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Transcript
ELECTROSTATIC 1. The study of electrostatic introduces us to the vast world by electricity, magnetism, current, working of circuits and charges. Right from the attraction between a proton and neutron to the working of large high voltage generator can be explained by the simple concepts of electrostatics, the most basic of phenomenon and yet the most interesting of all. Introduction: Electrostatic is the study of charges that are at rest and their interactions. Electric Charge: It is an intrinsic property of protons and electrons, proton and electron are smallest unit of positively and negatively charge. The S.I.unit of charge is Coulomb. 1) 2) Like Charge Some charges is quantized properties repel multiple of of while charge unlike on an charges: attract. electron. 3) Net amount of charge in a system is conserved. Discuss this topic Coulomb’s 1) Force Where And k = Law: between r = where the two separation charges between = q1 and q 2. = permittivity of the medium. Vector form- Important points: * Principle of superposition: Electrostatic force between any two charges is independent of other charges. 2) For a dielectric (insulator) medium Where k= e = permittivity of the medium dielectric Note: Electric force is an action–reaction pair and conservative in nature. constant. Illustration: Two charges are placed at a distance 1.0 apart what is the minimum possible magnitude of electric force. Question: Q-1: Why solution: Since q1, q2 should be minimum and q1min = q2min = e = C Q-2: Five point charges of values +q are arranged on the vertices of a regular hexagon as shown in the figure, and a –q charge is placed in the center. Find the force on –q charge. Fig (1) solution: Imagine two charges of on the vacant vertex, the net force on center is now only due to -q as all other forces cancel out (principle of superposition). Fig (2) Electric Field: It is the electrostatic force per unit charge. The space around a charge where its electric force can be experienced by another charged particle. Important 1) points: Field due to point charge = 2) Principle of superposition: The electric field due to multiple no of charges gets vectorially. added Note: For continuous charge distribution Where dq is charge due to infinitesimally small element. Electrical Lines of These are imaginary lines drawn to aid our visualization of properties of electric field. forces: Question: Q-1: Why electric field lines never intersects? Ans: Intersection of electric field lines implies that there are two direction of electric field at point which is impossible, therefore they never intersects. that Q-2: Why are electric lines always normal to the surface of a conductor? Ans: If electric lines are not normal then there is a component along the surface which means that change of potential along the surface which is not acceptable. Q-3: Why do electric field lines start from positive and end on negative charge? Ans: Because the direction of electric field is from positive to negative i.e. the force on a unit positive charge is away from positive and towards negative charge. Field lines from +ve Pointcharge. Field lines towards -ve Point charge. Equal +ve charge.Two equal but oppositecharged particle Illustration: Q-1: A field of 103N/C acts at a point. What will be the force on a charge of 3´10-6C and -2´10-6C, kept there? solution: For +3MC For -2MC Discussion: The direction of force changes for different charges. For a positive field(due to +ve charge) 1) The force is in direction of field on the +vecharge i.e. repulsive. 2) The force is opposite to field direction onthe –ve charge i.e. attractive. Similar inference can be drawn for negative charges. Electric field due to: a)Charged ring of charge Q and radius R. 1) The Center=0. Derivation: Linear Consider charge the field density at center But the field due to point diametrically opposite = (l) due on to any ring element = = in opposite direction. \Net field at center = 0 (By symmetry) 2)On the axis = On axis ofring at distance x. Derivation: Fig (5) As obviousfrom the diagram the field component along the line gets added due to oppositeelement. *)By substituting x=0 in 2ndresult, we can get the firstresult. *) Student should verify that the graph is of the followingmanner. Fig (6) b)Due to a straight charged rod of length 2L with charge per unit length ‘ ’ ata distance ‘a’ on its E perpendicular bisector. = Derivation: The rod is divided into infinitely small elements andthe field due to symmetrically; opposite part add up as shown in figure (7). Net field at P. E= dE = Fig(7) Useful Tips: 1) If x>>a, E = 2) If L like a point charge. i.e. for infinite length Question: 1) How did Ans: the electric field cancel inone direction and add in another? Observe the direction of electric field of two points which are symmetricallyopposite. Along the Along the perpendicular, axis, net field = Net field = Similarly all fields along the perpendicular to axis cancel out. Flux and Gauss Law: Flux-It literally means the no of field lines passing through it. Itis defined as. Gauss law- The electric flux enclosed by thesurface. through any closed surface is equal to times the net charge Illustration: A square frame of edge 5cm is placed such that a uniformelectric field of 10 V/m makes 30 0with positive normal. Application of gauss law: 1)Uniformly charged sphere (solid) with charge Q and radius R. Fig (9) Case-1: (r>R) Fig(10) Charge enclosed = Q Case-2: (r<R) Question: Q-1 How? Ans: The field lines should be radially outward and thereforecollinear with outer normal. Cos 2)Infinitely long linear charge distribution: Derivation: Consider the Gaussian surface shown: Fig (11) Qenclosed= ll For surface (1) and (3) =1 As electric field perpendicular to area vector for curved surface (2) By Gauss’s law E (2pRl) = 3)Field due to infinite charged plane sheet with charge density ( )= = Derivation: Consider a cylindrical Gaussian surface as shown. Flux through curved surface = 0 Fig (12) Conductors: The electric field inside conductorsis zero in electrostatic equilibrium. Cavity in a conductor: Consider the metallic shell with a postcharge in the center. Since E = 0 inside a conductor qenclosedbyG.S = 0 Þ -q charge is induced on inner surface, as shown in figure(13). Since shell is neutral, +q is induced on outer surface as shown infigure (14). Fig (13) Fig(14) Question: Explain why electric field inside a conductoris zero, with the help of induction. Solution:A conductorhas large no of free electrons. When it is kept in an electric field,electrostatic force acts on these electrons which draw most of the electron toone side causing negative charge on that side and positive charge on oppositeside due to deficiency of electrons. The field due to these induced chargesbalances the external electric field and thus there is no electric field in themetal. Fig (15) Enet=E+Ein= 0 On the surface of charged Conductor: Electric field means a Conductor = Force per unit area or electrostatic pressure experienced by a chargedconductor = (The derivation of above formula is not very important for IIT JEEpreparation) Energy density of electric field = Electric It is the work needed to bring a unitpositive charge from infinity to the given point. (Because, since the field is conservative.) Q-1: The value of integral depends only on end points A and B why? Ans: (As it is opposite to electrostatic force) Note: 1)It is a scalar quantity and has units Joules per coulomb (J/C) called Volt. 2)Potential difference between two points A and B is V B-VA= Where W ABis work done in moving from A to B. Electric potential due to a point charge = 2. Principle of superposition: Potential: Net potential due to InCartesian Co-ordinate- a system of charge is, V =V 1+V2+V3+V4+---------------- Illustration: 1) Three charges q1, q2,q3are placed at (r, 0, 0), (0, 3r, 0), (0, 0, 2r) respectively. Findpotential at origin. Ans: Net electric potential = Vq1+Vq2+Vq3 = = Equipotential surface: If a surface is drawn in such a way that electric potential is same at allpoints of the surface it is called equipotential surface. Workdone in moving a charge on equipotential surface = 0. Electric potential on – Derivation: Divide the ring into infinite elements and potential due to eachelement. Fig (16) If r = 0, V at center = as shown in figure (17) Fig (17) 2) axis of uniformly charged disc = Derivation: Consider disc having uniform surface charge6(c/m 2) Fig (18) Potential due to ring of radius r = Question: Q-1: How dq = Ans: dq = 6 x Surface area = 2 r.dr= area of the ring. 3) Due Derivation:Shell of radius R has charge Q Case-1: (r>R) Case-2: (r<R) Einside= 0 to a Shell: Electric Potential Energy: 1) Potential energy between two charges q 1and q2separated by distance r = . 2) P.E due to System of charge = Change should be replaced with sign. Question: Q-1: Why ½ in the above expression? Ans: If ½ were not produced each pairwould be counted twice. For example- For q1and q2 3) Illustration: Two point charges are located on x-axis q1=3microcoulomb at x=0 and q2= -1 microcoulomb at x=2m. a) Findwork to be done to bring q 3= 1microcoulomb from infinity to x=1m. b) Find total P.E of the system. solution: =Uf-Ui Work done = difference in P.E. Earthing:Potential of earth is zero. If a conductor is earthedits potential becomes zero. Tips for Concentric conducting sphericalshells: 1) Net charge inside a closed Gaussian surface drawn in anyshell is zero. 2) Potential of the earthed conductor is zero. 3)Two connected conductors are at same potential. 4) Charge remainsconstant in all conductors except those which are earthed. 5) Charge oninner surface of innermost shell = 0 (if no charge is keptinside) Charge on outer surface of outermost 6) Equal and opposite charge appear on oppositefaces. shell = totalcharge enclosed. Illustration: Figure (19) shows two conducting thin concentric shells ofradii r and 3r. The outer shell carries charge q. Inner shell is neutral. Findthe charge that will flow from inner shell to earth after the switch S isclosed. Fig (19) solution: Let q be the charge on inner shell when it isearthed V inner= 0. i.e. charge will flow from inner shell to earth. Electric dipole:An arrangement of two equal and oppositecharges separated by small distance is known as electric dipole. Fig 1) 2) 3) (20) It The It is line is a directed from –ve joining the charges a)Electric field on the axis (l<<r) = Derivation: Fig (21) b) On perpendicular bisector = Derivation: is vector charge to+ve the axis of quantity. charge. dipole. Fig(22) The component Alonghorizontal. Enet= of E cancels E+qCosq out in vertical + direction. E-qCosq (-ve indicates direction opposite todipole moment) If x>>1 c) Electric field at any point A(r,0) Fig (23) Dipole 1) 2) is an external Force uniform electric field ondipole=0. 3) Potential energy = -PECos q = -P.E Fig (24) Q= 0, U will be Q=1800, U will bemaximum, unstable equilibrium. minimum, stable equilibrium. Illustration: 1) A dipole of dipole moment P lies in uniform electricfield with its dipole moment along E. If dipole is rotated through900, Find work done? Ui= -P.E = -P.E Cosq =-P.E. Uf= -PECos900= 0 Work done = -DU =Ui-Uf=-P.E. 2) A dipole with dipole moment Cm is aligned at 800with the direction ofa electric field of magnitude 104N/C Calculate magnitude oftorque. |Z| = PESinq Examples: 1) A charge q = 2´10-6C is placed at (1m, 1m, 1m) and electricfield at P (0m,-2m, 1m). solution: 2) Two free particles with charges q and 4q are a distance L apart. Athird charge is placed so that system is in equilibrium. Find location,magnitude and size of third charge. solution: Let new chargebe +Q, It will be along the line joining the two charges to balance the forces.Let it be located as shown. Fig (25) Since For q: For 4q: they are in equilibrium (-ve x means charge will be in middle of the two charges.) 3) A charged particle of mass m and charge q is released from rest in an electric field of constant magnitude E then the KE of the particle after time t is. solution: Force = qE Acceleration = . 4) P and Q are two concentric metallic spheres. P is positively charged and Qis earthed then a) Charge density on Q is b) Electric field between P c) Electric potential inside d) Electric fieldoutside Q is zero. same and P as Qis is on P. uniform. zero. Fig(26) solution: VQ= 0 Let charge on P be q1 and on Q be q2 i.e. (a) is false. (b) Is false as there is potential difference between the two shells and theelectric field lines separate outwards. (c) is false . (d) is true as q enclosed by Q surface = q2+q1=0. 5) A certain charge Q is divided into two parts q and Q-q. For the maximumCoulomb force between them the ratio (q/Q) is: 1) 1/6, 2) 1/8, 3) 1/4, 4) 1/2 solution: Ans = 4). 6) Three large conducting plate sheets are kept as shown in figure (27). Findelectric field a) between (1) and (2), b) between (2) and (3). Fig (27) solution: a) Between (1) and (2) Field due to a plate = . b) Between (2) and (3) 7) A charged particle of mass m=2kg and q=1microcoulomb is thrown from groundat q=30 0with speed 30 m/s. There is a horizontal electric fieldE=10 7V/m exists. Find the range on horizontal ground of theprojectile. solution: 8) A charge q is placed at l/2 distance above a square of length l as shownin figure (28). Find flux through the square. solution: Consider the charge to be kept in a cube of sidel. Net flux = Fig (28) By symmetry flux through each face = . 9) An alpha particle of KE = 6MeV is heading towards a stationary nucleus ofatomic no 30 calculate distance of closet approach. By conservation of energy: ½ mV2= Uf-Ui Dump Question: Ans: Since charges are atinfinity Why Vi = 0? Medium: Two parallel conducting plates contain charge Q1and Q2 as shown. Find distribution of charges on four surfaces. Fig (29) Consider the Gaussian surfaceshown. solution:Two faces lie in theconductor where field is zero, while other two are perpendicular to fielddirection. Net charge enclosed = 0 So, the facing inner surface have equaland opposite charges. Fig (30) Let C be a point in the conductor – Question: Why net flux through Gaussian Surface = 0? Fig (31) There Field \E.dA = 0 is at no 2 So, net Flux = 0 Note: Short cut for parallel plates field and 4 at are 1 and perpendicular 3 toE. 1)Charge on facing faces of two plates are equal and opposite 2)Charge on outermost face = . Q-2: A curved line charge of density l forming a semi-curve as shown. Findelectric field at center. Fig (32) Consider The the two opposite field elements as cancels shown. alongx-direction. Consider an infinitely small element substending angle d Question: 1)What would change if the curve were negatively charged? Ans: The net fieldwill be upwards. 2)If half of the curve were negatively charged? Ans: Virtual components cancelwhile horizontal remains. (Observe yourself) Q-3: A uniform line charge exist from x=-a to x= a. Find electric field at point P a distance r along the perpendicularbisector. solution: Fig (33) Considering the field due to two opposite parts, the field cancels along thehorizontal. Along Vertical: Q-4: A charge Q is uniformly distributed over a spherical volume of radius R.Obtain an expression for the energy of the system. 1)Inside the sphere(r<R) 2)Outside the sphere (r>R) = Question: solution:Consider a G sphere Q-5: Find the potential on the edge of a disc P with surface charge density6. Ans: Fig (34) To calculate the potential at P, the disc is divided as rings with P ascenter as shown in the figure. Potential due to element between r andr+dr. Now r = dr =-2RSinqdq. 2RCosq (as diameter subtends 900at the edge) (Applying Integration by parts) Question: 1) Why were integration limits from p/2 to 0? Ans: Integration limits werefrom p/2 to 0 because at p/2, the area around the point would be covered andwhen angle is zero, the other end of diameter gets covered. Therefore the wholedisc is covered. Q-6: The intensity of an electric field depends onco-ordinates x and y as follows, Where P is a constant, find the charge within asphere of radius R with the center at the origin. solution: At any point A(x, y, z) on the sphere, a unitvector is perpendicular to the sphere radially outward. Fig(35) Now can begot easily by integrating, for it is independent of x, y, and z. Note: The question although easy may seem very difficult during exam becauseof the use of Cartesian Co-ordinates. So student should go through the aboveexample carefully. Q-7: A non-Conducting disc of radius R and Surface charge density isplaced on the ground with its axis vertical. A particle of mass m and positivecharge q is dropped, along the axis of disc from height H with zero initialvelocity. The particle has q/m= 4 e 0g/ a) Find value of H if the particle just reaches the disc. b) Sketchthe potential energy of the particle as a function of its height and find itsequilibrium position. solution: a) As proved earlier Vi= potential on axis of a disc Vf= potential at center By conservation Decrease in gravitational potential energy KEi= KEf=0 (As particle is mgH = q (Vf-Vi) of energy: =Increase in electrostatic potential energy at rest in both position) b) Net potential energy = electrostatic P.E+ gravitational P.E Fig (36) Note: 1) Potential energy includes both electrostatic potential energy andgravitational potential energy. Question: 1) How is K.E=0 in both position? Ans:Initially it has zero initial velocity K.Ei=0, finally it is justable to reach the disc. Final velocity = 0 Þ K.Ef=0. Hard Q-1: A non conducting sphere of radius R has a positive charge which isdistributed over its volume with density per unit volume, where x is distance from the center.If dielectric constant material the sphere is K=1, Calculate energy stored insurrounding space and total self energy of the sphere. solution: If strength of electric field at a point is E,energy stored per unit volume of dielectric at that point is hence to calculate density of electrostatic energy insurrounding space, first electric field strength is to be calculated. But forits calculation, total charge on sphere must be known. Hence, first considera thin spherical shell inside the sphere as shown in figure. Fig (37) Let radius of the Totalcharge shell on be x and the let radical thickness shell be dx. is Now consider a concentric spherical shell (in surrounding space) of radius r(>R) and radical thickness dr. Electric field at surface of thisshell. Energy stored in the shell, Energy stored in surrounding space, Self energy of the sphere is total electrostatic energy stored in surroundingspace and inside the sphere. To calculate energy stored inside the sphere,consider a concentric spherical shell of radius x and radical thickness dx.Electric field at surface of this shell is due to charge on inner concentricsolid sphere of radius x. Charge on this concentric sphere is Electric field at surface of this shell, Energy stored in the shell, Total energy inside the sphere is Self energy of the sphere = Us+Ui Q-2: Suppose in an insulating medium having dielectric constant K=1, andvolume density of positive charge varies with coordinates according to law . A particle of mass m having positive charge q is placed in themedium at point A (0, y0) and projected with Fig (38) Neglecting gravity and frictional resistance of the medium and assumingelectric field strength to be zero at y=0, Calculate slope of trajectory of theparticle as a function of y. solution: Since, charge density varies with y coordinatesonly, therefore in medium an electric field exists which is directed along yaxis. To calculate this electric field E(y). Consider a thin layer of mediumhaving thickness dy at a distance y from x-axis as shown infigure. Fig (39) Let strength of electric field at positions y and (y+dy) be E and (E+dE)respectively. Then for area A of the layer, According to Gauss law Integrating At above equation y=0,E=0 and with y=y, at limits, E=E Since frictional resistance of the medium isnegligible, therefore there is no force on particle along xaxis or velocityV0of particle along x-axis remains constant. Due to Electricfield E, particle accelerates along positive y, this acceleration = Let y-component of velocity of particle bev Hence, But at point A(y=y0) y component ofvelocity is v=0 Question: 1) Why there is no force inx-direction? Ans: The applied electric field is in the y-direction only andit has been mentioned in the question that friction, gravity have to beneglected. Thus there is no force in x-direction. Q-3: Three identically charged, small sphere each of mass m are suspendedfrom a common point by insulated light strings each of length l the spheres arealways on vertices of an equilateral triangle of length of the sides x(<<l). Calculate the rate with which charge on each sphere increases if rate ofincrease of length of the sides of the equilateral triangle varies as solution: Since x. is given as therefore to calculate . first q is to be calculated in terms of Since side of equilateral triangle is x, therefore electrostatic forcebetween two particle is Positions of particle A, B, C with respect to each other are as shown infigure (40)-A. Fig(40)-A Fig (40)-B Resultant electrostatic force on particle A is F = 2F1Cos300 In figure (A) G is centroid of triangle. Distance of centroid G from every particle is Since G is centroid therefore it always lies vertically below the point ofsuspension. Fig (B) shows a vertical plane passing through particle A and centroid G. Allof the force acting on particle A is shown in figure (B) Considering horizontal forces on the particle A, TSin = F---------- (1) For vertical Dividing forces eq TCosq (1) =mg by Since x<<l therefore inclination q of each thread with the vertical isvery small. From equation (3) ----------(2) eq (2)