1019 C, mass of the electron = 9.1 x 1031 kg] a) 5.57x1010 N/C b) 5.57 x 109 N/C 8 c) 5.57 x 10 N/C d) 5.57 x 1011 N/C 76. A and B are two small spheres with charges 9 C and 16 C. The distance between them is 0.28 m. How far from A along the line AB, will the intensities due to the two charges be equal? a) 0.12 m from A in between A and B b) 0.84 m from A outside AB c) both 1 and 2 d) neither 1 nor 2 77. Two equal charges repel each other with a force of 0.1 N, when situated 0.45 apart. The medium in 20 SARVAJNYA 840942174 between the charges has dielectric constant 9. The magnitude of each charge is. a) 4.5 C b) 4.5 mC c) 4.5 C d) 4.5 nC 78. Three small spheres each carrying a charge q are placed on the circumference of an equilateral triangle of side ‘a’. The potential at the centre of the triangle is. a) zero c) 3 3q 0 a b) 3 3q 2 0 a d) 3 3q 4 0 a 79. A charged oil drop remains stationary when sitauted between two parallel horizontal metal plates 25 mm apart and a p.d. of 1000 V is applied to the plates. Find the charge on the drop if it has a mass of 5x1015 kg (g=10 m.s2) a) 1.25x1017C b) 1.25x1016C c) 1.25x1019C d) 1.25x1018C 80. An isolated sphere of radius 0.1 m has lost 1012 electrons. The intensity of the electric field on the surface of the sphere is. a) 1.44x1016 N/C b) 1.44x105 N/C 4 c) 1.44x10 N/C d) 1.44 N/C 81. The radius of a spherical conductor which will have a potential of 6000 V when surrounded by an oil of dielectric constant 1.5 and charged with 5x108C is. a) 0.5 m b) 0.005 m c) 5 m d) 0.05 m 82. A hollow spherical conductor is charged to 2x108C. The potential inside the sphere is. a) 0 b) 900 V c) 1800 V d) 600 V 83. To what potential must an insulated sphere be charged so that its surface density of charge is 1012 C/cm2? [Radius of the sphere = 7 cm] a) 7920 V b) 792 V c) 79.2 V d) 7.92 V 84. Two metal balls of radii 0.05 m and 0.04 m are charged to the same potential. The surface densities of charge are in the ratio. a) 5:4 b) 4:5 c) 25:16 d) 16:25 85. Two point charges, one ten times as strong as the other, repel each other with a force of 81x104N when separated by a distance of 0.02 m in air. The charges are. a) 9 nC, 90 nC b) 6 C, 60 C c) 6 nC, 60nC d) 9 C, 90 C 86. Two charged spheres having the same radius and charges 9 nC and +5 nC are separated by a certain distance in air. They are brought into contact and then replaced in their original positions. The ratio of the forces between them before and after contact is. a) 9:5 b) 5: 9 c) 45:4 d) 45:4 87. The separation between two pints in order that the electrostatic force of repulsion acting on either of them may be equal to its weight is. a) 12 cm b) 12 mm c) 1.2 m d) 12 m Charge of a proton = 1.6 x 1019C, mass of a proton = 1.67 x 1027 kg, g =9.8 m/s2] 88. Charges equal to q1 are palced at two diagonally opposite corners of a square and charges equal to q2 are placed at the remaining corners. If the resultant force on q2 is zero, then q2= a) + 2 q1 b) 2 q1 c) +2 2 q1 d) 2 2 q1 89. Two point charges +9x108C and 9x108C are placed 0.5 m apart in air. The magnitude of resultant intensity at a point located 0.5 m from either charge is a) 3.24x104 N/C b) 3.24x103 N/C c) 32.4 N/C d) zero 90. Point charges 2nC and 2nC are placed at the points (3, 4) and (3, 4) in the X-Y plane. Find the electric intensity at the origin. All the co-ordinate distances are in metres. a) 0.12 NC1 b) 12 NC1 c) 1.2 NC1 d) 120 NC1 91. ABC is an equilateral triangle of side 3 cm. Charges of +5 nC and 5nC are placed at the corners A and B. The magnitude of the electric intensity at C is a) 5 NC1 b) 50 NC1 1 c) 500 NC d) 5x104 NC1 92. ABC is a triangle with sides AB=3m, BC=4m and B=90º. Charges of magnitude 9x105C and 16x105C are placed at the corners A and C respectively. the magnitude of the electric intensity at B is. a) 1.273x104 NC1 b) 1.273x105 NC1 3 1 c) 1.273 x 10 NC d) 1.273 NC1 93. An infinite number of charges each equal to 1 nC are placed along the X-axis at x=1 m, x=2 m, x=4 m, x=8 m…. The electric intensity at x=0 due to this set of charges is. a) b) 0 c) 12 NC1 d) 120 NC1 94. Volt/ metre is a unit of. a) potential b) potential difference c) distance d) electric intensity 95. A metal sphere of radius 0.05 m has a charge of +1 nC. The surface density of charge is. a) 3.18x108 C.m2 b) 3.18x107 C.m2 c) 3.18x107 C.m2 d) 3.18x108 C.m2 96. A point charge of +10 C is at the centre of a cubical gaussian surface of side 0.1 m. The flux of the electric field from the surface of the cube is. a) 1.13 x 105 NC1 b) 1.31x106 NC1 c) 1.13x106 NC1 d) 1.31 NC1 97. ABC is a right angled triangle with AB = 0.2 m, BC = 0.6 m and B = 90º. A metal sphere of radius 2 cm is charged to a potential of 9x104V and is placed at B. The work done is carrying a charge of 1 C from C to A is. a) 60 J b) 6 J c) 600 J d) 6000 J 98. ABCD is a square of side 2 m. Charges of 5 2 nC, +10 nC and 5 2 nC are placed at the corners A, B and C respectively. The work done in transferring a charge of 5 C from D to 0 (the centre of the square) is. a) 2.25x103 J b) 225 J c) 2.25 x 104J d) 0.225 J 21 SARVAJNYA 840942174 99. An infinite number of charges each equal to 1 nC are placed along the X-axis at x=1, x=2, x=4…. The distances are in m. The electric potential at the origin due to this set up of charge is. a) 6 V b) 12 V c) 18 V d) 24 V 100. A work of 16x1016 J is done to transfer an electron between two points in an electric field. The potential difference between the two points is. a) 1 KV b) 10 KV c) 100 KV d) 1000 KV 101. An electron is accelerated from rest in a uniform electric field of intensity 10 KV.m1. Its velocity at the end of 1 ns is. a) 1.76x104 ms1 b) 1.76x105 ms1 6 1 c) 1.76x10 ms d) 1.76x108 ms1 102. The potential energy of a charge 1 C in an electric field is 3x104 J. The potential of the field is. a) 3 V b) 30 V c) 300 V d) 3000 V 103. A spherical conductor of radius R, placed in air, is given a charge Q. then the potential at a point inside the conductor and at a distance R/2 from its centre is (CET 1997) a) V =40R b) V=4R c) V = 1 Q 4 0 2R d) V= 1Q 4 0 R 104. Two spherical conductors of radii 4 m and 5 m are charged to the same potential. If 1 and 2 be the respective values of the surface density of charge on the two conductors, then the ratio 1 is(CET 1998) 2 a) 1.5 b) 0.5 c) 1 d) 2 105. A point charge A of charge +4 C and another point charge B of charge 1 C are placed in air at a distance 1 metre apart. Then the distance of the point on the line joining the charges and from the charge B, where the resultant electric field is zero, is (metre) (CET 1998) a) 1.5 b) 0.5 c) 1 d) 2 106. When a positvely charged conductor is earth connected (CET 1998) a) electrons flow from the earth to the conductor b) protons flow from the conductor to the earth c) no charge flow occurs d) electrons flow from the conductors to the earth 107. Electric charges +10 C, +5C, 3C and +8C are placed at the corners of a square of side potential at the centre of the square is 1999) a) 18x105 b) 1.8x106 c) 1.8 d) 1.8x105 108. When 1019 electrons are removed from metal plate, the electric charge on it is 1999) a) 10+19 b) +1.6 2 m. The (V) (CET a neutral (C) (CET c) 1.6 d) 1019 109. Two metal spheres of radii R1 and R2 are charged to the same potential. The ratio of the charge on the two spheres is. (CET 1999) a) R1/R2 b) 1 c) ½ d) R1 – R2 110. A cube of side b has charge q at each of its vertices. The electric field at the centre of the cube (CET 2000) a) q/b2 b) q/2b2 c) 32q/b2 d) zero 22 SARVAJNYA 840942174 ELECTRIC CHARGE AND ELECTRIC FIELD & ELECTRIC POTENTIAL QUESTION BANK: One marks type questions: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. Define charge. What is electrification? How many electrons constitute a charge of 1 coulomb? A body is at positive potential. What does it mean? A body is at negative potential. What does it mean? What is the relation between curvature of the surface and the concentration of charge on it? Define the term linear density of charge. What is meant by electric permittivity of the medium? Write the SI unit of electric field strength. Define the term, dielectric strength of a medium. What is the difference in charging a conductor & an insulator? What is the minimum amount of charge that can be given to a body? Define an electric line of a force. What is the field inside the charged conductor? What is the direction of an electric line of force on an equipotential surface? What is electrostatic shielding? Write the SI unit of electric permittivity. Define an electron volt. Write an example of equipotential line. A charge is moved on equipotential surface, what is the net work done? What is electric dipole? Define electric dipole moment. An electron is kept in a field of strength (1/1.6x10-19) N/c. What is the force experienced by it? An electron with KE of 2J enters a region of Pd ( 2/e) Volts. What will be its new KE? A charge of 2C is enclosed by a cubical Gaussian surface, is transferred to the spherical box. What is the change electric flux through the surface? What is the potential of earth? What is the direction of field due to a positive charge? What is the direction of field due to a negative charge? What is the direction of electric dipole moment? If a slab of dielectric material is introduced between two charged bodies, the force between them reduces to half. What is the dielectric constant of the slab? Write the dimensional formula for electric flux. Who classified the charges as positive and negative? Which of these two cannot be charged by induction? A glass rod, a metal sphere. Charged bodies can attract light uncharged bodies. State true/ false. What is meant by discharging action of points? How is the lighting caused? Write a limitation of Coulomb’s law. Write an example of polar molecule. Write an example of non-polar molecule. What is meant by induced dipole moment? 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. Define strength of electric field. Is it a scalar or a vector? Define the uniform and non-uniform electric field. Obtain an expression for electric field at a point due to a point charge. Write any four properties of electric lines of force. Define electric flux. Write its SI unit. Derive the relation between electric field and electric potential. Define the terms electric potential and potential difference. Write any four properties of electric charge. Write a note on charging by friction. Write a note on charging by conduction. Write the formula for electric field on the axis of electric dipole and define the terms. Write the formula for electric field on the equatorial line of electric dipole and define the terms. 26. 27. 28. 29. 30. Two marks type questions: 23 840942174 53. 54. 55. SARVAJNYA Write the formula for the potential energy of system of charge, containing n number of charge. Define the terms used in the formula. Write the note on the potential energy of the system of charges. Write the formula for electric field near charged plate & define the terms. Five marks type questions: 56. 57. 58. 59. 60. 61. What is charging? Explain charging by induction. State and explain Coulomb’s law in electrostatics. Explain the distribution of charge on a charged body. Define surface density of charge. State Gauss theorem. Show that electric field at a point near a charged conductor is directly proportional to the charge density below the point. Define electric potential. Derive an expression for electric potential at a point due to an isolated point charge. What is a dielectric material? Explain the phenomena of dielectric polarization. NUMERICAL BANK: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. Two point charges +3C are and 6C located 1 m apart in air. Find the resultant electric intensity midway between them. [3.24x105 N/C] Two point charges 9 nC and +18 nC are separated by a distance of 0. 1m in air. Find the resultant electric intensity at a point between them and 0.01 m from 9nC. [83x104 N/C] Two point charges of +6x108 C and 6x108C are 0.4 m apart in air. Find the resultant electric intensity at a point 0.4 m from either charge. [3.375x103 N/C] Two point charges of +1 nC and +4 nC are located 0.1 m in air. Find the position between them and along the line joining them at which resultant electric intensity is zero. [0.0333 m] Two point changes +2nC and 18 nC are located 1 m apart in air. Find the position along the line joining the two charges at which resultant electric intensity is zero. [0.5 m] ABC is an equilateral triangle of side 0.1 m. Point charges of +3 nC and 3 nC are placed at corners A and B respectively. Calculate the resultant electric intensity at C. [27x102 N/C] ABC is a right angled triangle with sides AB=0.3 m, BC= 0.4 m and B=90º. Point charges of +18 nC and +32 nC are placed at corners A and C respectively. Find the resultant electric intensity at B. [2.546 x 103 N/C] ABC is triangle with sides AB = 3m, BC = 4 m and ABC = 90º. Charges of +9x1010 C and 16x1010C are placed at corners A and C respectively. Find the resultant at B. [0.9 N/C] ABC is right angled triangle with sides AB = 0.2 m, BC = 0.4 m and ABC = 90º. Point charges of +4 x1012 C and +32x1012 C are placed at corners A and B respectively. Calculate the resultant electric intensity at B. Suppose a point charge of 5C is placed at B, what force would it experience? [2.102 N/C, 10.06x106 N] ABCD is a square of side 1 m. Point charges of +4C, +8C and +4C are placed at corner A, B and C respectively. Find the resultant electric intensity at D. [8.691x104 N/C] ABCD is a square of side 2 m. Point charges of 2x109 C, +3x109 C, +4x109 C and +5x109 C are placed at corners A, B, C and D respectively. Calculate the magnitude of resultant electric intensity at O, the point of intersection of diagonals. [25.45 N/C] An electric dipole with a dipole moment 5x109 C-m is placed in a uniform electric field of magnitude 4x104 N/C at (i) 30º and (ii) 45º. Calculate the magnitude of the torque acting on the dipole in each case. [10x105 N-m, 1.414x104 N-m] o An electric dipole consisting of an electron and a proton separated by a distance of 5 A is located in an electric field of intensity 4x105 N/C at an angle of 30º with the field. Calculate the dipole moment and the torque acting on it. Given: Charge on electron=1.602x1019 C. [8.01x1029 C-m, 1.602x1023 N-m] An electric dipole consists of two opposite point charges each of 2C separated by 1 cm. The dipole is placed in an external uniform electric field of 3x10 5 N/C. Find (i) maximum torque exerted by the field on the dipole and (ii) the work done in rotating the dipole through 180º starting from the position =0º. [6x103 N-m, 3600 J] 12 12 Two point charges of 0.1x10 C and 0.1x10 C are separated by a distance of 108 m. Determine the electric field (i) at an axial point distant 0.1 m from the mid point of the dipole and (ii) at an equatrial point distant 0.15 m from the dipole. [1.8x108 N/C, 2.67x109 N/C] A cubical surface encloses a system of 3 charges +4nC, 1 nC and +2nC. What is the electric flux over the cubical surface? [564.7 N-m2/C] A cube of side 1 m enclose a point charge of +2nC at its centre. 24 SARVAJNYA 840942174 (i) Find the total flux emanating from the cubical surface. (ii) What is the flux through each surface of the cube? [225.9 N-m2/C, 37.65 N-m2/C] 8 A spherical shell of radius 0.1 m is charged with 10 C of electricity. Find the potential at (i) the surface of the shell and (ii) at a distance of 0.5 m from the surface of the shell. [900 V, 150 V] The following figure shows variation of electric potential with distance due to an isolated charge. From the figure calculate the electric intensity at (i) x=1 m, (ii) x=3 m and (iii) x = 5 m.[2 V/m, 1.33 V/m, +2 V/m] Electric potential due to a point charge varies according to V = 3x 2+2. Calculate electric intensity at (i) x = 0.1 m and (ii) x=2 m from the charge. [0.6 V/m, 12 V/m] Two point charges of +2 C and 8 C are placed 2 m apart in air. Find the positions along the line joining the two charges at which resultant electric potential is zero. [0.4 m, 0.67 m] Two point charges of +1 C and 9C are placed 1 m apart in air. Find the positions along the line joining the two charges at which the resultant electric potential is zero. [0.1 m, 0.125 m] ABC is a right angled triangle with AB = 0.3m, BC = 0.6 m and ABC = 90º. A metal sphere of radius 1 cm is charged to a potential of 5x104 V and is placed at B. What is the work done in moving a charge of 1 C from C to A? [832.5 J] ABCD is a square of side 1 m. Point charges of +1x1010 C, 2x1010 C and +5x1010 C are placed at corners A, C and D respectively. Find the resultant electric potential at B. [2.2815 J/C] ABCD is a square of side 2 m. Point charges of +1 nC, +2 nC, 3 nC and +4 nC are placed at corners A, B, C and respectively. Find the resultant electric potential at the point of intersection of the diagonals. [36 J/C] 8 8 8 ABCD is a square of side 2 m. Point charges of 2x10 C, 4 x10 C and 5x10 C are placed at corners A, B, C and D respectively. Calculate the work done in moving a point charge of 2C from B to be point of intersection of diagonals. [7.664x104 J] n identical drops, each charged to a potential of V, combine to form a bigger drop. What is the potential of the bigger drop? [n2/3 V] Three point charges are arranged at the three vertices of an equilateral triangle of side 0.1 m as shown. Calculate the electrostatic potential energy of the system. Given: q=10 9 C. [9x107 J] Three point charges +2 nC, +4nC and +8nC are placed at the corners of an equilateral triangle of side 0.2 m. What is the potential energy of the system? What is the work done to remove 8 nC from the triangle to infinity? [2.52x105 J, 3.6x106 J] Two positive charges 8C and 4C are 0.1 m apart in free space. Calculate the work done in bringing them 0.02 m closer. [0.72 J] 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. Electric potential energy of electric dipole in uniform electric field or work done in rotating and electric dipole in uniform electric field: Consider an electric dipole of dipole moment p placed in a uniform electric field E . Let the dipole moment of the dipole make an angle to the direction of electric field. Then torque acting on the dipole is given by; = pE sin If the dipole is rotated through a very small angle d against this torque, then small amount of work done is. dW = d = pE sin d Total work done in rotating the dipole from 1 to 2 is W 2 pE sin d 1 Since p and E are constant, they can be taken out of the integral. W = pE 2 1 sin d = pE cos 2 = pE (cos 2 - cos2) 1 Special cases: (i) When =0º; U = pE cos 0º = pE In this position, the dipole has minimum potential energy and hence it is in stable equilibrium. The dipole has more potential energy in all other positions. (ii) When =90º; U = pE cos 90º = 0 In this position, the potential energy of the dipole is zero. (iii) When =180ºl U=pE cos 180º = +pE In this position, the dipole has maximum potential energy and is in unstable equilibrium.