* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Document
Survey
Document related concepts
Magnetic monopole wikipedia , lookup
Electrical resistivity and conductivity wikipedia , lookup
Speed of gravity wikipedia , lookup
Potential energy wikipedia , lookup
Fundamental interaction wikipedia , lookup
Casimir effect wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
History of electromagnetic theory wikipedia , lookup
Electromagnetism wikipedia , lookup
Maxwell's equations wikipedia , lookup
Field (physics) wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Lorentz force wikipedia , lookup
Transcript
1 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 1 COME FOR SUCCESS Contents: 1. COULOMB’S LAW, 2. ELECTRIC FEILD, ELECTRIC INTENSITY AND CALCULATION. 3. ELECTRIC FLUX AND CALCULATION FLUX DUE TO CHARGE. 4. GAUSS’S LAW ANDTHREE APPLICATIONS. 5. ELECTRIC POTENTIAL ENERGYAND POTENTIAL DIFFERENCE. 6. ABSOLURE POTENTIAL AND ABSOLUTE ENERGY 7. KINETIC ENERGY AND LAW OF CONSERVATION OF ENERGY 8. ELECTRON VOLT , RELATION BETWEEN ELECTRIC FIELD AND POTENTIAL , AND EQUIPOTENTAL SURFACE 9. CAPACITOR 10. CAPACITANCE OF PARALLEL PLATES CAPACITOER 11. COMBINATION OF CAPACITORS 12. VARIABLE CAPACITORS 13. EQUATIONS. 14. DIMENSIONS 15. SHORT QUESTIONS AND ANSWERS Electrostatics: It is the branch of physics explaining phenomena arising due to the existence of electric charges, which do not move means, they are static. Electrostatic phenomena arise from the forces that electric charges exert on each other. Electroscope is device used to detect static charge. Electrostatic induction: It is a redistribution of electrical charge in an object, caused by the influence of nearby charges. Conductors: Materials which contain movable charges that can flow with minimal resistance. Insulators: Materials with few or no movable charges, or with charges which flow with extremely high resistance. Semiconductors: Materials whose behavior ranges between that of a conductor and that of an insulator under different conditions. Their conducting behavior may be heavily dependent on temperature. They are useful because we are able to change their conducting behavior to be dependent on many other factors. The Atom: An atom contains a positively charged nucleus and one or more negatively charged electrons. The atom exists in three states: neutral, positively charged, and negatively charged. A neutral atom has the same number of electrons and protons, a positively charged atom has more protons than electrons and a negatively charged atom has more electrons than protons. Charge : charge, property of matter that gives rise to all electrical phenomena. The basic unit of charge, usually denoted by e, is that on the proton or the electron; that on the proton is designated as positive (+e) and that on the electron is designated as negative (-e). There are two kinds of charge, positive and negative like charges repel, unlike charges attract. Charge is conserved Positive charge comes from having more protons than electrons; negative charge comes from having more electrons than protons Charge is quantized, meaning that charge comes in integer multiples of the elementary Charles Augustine de Coulomb did an experiment in the 1780's, in which he measured the dependence of the electrostatic repulsion between two charged spheres as a function of distance between the spheres using a sensitive torsion balance. Name.- - - - - - - - - - - - - - - -- - S/O , D/O- - - - - - - - - - - - - - - belongs to - - - - - - - - - - - - College - - - - GROUP TIME - - - - - - -GROUP #---------------- Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 1 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #1 CHAPTER # 11 PTCL # 022-2670019 2 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 2 COME FOR SUCCESS The electrostatic force between two point charged particles is directly proportional to the product of the charges on these particles and inversely proportional to the square of the distance between them and force acts along a straight line joining the centers of these particles. q q F 12 2 r Explanation: Consider two point charges q1 and q2 create equal electrostatic force on each other which is directly proportional to the magnitude of each charge. F q1 q 2 And electrostatic force on each other which is inversely proportional to square of the separation between their centers. F 1 r2 Mathematical Derivation: q1 q 2 r2 q q F = K 1 2 2 r In this equation, “K” is called “Coulomb’s Constant”, its value depend on the medium in which the electric charges interact. 1 =9×109 N m 2 c-2 That, K = 4 π εo According to the statement of Coulomb’s law, F The “o” is called “Permittivity of free space”; its value is 8.854210-12 C² N-1 m-², when the charges are placed in vacuum. Thus, the electrostatic coulomb’s force can be written as, 1 q1 q 2 , For vacuum. F= 4 π ε o r 2 Effect of medium: When a dielectric medium is completely filled between charges the force between the same two charge decreases. A dielectric substance has relative permittivity “r”. 1 q1 q 2 For air, r = 1(least value). , For any dielectric medium F' = 4 π ε o ε r r 2 Vector form: As we know that electrostatic force is a vector quantity, then it can be written as, F' = 1 q1 q 2 r 4 π ε o ε r r 2 Relative permittivity F ε r = It is defined as the ratio of the force in vacuum to the force in any other medium between the same F' pair of charges separated by the same distance r. The relative permittivity εr is greater than 1 for any medium other than vacuum or air. Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 2 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #2 CHAPTER # 11 PTCL # 022-2670019 3 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 3 COME FOR SUCCESS Coulomb: The coulomb is defined as the quantity of charge that has passed through the cross-section of an electrical conductor carrying one ampere within one second. 1 Coulomb = 6.3x1018 elementary charges 1 elementary charge = 1.6x10-19 Coulomb. The electric field can be represented with electric field lines. Their density is a measure of the strength of the electric field at that point. 1. They start from a positive charge and end in a negative charge 2. Field lines are smooth continuous curves without any break. 3. The tangent to it at any point gives the direction of electric field at that point. 4. Electric field lines never intersect each other. 5. They always enter or emerge normal to the surface of a conductor. 6. They tend to contract longitudinally (on account of attraction between unlike charges) 7. They tend to expand laterally (on account of repulsion between like charges) 8. Stronger fields have closer lines. 9. Field strength and line density decreases as move away from the charges. 10. The negative flux just equals in magnitude the positive flux, so that the net, or total, electric flux is zero. Electric field of a charge is an area in which the charge has electric force. An electric field exists in a region if an electric charge at rest. Since an electric charge experiences a force if it is in the vicinity of a charged body, there is an electric field surrounding any charged body, the force due to the charge is detectable. The Electric field is represented by electric lines of force. The strength of electric field at a particular point is called as Electric Field intensity. In other words, the force per unit positive charge situated at that point is called Electric field intensity. E = F qo Explanation: Let's suppose that an electric charge “q” creates an electric field; as the source charge. The strength of the source charge electric field could be measured by any other charge placed in its surroundings as a test charge since it is used to test the field strength. The test charge has a quantity of charge qo, when placed within the electric field. The magnitude of the electric Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 3 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #3 CHAPTER # 11 PTCL # 022-2670019 4 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 4 COME FOR SUCCESS field is simply defined, as the force per charge on the test charge. The electric field strength is denoted by the symbol E, then the equation can be rewritten as, F qo The unit of electric intensity is NC-1 or Vm-1.Electric field strength is a vector quantity; it has both magnitude and direction. F = E qo , is the magnitude of electrostatic force. The direction of force is the direction of electric intensity. E = Explanation: Let's suppose that two point charges, “q” and “qo” are placed at “r” distance apart as the source and test charge respectively. The electrostatic force experienced by the charges on each other is given by according electrostatics coulomb’s force. Mathematical Derivation: 1 q qo ^ As we know, F= r 4 π ε o r 2 F 1 q qo ^ 1 = r qo 4 π ε o r 2 q o Divide both sides by q o we get, 1 q ^ r 4 π ε o r 2 1 q , is the magnitude of electric intensity. E= 4 π ε o r 2 We get, E= Conclusion: The magnitude of electric intensity is directly proportional to the charge and inversely proportional to the square of the distance, from the charge to that point. The electric flux is property of an electric field that is the number of electric lines of force passing through area. It is the measurement of flow, of electric lines of force through area. The dot product between electric intensity E and area A is called “Electric Flux”. = E .A Explanation: Let's suppose that surface of area A be so small that is in the uniform electric field E . The electric flux passing thorough the surface is as scalar product, Or = E .A Its magnitude is given by, = E A Cos Electrical flux has units of volt metres (V m), or, N m2 C−1= kg m3 s−3 A−1. Special cases: i) The surface is placed perpendicular on to the electric field, the direction of electric intensity and area are parallel, say = 0o Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 4 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #4 CHAPTER # 11 PTCL # 022-2670019 5 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 5 COME FOR SUCCESS = E A Cos 0o = E A(1) Or = E A It means, “Maximum” Electric flux passing through the surface. ii) The surface is placed 45o on to the electric field, the direction of electric intensity and area are making45o, Therefore, = E A Cos 45o = E A(0.707) Or = 0.707 E A iii ) The surface is placed parallel to the electric field the area is perpendicular onto the electric intensity, say = 90o Therefore, = E A Cos 90o =EA(0) = 0 It means “minimum” or zero Electric flux passing through the surface. iv) The surface is placed perpendicular on to the electric field, the direction of electric intensity is opposite to the area, say = 180o Therefore, = E A Cos 180o = E A(-1) Or = -E A Therefore, Explanation: Let's suppose that a positive charge “+q” is placed at the center of sphere. It is surrounded by a closed spherical surface of radius “r”. The uniform electric intensity E produced due to a charge with in 1 q ^ spherical region. E= r 4 π ε o r 2 The electric lines of force are passing perpendicular through a surface of sphere. The magnitude of electric intensity is same at “r” distance on the spherical surface area. Mathematical Derivation: The electric flux “” through such smaller vector area is given by, φ = E . A Δφ = E ΔA Cosθ Δφ = E ΔA Cos0o Δφ = E ΔA 1 Δφ = E ΔA 1 q φ A 4 π εo r 2 N Δφ= i =1 Because, N ΔA =A = 4 r2 i 1 q N ΔA 4 π ε o r 2 i=1 is the total are of sphere. i=1 Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 5 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #5 CHAPTER # 11 PTCL # 022-2670019 6 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 6 COME FOR SUCCESS 1 q 4 π r2 2 4 π εo r q We get, electric flux due to charge, φ εo Conclusion: The electric flux due to charge placed at the center of sphere is independent of size. When, q = 0 (no charge), then, = 0 (no electric flux). Thus, φ Total electric flux through a closed surface enclosing a charge is 1 times εo the magnitude of the charge enclosed e = Q 1 εo Proof: Let us consider charge is uniformly distributed on the surface, where q = q1+ q2 + - - - + qn . The 1 1 1 electric flux due to the individual charge is given by, 1 = q1 , 2 = q2 , - - -, n = qn εo εo εo We know that, electric flux is scalar quantity; the total electric flux e is given by, e = 1 + 2 + 3 +- - + n q q q q φe = 1 + 2 + 3 + - - - + n εo εo εo εo 1 φe = q1 + q 2 + q 3 + - - - + q n or εo 1 Hence, e = ( q ) εo 1 or e = Q This is called Gauss’ law. εo FIRST APPLICATION : Let us consider a spherical shell of radius “r” uniformly charged by charge “Q”. The intensity of electric field at a point due to this shell depends onto the position of the point with respect to this shell. The intensity of electric field at different positions of point is calculated as follows: a) Electric field of a uniform spherical surface charge at a distance “r” from the center: Consider a uniform spherical distribution of charge on a conductor would be free to move and would end up on the surface. This charge density is uniform throughout the sphere. Charge Q is uniformly distributed Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 6 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #6 CHAPTER # 11 PTCL # 022-2670019 7 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 7 COME FOR SUCCESS throughout a sphere of radius “a”. First consider r > a; the electric field is at a radius r. That is, find the electric field at a point outside the sphere. Just as for the point charge with Gauss's Law. we find =E A Mathematical Derivation: =E (4 r2) Q and we know = εo Q =E(4 r2) εo 1 Q E= 4πε o r 2 That is, the electric field outside the sphere is exactly the same as if there were only a point charge Q. b).When a point is on to the surface of the spherical shell (outside the sphere): Suppose a point “P” is lying on the spherical shell .The electric intensity, of the electric field, is to be calculated at the distance “a” from the center at the given shell. Where, a > r, at every point of this inner sphere, the electric field will be equal and normal on to the surface. According to the gauss’s law, Mathematical Derivation: q φ= εo Hence, q εo q E ( 4πa² ) = εo E.ΔA = (4πa²) σ εo q From the definition of charge density, σ= A Therefore, q = ( 4 a² ) σ E= εo This shows that, the electric field intensity at a point outside the sphere will be same as if the whole charge is concentrated at the center of sphere. c). When a point lies inside the charged hallow sphere: Consider a hollow conducting sphere with radius R. Let q be the charge on this sphere. To find the field at a point P, draw a Gaussian surface (dotted circle) of radius r. Since, this surface does not enclose any charge, we have. Mathematical Derivation: We get, E ( 4πa² ) = Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 7 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #7 CHAPTER # 11 PTCL # 022-2670019 8 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 8 COME FOR SUCCESS According to the Gauss’s law, =E A Where, q = 0, this result is independent of the radius, provided that it is less than that charged sphere. Therefore, E = 0, And in terms of charge density, = 0 Conclusion: Intensity of electric field at every point inside the uniformly charged hollow empty sphere is zero. SECOND APPLICATION : Explanation: Let's suppose an infinitely large and thin plane sheet which carries a positive charge a per unit area. The electric field is perpendicular to the plane. q σ= - - - - - - -► eq. # 1 A Let be the “charge density” of an infinite plane sheet. For Gaussian surface we can choose a point P, is right circular cylinder with faces parallel to the plane of charge. The field lines are parallel to the sides of cross sectional area of the cylinder. It has two faces P and P’ will contribute to the flux as the electric field lines parallel to them. Mathematical Derivation: q Total charge By putting “q “ from φe Total electric flux = εo equation # 1 we get, According to the Gauss’s law, σA φ e Total electric flux = εo The flux through the P surface is, p = E A Cos p = E A Cos 0o Or p = E A( 1 ) p = E A and similarly the flux through the P’ surface is, p’ = E A The total electric flux, “e” through the surface will be, e = P + P’ or e = E A + E A e = 2 E A σA By putting the value of “e” in equation # 2, we get, 2 E A = εo σ E= or 2 εo We know that, electric intensity is a vector quantity, its direction given by unit vector r ^. Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 8 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #8 CHAPTER # 11 PTCL # 022-2670019 9 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS Hence, 9 COME FOR SUCCESS E = σ ^ U 2 εo Conclusion: The electric intensity is independent of distance. That is because the field lines everywhere are straight, parallel, and uniformly spaced. This is because, the sheet is infinitely large. THIRD APPLICATION : Explanation: Consider two parallel conducting plates are equal size it is assumed that each plate has equal and opposite charge. One plate has positive charge with charge density “+”, while the other plate negatively charged with charge density “-”. Imagine a point P is between the two parallel oppositely charged plates. The electric field due to the charged plate is uniform in the space between them. The electric intensity due to positively charged plate is given by, σ ^ E1 = + U 2 εo σ ^ U And due to the negatively charged plate is, E 2 = + 2 εo The total electric intensity due to the two oppositely charged plates will be, E = E1 E 2 σ ^ σ ^ E = + U + U 2 εo 2 εo σ ^ E = +2 U 2 εo σ ^ E= U εo or The electric intensity due to two oppositely charged plates is double the electric intensity due to single charged plate. Conclusion: The amount of workdone, in displacing the charge from one point to another against the electric field, is called “Electric Potential energy”. Explanation: Let us consider a charge “+q”, produced an electric field of intensity “ E ” another test charge “+qo” is placed in its field at point “P”. A test charge is moved to wards the source charge through smaller displacement r , in the opposite direction of electric field. The work done Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 9 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #9 CHAPTER # 11 PTCL # 022-2670019 10 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 10 COME FOR SUCCESS between two points is the change of potential energy U. Mathematical Derivation: Work = F . d F qo The magnitude of electrical potential energy is given by, U = E qo r Cos U = E qo r Cos180o U = -E qo r UO - UP = E qo . r Because, E = The work done per unit charge in moving a test charge from point to another is the electrostatic potential difference between the two points. Explanation: Let us consider a charge “+q” produced an electrostatic field E . The charge is to be moved from P to O, the work be to be done in the opposite direction of electric field. Mathematical Derivation: U = E q o . r We know that, E qo . r U = qo qo U =E . r qo UO U - P= E . r qo qo The change in potential energy per unit charge between two fixed points is called “potential difference”. VO - VP = E . r V = E . r Thus Potential difference is, “The dot product between electric intensity and displacement”. It is scalar quantity, denoted by “V” having magnitude, V = E r Cos The unit of potential difference is “Volt”, denoted by “V”. Volt: When one-Joule work is to be done in moving one coulomb charge against the electric intensity, then potential difference between two points is one volt. work Potential difference = charge 1 Joule 1 Volt = 1 Coulomb The potential at a point in a field is equal to the work done per coulomb in moving a Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 10 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #10 CHAPTER # 11 PTCL # 022-2670019 11 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 11 COME FOR SUCCESS positively charged particle from infinity to that point in the field. Explanation: Let us suppose a positive charge “+q”, produced an electric field of intensity E . It is required to calculate the potential at an isolated point “B” in the electric field at a distance “r”. Let us calculate work done in carrying the charge “qo” from infinity point “B” to that point A. The displacement is to be divided into very smaller equal displacement elements“ r ”. Mathematical Derivation: Where, r = rB - rA Work done = F . r Or Work = E q o . r Or Work = E qo r Cosθ Work = E q o r Cos 180o Hence, Work = - E qo r We know that, E = 1 4 π εo q r 2 1 q q o r - - - - - - - -►eq. # (1) 4 π ε o r 2 BA r +r The average distance between the these two points is given by, rave = A B 2 2 By simplifying and squaring we get, r = rA rB , equation # 1 can be written as, Work =- Or Work B A Or Work BA We get, Work BA 1 q 4 π ε o rBrA 1 =q qo 4 π εo =- qo rB - rA rB rA rB 1 1 =q qo 4 π εo rA UA + UB = - rA rA rB 1 rB 1 1 1 q qo 4 π εo rA rB 1 UA U 1 1 + B =q - qo qo 4 π εo rA rB The electric field at point B is zero, thus electric potential energy at that point also zero, Say, U B = 0 and rB = . Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 11 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #11 CHAPTER # 11 PTCL # 022-2670019 12 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS Therefore, 12 COME FOR SUCCESS 1 UA 0 1 1 =q - qo qo 4 π ε o rA 1 q VA + 0 = 0 4 π ε o rA VA = - 1 q 4 π ε o rA If, VA = V then, rA = r Thus, absolute potential t any isolated point is given by, V= - 1 4 π εo q r Conclusion: The potential is directly proportional to the charge and inversely proportional to the distance at that point. “The amount of work done in bringing the positive charge from infinity to that point, in the electric field against the electric intensity, is called “absolute potential energy”. Mathematical derivation: 1 q We know that , V = 4 π ε o r work The definition of electric potential energy is, Potential = charge Work = Potential at distance ( r ) Charge ( qo ) 1 q Or Electric absolute potential energy = qo 4 π ε o r 1 q qo Electrical potential energy U = 4 π ε o r Explanation: Suppose a charge “q” of mass “m” is moving under the influence of electric field of intensity “E” from one plate to another with the velocity “v”. Hence, a charge acquires kinetic energy, from the state of rest. It is given by, Electrical kinetic energy = ½ m v2 According to the law of conservation of energy. Loss of electric potential energy = Gain of electric kinetic energy 1 V q = m v2 2 1 V e = m v2 or 2 The modern unit of the electrical energy is “electron volt”. An electron volt is a measure of energy. An electron volt is the kinetic energy gained by an electron passing through a potential Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 12 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #12 CHAPTER # 11 PTCL # 022-2670019 13 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 13 COME FOR SUCCESS difference of one volt “An electron volt (eV) is the energy that an electron gains when it travels through a potential of one volt”. Conversion: Work We know that, V = q Energy (Work) = Vq 1 electron volt (eV) = potential difference charge of an electron. 1eV = V (1.6 10-19 C) J 1eV= (1.610-19 C ) C 1eV = 1.6 10-19 J An equipotential surface refers to a surface where the potential is constant. The intersection of an equipotential surface and a plane results into a path called an equipotential line. No work is done in moving a charge from one point to the other along an equipotential line or surface i.e. VA – VB = 0 the electric flux lines and the equipotential surface and normal to each other. A potential gradient is the rate of change of the potential with respect to displacement. V Potential Gradient = r Units are volts per meter (V/m). The electric field is the same as the potential gradient but with opposite sign. Relation between Electric Potential Gradient and Electric Intensity: The relation between potential difference and electric intensity is given by an equation, V r = - E The change in potential per unit distance in an electric field is called “Potential gradient”. V Potential Gradient = r Potential gradient = - E This is the relation between potential gradient and electric intensity. The electric field is directed from higher potential to lower potential point while the potential gradient is directed from lower to higher potential. A capacitor is a device that stores electric charge on to parallel plates. Construction: Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 13 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #13 CHAPTER # 11 PTCL # 022-2670019 14 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 14 COME FOR SUCCESS A capacitor get's charged when a battery is connected across the two parallel plates. The plate A attached to the positive terminal of the battery get's positively charged +Q and the B plate joined to the negative terminal get's negatively charged -Q. Once capacitor get's fully charged, flow of charge carriers stops in the circuit and in this condition potential difference across the plates of capacitor is same as the potential difference across the terminals of battery . Explanation: Suppose total charge Q is stored on the parallel plates and the potential V of the battery produced across the plates of capacitor. The charge increases to the potential difference increases applied across the plates or vice versa. Mathematical Derivation: Q V Q=CV Thus, Capacitance “C” is a measure of the amount of electric charge stored for a given electric potential V. The capacitance of parallel plate’s capacitor is, Q C V The amount of charge stored between the two plates for a potential difference existing across the plates. It is called “Capacitance of parallel plates capacitor”. The basic unit of capacitance is a farad. The electric charge one coulomb stored on the parallel plates of a capacitor per one Q volt electric potential, the capacitance is called “One farad”. C = V 1 coul. 1Farad = 1Volt Explanation: A capacitor consists of two parallel conducting plates, each of area” A”, separated by a distance “d”. When the capacitor is charged, the plates carry equal amounts of charge, one plate carries positive charge + Q, and the other carries negative charge – Q on the surface. The electric field lines for a parallel-plate capacitor that the field is uniform in the central region between the plates. Mathematical Derivation: The electric intensity due to two oppositely charged pates is given by the Gauss’ law, σ E= εo Q σ= We know that, A Q E= Therefore, - - - - - - - - -► eq. 1 A εo The potential difference “V” is applied between the plates, then V = E d Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 14 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #14 CHAPTER # 11 PTCL # 022-2670019 15 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 15 COME FOR SUCCESS V - - - - - - - - -►eq. 2 d Q V = Compare eq.1 and eq.2 we get, A εo d We know that, Q = C V CV V = Therefore, A εo d A εo C= We get, d Conclusion: The capacitance of a parallel-plate capacitor is directly proportional to the area of its plates and inversely proportional to the plate separation between the plates. Effect of Dielectric: If the dielectric completely fills the space between the plates, the capacitance increases by r, which is called the dielectric constant known as relative permittivity. The capacitance of the capacitor will be, A εoε r C' = d Thus, we see that a dielectric provides the following advantages: • Increase in capacitance • Increase in maximum operating voltage Or, E= There are two methods to find out the equivalent capacitance using methods described.1.parllel and 2. Series combination Description: Two or more capacitors connected as, the left plates of all the capacitors are connected by a wire to the positive terminal of the battery and right plates to the negative terminal of the voltage source. Thus, the potential difference across each capacitor is same and equal to the potential applied. Hence, such network connection is called “parallel combination”. Explanation: Suppose two capacitors C1 and C2 are connected as shown in the figure. One plate of both capacitors is connected at junction A, while the other plate at the B, the junctions is connected with the positive and negative terminals of source respectively. The voltage V across the capacitors is equal to that across the battery terminals. The charges on each capacitor is Q1 and Q2 while. However, different charge on each capacitor is given by, Mathematical Derivation: The total charge Q of the combination capacitor, is Q = Q1 + Q2 Q1 = V C1, and Q2 = V C2 Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 15 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #15 CHAPTER # 11 PTCL # 022-2670019 16 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 16 COME FOR SUCCESS Or Q = V C1 + V C2 Q =V (C1 + C2) Q = C1 + C 2 V These two capacitors are replaced by one equivalent capacitor having a capacitance Ce. Q Ce Therefore, V Thus, Ce= (C1 + C2) Conclusion: The equivalent capacitance of a parallel combination of capacitors is equal to the sum of individual capacitances of capacitors. Description: Two or more capacitors connected as the left plate of first capacitor and the right plate of last capacitor is connected to the terminals of a battery. The remaining plates are connected to each other. The charges on each capacitor connected are same. Such type of network is called “Series combination of capacitors”. Explanation: Let’s suppose two capacitors C1, and C2 are connected as shown. If +Q charge are given to left plates of C1, then by induction Q charge appears on the right plate. +Q appears on the left plate of capacitor C2 and so on. Thus, same charge Q appears on each capacitor. The potential difference across each capacitor however is different. Therefore, Mathematical Derivation: The total potential V of the combination capacitor, is We know that, Q = C1V1 and Q = C2 V2 Q Q V1 = , and V2 = C2 C1 Q Q V= C1 C2 1 1 V=Q C1 C2 V 1 1 = Q C1 C2 These two capacitors are replaced by one equivalent capacitor having a capacitance Ce. V 1 Therefore, = Q Ce 1 1 1 = + Thus, Ce C1 C2 Conclusion: Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 16 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #16 CHAPTER # 11 PTCL # 022-2670019 17 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 17 COME FOR SUCCESS This implies that when the capacitors are connected in series, the reciprocal of the equivalent capacitance equals to the reciprocals of the individual capacitance. 1) Variable capacitor: Dielectric Capacitors are usually of the variable type such as used for tuning transmitters, receivers and transistor radios. Variable capacitors are used for adjustment etc. of frequency. In mechanically controlled variable capacitors, the distance between the plates, or the amount of plate surface area which overlaps, can be changed. The most common form arranges a group of semicircular metal plates on a rotary axis (“rotor”) that are positioned in the gaps between a set of stationary plates (“stator”) so that the area of overlap can be changed by rotating the axis. Air or plastic foils can be used as dielectric material. By choosing the shape of the rotary plates, various functions of capacitance vs. angle can be created, e.g. to obtain a linear frequency scale. Various forms of reduction gear mechanisms are often used to achieve finer tuning control, i.e. to spread the variation of capacity over a larger angle, often several turns. 2)Trimmer capacitors Trimmer capacitors (trimmers) are miniature variable capacitors. They are designed to be mounted directly onto the circuit board and adjusted only when the circuit is built. A small screwdriver or similar tool is required to adjust trimmers. The process of adjusting them requires patience because the presence of your hand and the tool will slightly change the capacitance of the circuit in the region of the trimmer. Trimmer capacitors are only available with very small capacitances, normally less than 100pF. It is impossible to reduce their capacitance to zero, so they are usually specified by their minimum and maximum values, for example 2-10pF 3) Multiple capacitors: This type of capacitor consists a mica dielectric. The capacitance is “n” times the capacitance between two successive plates, where “n” is the number of dielectrics between all the plates .The whole arrangement is sealed into plastic case. Very often, multiple stator/rotor sections are arranged behind one another on the same axis, allowing for several tuned circuits to be adjusted using the same control, e.g. a preselector, an input filter and the corresponding oscillator in a receiver circuit. Capacitors with multiple sections often include trimmer capacitors in parallel to the variable sections, used to adjust all tuned circuits to the same frequency. 4) Paper capacitor: The film and foil types of capacitors are made from long thin strips of thin metal foil with the dielectric material sandwiched together which are wound into a tight roll and then sealed in paper or metal tubes. These film types require a much thicker dielectric film to reduce the risk of tears or punctures in the film, and is therefore more suited to lower capacitance values and larger case sizes. Metallized foil capacitors have the conductive film metallized sprayed directly onto each side of the dielectric which gives the capacitor selfhealing properties and can therefore use much thinner dielectric films. This allows for higher capacitance values and smaller case sizes for a given capacitance. Film and foil capacitors are generally used for higher power and more precise applications. Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 17 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #17 CHAPTER # 11 PTCL # 022-2670019 18 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 18 COME FOR SUCCESS 5) Electrolytic Capacitor: Aluminum is used for the electrodes by using a thin oxidization membrane. Large values of capacitance can be obtained in comparison with the size of the capacitor, because the dielectric used is very thin. The most important characteristic of electrolytic capacitors is that they have polarity. They have a positive and a negative electrode.[Polarised] This means that it is very important which way round they are connected. If the capacitor is subjected to voltage exceeding its working voltage, or if it is nnected with incorrect polarity, it may burst. It is extremely dangerous, because it can quite literally explode. Make absolutely no mistakes. . F 1 q1 q 2 1 q q q 1. F=K 1 2 2 = 2. E = 3. F= E q o 4. E = 5. φ = E . A 6. φ = E A Cos θ 2 qo 4 π ε o r 2 r 4πε o r q Total charge 1 7. φ = q 8. φMax. Flux = E A 9. φMin.Flux = 0 10. Charge density σ = A o work energy σ 1σ 11. E = 12. Potential V = 13. E = 14. V = E . r 15. ΔV = E Δr Cosθ q charge 2 εo εo 1 C' 1 q 21. Ve = m v 2 = ε r 19. Potential energy U = E q . r 20. V = 2 C 4πε o r A εo εr 1 1 1 1 = + 23. C'for dielectric = 24. Ce C1 C2 . 25. d Ce C1 C 2 C3 16. U = V e 17 Q = C V 18. 22. C = A εo d PHYSICAL QUANTITY & SYMBOL Electric Charge, (Q = I t) Electric Intensity, E = F qo Electric Potential, (V =E d ) Electric Flux, ( = E A ) DIMENSION A T UNIT C = amp.sec M L T -3 A -1 N C-1 = kg.m.sec-3.amp-1. M L2 T -3 A -1 M L3 T -3 A -1 J-coul -1 = V= kg.m2.sec-3.amp-1. N - m2 C-1 = Kg. m3. sec-3. amp-1. A εo =[Charge]/[Potential] M -1 L-2 T 4 A 2 Farad = kg-1.m-2sec4.amp2 =A2 C.V-1 d M L2 T -2 Electric potential energy ,U =Ve Joule or electron volt = kg.m2. sec-2 Capacitance, C = A T L Electric dipole moment coul.meter = amp.sec. m. Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 18 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #18 CHAPTER # 11 PTCL # 022-2670019 19 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS Dielectric constant, k = ε εo dimensionless Charge Density, σ = q A Energy, U = V e A T Permittivity of Free Space, ε o = q1 q 2 4 π F r2 F qo Electric Flux, ( = E A ) A εo d Electric potential energy ,U =Ve Electric force, F= E q o A 2 T 4 M -1 L-3 C N -1 m-2 = amp2.sec4.kg-1.m-3 N C-1 = kg.m.sec-3.amp-1. M L2 T -3 A -1 M L3 T -3 A -1 N - m2 C-1 = Kg. m3. sec-3. amp-1. M -1 L-2 T 4 A 2 Farad = kg-1. m-2 sec4.amp2. M L2 T -2 A T L Electric dipole moment ε εo C = amp.sec [M L T-3A-1] Electric Potential, (V =E d ) Capacitance, C = Coul. / m2 q1 q 2 A 2 T 4 M -1 L-3 C N -1 m-2 = amp2.sec4.kg-1.m-3 2 4πFr M L2T -2 A -1 T -1 A T = M L2 T -2 kg.m2.sec2. = J Electric Charge, (Q = I t) Dielectric constant, k = N A T L-2 Permittivity of Free Space, ε o = Electric Intensity, E = no unit MLT -2 C-1 C = M LT -2 Electric force, F= E q o 19 COME FOR SUCCESS J-coul -1 = V= kg.m2.sec-3.amp-1. Joule or electron volt = kg.m2. sec-2 coul.meter = amp.sec. m. dimensionless MLT -2 C-1 C = M LT -2 q A T L-2 A Charge, molar [Charge]/[Quantity] [T.A.mol-1 ] Energy flux [Energy]/[Time]. [ M L2 T 3 ] Charge Density, σ = no unit N m-3.s.A = Coul. / m2 C.mol-1 J.s-1= W Same as power. Q.No.1 What happens to the force between two charges, if the distance between them is halved? q q Answer We know that, F 1 2 r² q q r If, r = then, we know that, F 1 2 2 2 r Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 19 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #19 CHAPTER # 11 PTCL # 022-2670019 20 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS Therefore, F' 20 COME FOR SUCCESS q1 q 2 2 r 2 q q F' 1 2 2 r 4 4 q1 q 2 F' Or r2 F' 4 F This shows that, if the distance between two charges is halved then force increases four times the initial value. Q. No.2 What happens to the force between two point charges if each charge is tripled and the distance between them is made one third? qq Answer We know that, F = K 1 2 2 r 3q 3q F' = K 1 22 1 3 r 9q q F' = K 1 2 or 1 2 r 9 81q1q 2 F' = K r2 F' = 81F . This dhows that force increases 81 times when each charge is tripled and the distance between them is made one third. Q.No.3 Two identical metal spheres carry positive charge and negative charge; they touched together and again separated. What are their charges? Answer The identical charged two spheres carrying positive and negative charge are touched together. They cancel their charge. Hence, they have no any charge after separation. Q. No.4 Explain why it is so much easier to remove an electron from an atom of large atomic weight than it is to remove proton? Answer The outer electrons of large atomic weight have less force of attraction with protons. Thus an atom contains number of orbits. So that, it is easier to remove electron from outer most shell an atom of large atomic weight than proton. Q.No.5 What will be the flux through a surface placed at right angle to the electric field? Answer We know that, = E A Cos = E A Cos 0o = E A, The flux will be maximum through the surface. Q. No.6 Show that, N C-1 = V m-1 Answer N C-1 = [N.m.C]. m-1 N C-1 = [J.C]. m-1 N C-1 = V.m-1 Hence shown Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 20 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #20 CHAPTER # 11 PTCL # 022-2670019 21 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 21 COME FOR SUCCESS Q. No.7 What will be flux through a closed surface, if it is enclosed with no charge? q Answer We know that, φ = εo If, q = 0, then, =o Hence, the flux will be zero through a closed surface. Q. No.8 Convert 1 kWh in to joules. Answer We know that, Work = 1kilowatthour 1kwh = 1000 w x3600 sec 1kwh = 36x 10 5J.sec / sec 1kwh = 3.6 x10 6J Q. No.9 Repulsion is a sure test for electrification. Explain? Answer Attraction is possible even when one of the bodies is charged. But repulsion is possible only when BOTH the bodies are electrically charged. The force exerted by one charge over another, to keep away from one another. Hence, repulsion is the sure test of electrification. Q. No.10 What are equi potential surfaces? Answer The surfaces on which electric lines of force are perpendicular and the value of potentials are same known as “equi potential surfaces”. Q. No.11 The work done in moving the one coulomb charge between two points in an electric field is 10 joules. What is the potential difference between them? Answer From the definition of potential, work Potential = charge 10J Potential = . hence , Potential = 10 volt 1coul Q. No.12 What kind of energy a charged capacitor has got? Answer A charged capacitor has got electrostatic potential energy. Q. No.13 Convert 1eV in to joules. Answer 1eV = 1.6 x 10-19C J / C 1eV = 1.6 x 10-19 J Q. No.14 Define electron volt? Answer It is the electrical energy supplied by the unit potential on to one electronic charge in electric field. Q.No.15 How three capacitors, each of 3 micro farad are arranged to obtain minimum capacitance? Answer The capacitors are arranged in series to obtained minimum valued capacitance. 1 1 1 1 = + Ce C1 C 2 C3 3 1 1 1 1 = = 3 Ce 3 3 3 1 3 = Ce 3 Ce = 1 micro farad Q.No.16 Name some examples in which capacitors are used? Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 21 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #21 CHAPTER # 11 PTCL # 022-2670019 22 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 22 COME FOR SUCCESS Answer The capacitors are used to store charge and energy; they are used in i) flashguns ii) stabilizer iii) fans iv) motors v) computers vi) electronic devices, etc. Q. No.17 How two or more capacitors of equal values, should be connected to obtain maximum capacitance? Answer: To obtain maximum capacitance at constant potential, the capacitors are connected in parallel. Ce = C 1 + C 2 Q. No.18 A capacitor is connected across battery. Why does each plate receive a charge of same magnitude? Will it be true even if the plates are of different sizes? Answer When a capacitor is connected across the battery, pates of a capacitor receive a charge of same magnitude due to induction. This is also true even when, the plates are of different sizes. Q.No.19 If the absolute potential is zero at a point, must the electric field be zero there as well? Answer If the absolute potential is zero at a point, the electric field, intensity must be zero there as well. Q.No.20 Will solid metal sphere hold a large electric charge than a hollow sphere of the same diameter? Where does the charge reside in each case? Answer No, solid metal sphere will not hold a large electric charge than a hollow sphere of same diameter. The charge reside in each case the outer surface. Q. No.21 A parallel plates capacitor has a fixed charge on its plates .The plates are now pulled further apart. Some work must be done during this pull. Why? Does the potential difference change during the process? Answer Yes, work must be done in pulling the charge. And, we know that, V = E d σ The electric intensity due two parallel oppositely charged plates is, E = , εo σd Therefore, V = εo Q If, σ = , as charge density, A Qd Then, V = . A εo This shows that, if Q and A are kept constant then, V d Q. No.22 Can an electric potential exist point in region where electric field is zero. Can the potential be zero at a place where the field intensity is not zero? Answer No, an electric potential cannot exist at a point in a region where the electric field is zero. No, the potential doesn’t be zero at a place, where the electric field intensity is not zero. Q.No.23 How will the capacitor of parallel plates capacitor be affected, if the separation between its plates is doubled and size of the plates is reduced to half? A εo εr Answer We know that, C = d 1 C This shows that, and C A d It means, if the separation between plates is doubled then capacitance decreases to one half the initial value. Similarly, if the size is reduced to half then capacitance also decreases to half the original value. Q. No.24 Define absolute potential energy? Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 22 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #22 CHAPTER # 11 PTCL # 022-2670019 23 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 23 COME FOR SUCCESS Answer The Potential on a positive test charge placed in the electric field, at that point is called “absolute potential energy”. 1 q U= qo 4 π ε o r 2 U = V qo Q. No.25 Define electric intensity and give its unit in SI system. Answer The force per unit charge is called electric intensity. F E = . Its unit is N / coul. in SI System. q Q. No.26 Define coulomb’s law. Answer The magnitude of electrostatic force is directly proportional to the product of charges and inversely proportional to the square of distance between them, is called “Coulomb’s law”. 1 q1 q 2 F= 4 π ε o r 2 Q. No.27 Define volt. Answer One -joule work is to be done on to unit coulomb charge the potential is called “one volt.” 1J work 1volt = or Potential difference = 1coul charge Q. No.28 Define potential difference. Answer Potential Difference between two points is the difference in Potential between the two points and is equal to the work done per unit charge ion carrying a positive test charge from one point to another against the electric field and without any acceleration” Work (U) Potential Difference = Charge(q) V=E .d Q. No.29 Define flux . Answer The dot product between electric intensity and vector area is called “electric flux”. φ = E . A Q.No.30 What happens to the capacitance of parallel plate capacitor if, a) the area of each plate increased and b) the distance between the plates is increased Answer The capacitance of parallel plate’s capacitor, a) “Increased”, if the area of plates increased and b) “Decreases”, if the distance between the plates increased. Q.No.31 How the force will change between two points charges if each charge is double and the distance between them also double. Answer: There will be no change in electrostatic force. 1 2q1 2q 2 F= Because, 4 π ε o 2r 2 1 4 q1 q 2 4 π ε o 4 r 2 1 q1 q 2 Or, F = 4 π ε o r 2 F= Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 23 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #23 CHAPTER # 11 PTCL # 022-2670019 24 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 24 COME FOR SUCCESS Q. No.32 Give the advantages of dielectric medium used in capacitor. Answer If dielectric medium used in capacitor then capacitance increases. We know that, A εoεr C = d Q. No.33 How many electrons contain one coulomb charge? Answer We know that, Q = n e Q n= e 1coul. n= .Thus, n = 6. 25x1018, electrons contain by one coulomb charge. 1.6 10-19 coul Q.No.34 What will happen to the capacitance of a charged capacitor, if a dielectric completely fills the space between the plates? Explain. Answer If a dielectric constant completely filled the space between the plates of a capacitor then capacitance of a capacitor increases. Q. No.35 Why, it is logical to say that potential of an earth connected object is zero? Answer The potential of earth is zero. So that, it is correct to say logical that, potential of earth connected object is zero. Q. No.36 If the capacitor is charged to a certain potential difference then it is immersed in oil. What happens to its capacitance? Answer If the capacitor is charged to a certain potential difference then it is immersed in oil. The capacitance of a capacitor increases. Q. No.37 What is Gaussain surface: Answer Gauss's law is valid for symmetrical charge distribution. Gauss's law is very helpful in calculating electric field in those cases where electric field is symmetrical around the source producing it. Electric field can be calculated very easily by the clever choice of a closed surface that encloses the source charges. Such a surface is called "Gaussian surface". This surface should pass through the point where electric field is to be calculated and must have a shape according to the symmetry of source. Q. No.38 Define flux density . φ Answer Total electric flux passing through a unit area is called electric flux. E = , its unit is web.m-2. A Q. No.39 Can two equi potential surfaces intersect each other? Answer No, two equi potential surfaces can’t intersect each other, because, these surfaces are always perpendicular to electric field. The value of potential is same at every position. Q. No.40 Will the energy stored by three capacitors be greater if they were connected in series or in parallel? 1 Answer We know that, Energy = C V 2 2 This shows that, when capacitors are connected in parallel then energy will be greater. Q. No.41 A capacitor of 6µf is connected in parallel with the combination of 4µF capacitor and a 3µF capacitor that are in series. Find (a) the net capacitance of the entire combination. and (b) The Potential Difference across the 3µF capacitor when 20V is maintained across the 6µF capacitor. AnswerCapacitors 3 µF and 4 µF are in series Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 24 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #24 CHAPTER # 11 PTCL # 022-2670019 25 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 25 COME FOR SUCCESS let Cs = capacitance of a single capacitor which replaces the above two capacitors in series. Then, 1/Cs = 1/4 + 1/3 =(4+3)/12=7/12 therefore Cs = 12/7 (see Fig a ) let Cp = capacitance of the capacitor which replaces the capacitors in parallel on Fig b Cp = (12/7+6) µF = 54/7 µF = net capacitance 20 V potential difference is across 6 µF as well as 12/7 µF as they are in parallel Charge through 12/7 µF =12/7 µF x 20 = 240/7 µC Charge on 4 µF and 3 µF capacitors will be the same as they are in series.Potential difference across 3 µF = (240 µC) / (7 x 3 µF) = 80/7 V = 11.4 V Q: No: 42 Why is static electricity more apparent in winter? Answer: You notice static electricity much more in winter (with clothes in a dryer, or taking a sweater off, or getting a shock when you touch something after walking on carpet) than in summer because the air is much drier in winter than summer. Dry air is a relatively good electrical insulator, so if something is charged the charge tends to stay. In more humid conditions, such as you find on a typical summer day, water molecules, which are polarized, can quickly remove charge from a charged object. Q: No: 43 why there is no net charge on gaussian surface? Answer: There is no reason why there should be zero charge on a Gaussian surface. For example, imagine a conducting sphere on which I have placed a certain amount of charge. All the charge resides on the surface. If I say that Gaussian surfaces are concentric spheres, then the charged surface is one of them. Incidentally, any surface you choose to think about is a Gaussian surface as long as it is closed, that is encloses some volume; it just so happens that the most useful Gaussian surfaces are equipotentials. Q No: 44 Electric field lines never intersect each other. Why? Answer: If they intersect, two tangents can be drawn at the point of intersection indicating two different directions of electric field at the same position which is impossible. Q No: 45 Electric field lines are always normal to the surface of a conductor. Why? Answer: If the electric field line is not normal, it can be resolved into two components, a normal component and a tangential component. The tangential component will make the free charges of the conductor to move along the surface of the conductor. But no such movements are observed indicating that there is no tangential component which implies that the electric field lines are always normal to the surface of a conductor. Q No: 46 What is capacitor? Answer: Which stores charge is called capacitor. Capacitor stores capacity of charge Q No: 47 What is the Field between two Oppositely Charged Parallel Conducting parallel Plates? Answer: The field between the plates is basically uniformly spaced but there is a little fringing, at the edges. We are not concerned with the edges so we can assume it is uniform. These parallel plates are the basis of capacitors. These two plates can be treated as two sheets of charge of opposite charge and we can use Gauss's Law. Outside the two sheets, the components of the electric fields are of the same magnitude but oppositely directed so their resultant is zero. At any point between the plates the components are in the same direction so their resultant is the sum. Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 25 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #25 CHAPTER # 11 PTCL # 022-2670019 26 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 26 COME FOR SUCCESS Basically giving you the same result as that of the charged conducting plate: E= εo Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 26 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #26 CHAPTER # 11 PTCL # 022-2670019 27 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 27 COME FOR SUCCESS Electric Fields Also similar to gravity, the electrostatic force is a non-contact force. Charged objects do not have to be in contact with each other to exert a force on each other. Somehow, a charged object feels the effect of another charged object through space. The property of space that allows a charged object to feel a force is a concept called electric field. Although we cannot see an electric field, we can detect its presence by placing a positive test charge at various points in space and measuring the force the test charge feels. While looking at gravity, the gravitational field strength was the amount of force obsered by a mass per unit mass. The electric field strength is the amount of electrostatic force observed by a charge per unit charge. Therefore, the electric field strength, E, is the electrostatic force observed at a given point in space divided by the test charge itself. Electric field strength is measured in Newtons per Coulomb (N/C). Electric Field Lines Since we can't actually see the electric field, we can draw electric field lines to help us visualize the force a charge would feel if placed at a specific position in space. To help us visualize the electric field, we can draw electric field lines in space. These lines show the direction a positive charged particle would feel a force if it were placed at that point in space. The more dense the lines are, the stronger the force a charged particle would feel, therefore the stronger the electric field. As the lines get further apart, the strength of the electric force a charged particle would feel is smaller, therefore the electric field is smaller. By convention, we draw electric field lines showing the direction of force on a positive charge. Therefore, to draw electric field lines for a system of charges, follow these basic rules: 1. 2. 3. 4. 5. Electric field lines point away from positive charges, and toward negative charges. Electric field lines never cross. Electric field lines always intersect conductors at right angles to the surface. Stronger fields have closer lines. Field strength and line density decreases as you move away from the charges. Let's take a look at a few examples of electric field lines, starting with isolated positive (left) and negative (right) charges. Notice that for each charge, the lines radiate outward or inward spherically. The lines point away from the positive charge, since a positive test charge placed in the field (near the fixed charge) would feel a repelling force. The lines point in toward the negative fixed charge, since a positive test charge would feel an attractive force. Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 27 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #27 CHAPTER # 11 PTCL # 022-2670019 28 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 28 COME FOR SUCCESS If you have both positive and negative charges in close proximity, you follow the same basic procedure: Of course, electric field lines actually lie in three dimensions, as demonstrated in this video animation. Comparing Electrostatics & Gravity Because gravity and electrostatics have so many similarities, let's take a minute to do a quick comparison of electrostatics and gravity. Electrostatics Force: Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 28 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #28 CHAPTER # 11 PTCL # 022-2670019 29 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 29 COME FOR SUCCESS Field Strength: Field Strength: Electrostatic Constant: Charge Units: Coulombs Gravity Force: Field Strength: Field Strength: Gravitational Constant: Mass Units: kilograms The big difference between electrostatics and gravity? The gravitational force can only attract, while the electrostatic force can both attract and repel. Notice again that both the electric field strength and the gravitational field strength follow the inverse-square law relationship. Field strength is inversely related to the square of the distance. Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 29 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #29 CHAPTER # 11 PTCL # 022-2670019 30 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 30 COME FOR SUCCESS Electric Potential Difference When we lifted an object against the force of gravity by applying a force over a distance, we did work to give that object gravitational potential energy. The same concept applies to electric fields as well. If you move a charge against an electric field, you must apply a force for some distance, therefore you do work and give it electrical potential energy. The work done per unit charge in moving a charge between two points in an electric field is known as the electric potential difference, (V). The units of electric potential are volts, where a volt is equal to 1 Joule per Coulomb. Therefore, if you did 1 Joule of work in moving a charge of 1 Coulomb in an electric field, the electric potential difference between those points would be 1 volt. This is given to you in the reference table as: V in this formula is potential difference (in volts), W is work or electrical energy (in Joules), and q is your charge (in Coulombs). Let's take a look at a sample problem. Parallel Plates If you know the potential difference between two parallel plates, you can easily calculate the electric field strength between the plates. As long as you're not near the edge of the plates, the electric field is constant between the plates, and its strength is given by: You'll note that with the potential difference V in volts, and the distance between the plates in meters, units for the electric field strength are volts per meter [V/m]. Previously, we stated that the units for electric field strength were newtons per Coulomb [N/C]. It is easy to show that these units are equivalent: Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 30 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #30 CHAPTER # 11 PTCL # 022-2670019 31 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD. Let’s G¤ To HOUSE OF SUCCESS 31 COME FOR SUCCESS Question: The diagram represents two electrons, e1 and e2, located between two oppositely charged parallel plates. Compare the magnitude of the force exerted by the electric field on e1 to the magnitude of the force exerted by the electric field on e2. Answer: The force is the same because the electric field is the same for both charges, as the electric field is constant between two parallel plates. Equipotential Lines Much like looking at a topographic map which shows you lines of equal altitude, or equal gravitational potential energy, we can make a map of the electric field and connect points of equal electrical potential. These lines, known as equipotential lines, always cross electrical field lines at right angles, and show positions in space with constant electrical potential. If you move a charged particle in space, and it always stays on an equipotential line, no work will be done. Second YEAR Moving 2011 Life Forward 2012 ELECTROSTATICS 31 Theory attracts practice as the magnet attracts iron. — Carl Friedrich Gauss Heart collection, mind satisfaction, best time for carrier, hard working, try for unlimited Education ADDRESS: Behind ALFALAH BANK, Grain Market, Branch. Prince Ali Road, adjacent G.G. JAGHRANI MEDICARE CELL # 0333-2602675 PAGE #31 CHAPTER # 11 PTCL # 022-2670019