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2, Electrostatic potential and capacitance, , 1. Electrostatic potential Energy, 1. Electrostatic Potential energy difference, “Electrostatic potential energy difference, between two points is defined as the work done by, an external force in bringing the charge ‘q’ from, one point to another. (Without any acceleration,, against the electrostatic force, Fext = − 𝐅𝐄 ) ”, , Note 2: The work done depends only on the initial, and the final points and is independent of the path, taken. (It is the fundamental characteristic of a, conservative force.), , Note 3: The potential energy difference can be, defined in terms of the physically meaningful, Imagine that we bring a test charge q from a quantity work. So, The actual value of potential, point R to a point P against the repulsive force on energy is not physically significant; it is only the, difference of potential energy that is significant., it due to the charge Q., Note 5: Adding an arbitrary constant ‘α’ to, +q, potential energy at every point will not change the, potential energy difference., +Q, P, R, Work done by external forces in moving a, (𝐔𝐏 + 𝛂)-(𝐔𝐑 + 𝛂) = 𝐔𝐏 -𝐔𝐑, charge q from R to P is,, P, Note 4: There is a freedom in choosing the point, WRP=∫R Fext . dr, But Fext = − 𝐅𝐄, where potential energy is zero. A convenient, choice is to have electrostatic potential energy, 𝐏, ⃗⃗⃗⃗ ------(1), zero at infinity., 𝐖𝐑𝐏 =− ∫𝐑 ⃗⃗⃗, 𝐅𝐄 . 𝐝𝐫, When P is at infinity,, 𝐖∞𝐏 = 𝐔𝐏 -𝐔∞ = 𝐔𝐏 ( from eqn(2)), This work done is against electrostatic repulsive, force and gets stored as potential energy., 2. Electrostatic Potential energy, “Electrostatic potential energy of a charge ‘q’ at, But, Potential energy difference, a, point, is defined as the work done by an external, 𝐖𝐑𝐏 = ∆U = 𝐔𝐏 -𝐔𝐑 ------(2), force in bringing the charge ‘q’ from infinity to, Note: RHS of Eq. (2) depends only on the initial that point.”, Potential energy at ‘P’, 𝐔𝐏 = 𝐖∞𝐏, and final positions of the charge., , 2. Electric potential, “Electrostatic potential at a point is defined as, the work done in bringing a unit positive charge, from infinity to that point (without any, acceleration) against the electrostatic force. “, 𝐖, V= 𝐪, From figure 1,, 𝐏, ⃗ . ⃗⃗⃗⃗, 𝐕𝐏 -𝐕𝐑 = 𝐪 = − ∫𝐑 𝐄, 𝐝𝐫, 𝐏, 𝐖, ⃗⃗⃗⃗, ⃗ . 𝐝𝐫, 𝐕𝐏 -𝐕∞ = 𝐪∞𝐏 = − ∫∞ 𝐄, 𝐖𝐑𝐏, , 𝐅𝐄, , (∵ 𝐪 =E), , 𝐏, , ⃗⃗⃗⃗, 𝐕𝐏 = − ∫∞ 𝐄⃗ . 𝐝𝐫, Note: It is independent of test charge ‘q’, but, characteristic of the electric field associated with, the source charge., [W], , Dimension: [V] = V= [q] ;, , =, , ML2 T−2, =𝐌𝐋𝟐 𝐓 −𝟑 𝐈 −𝟏, IT, , Potential difference, “The potential difference between two points in, an electric field is defined as the work done in, bringing a unit positive charge from one point to, another without any acceleration against the, electrostatic force”, , 𝐕𝐏 -𝐕𝐑, , =, , 𝐖𝐑𝐏, 𝐪, , 𝐏, ⃗ . ⃗⃗⃗⃗, =− ∫𝐑 𝐄, 𝐝𝐫, , The work done is independent of the path taken., q1, R, P, q2, q3, It depends only on the initial and final positions.

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8, Electrostatic potential and capacitance, Problem 1:An electric dipole consisting of two charges +1𝛍C and -1𝛍C are separated by 2cm.It is placed, in an external electric field of 106 NC −1 .Calculate the work done to rotate from 0° to 180°., [U= 𝐩𝐄 (𝐜𝐨𝐬 𝛉𝟏 -𝐜𝐨𝐬 𝛉 𝟐 ); U= pE (cos 0-cos 180) = (q2a)E [1-(−1)] =(10−6x2x10−2)x106 x2 = 4x𝟏𝟎−𝟐 𝐉], , ELECTROSTAICS OF CONDUCTORS, 10. Important results regarding electrostatics of conductors, Behavior of a conductor in an electric field., ⃗= 𝛔 𝐧, ̂, 𝐄, 𝛜, 𝟎, , (at surface), , 1. Inside a conductor, electrostatic field is zero., (Electrostatic shielding, the defining, property of a Conductor), 2. The free electrons move against the direction, of the electric field., (The free charges have so distributed, themselves, that the electric field is zero, everywhere inside.), , 3. Charges reside only on the surface of the, conductor., (Consider any closed surface S bounding a, volume element V. Inside the surface electric, field is zero. Thus the total electric flux through, S is zero. Hence, by Gauss’s law, there is no net, charge enclosed by S.), 4. At the surface of the conductor, electric field is, perpendicular to the surface., ⃗ would have some, (If not perpendicular, E, non-zero component along the surface. Free, charges on the surface of the conductor would, then move which is not possible in static, situation.), 5. Potential is constant inside and outside the, surface of the conductor., ⃗ = 0 inside the conductor and has, (Since E, no tangential component on the surface, work, done is zero. That is, there is no potential, difference between any two points inside or on, the surface of the conductor.), , Note 1: If a conductor is charged, electric field normal to the surface exists; this means potential will be, different for the surface and a point just outside the surface, Note 2: each conductor is characterised by a constant value of potential., , 11. Electric field at the surface of a charged conductor., Prove that, for a conductor without any surface, Just inside the surface, the electrostatic field is, charge density, field is zero even at the surface., zero; just outside, ‘E.’, Thus, the contribution to the total flux through, Consider a pill box (a short cylinder) as the the pill box comes only from the outside crossGaussian surface ., sectionof the pill box, ⃗ ( θ=0), E, By Gauss’s law,, ⃗⃗⃗⃗, ds, Pill box, 𝟏, ds = 𝛜 q, ∮ ⃗E. ⃗⃗⃗⃗, 𝟎, , 𝟏, , Edscos θ = 𝛜 (𝛔ds), 𝟎, , Surface of conductor, The pill box is partly inside and partly outside, , 𝟏, E= 𝛜, 𝟎, , (𝛔), , 𝟏, , E= 𝛜 (𝛔), 𝟎, , (∵θ=0, cos 0=1)

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9, Electrostatic potential and capacitance, the surface . It has a small area of cross section ‘ds’ ∴ Ans: If 𝛔=0, E= 𝟏 (𝛔) =0, 𝛜𝟎, and negligible height., Vector form:, , ⃗= 𝛔 𝐧, ̂, 𝐄, 𝛜, , 1. For σ > 0, electric field is normal to the surface outward, , 𝟎, , 2. For σ < 0, electric field is normal to the surface inward., , 12. Dielectrics and polarisation, Materials which do not conduct electricity, but Differences in behavior of a conductors and, transmit electrical effects are called dielectrics., dielectrics in an external electric field., e.g. Mica,PVC,glass,water etc., In a dielectric, the free movement of charges is, Conductors, Dielectrics, not possible. the external field induces dipole, 1. Free charges move 1. Free movement is, moment by stretching or re-orienting molecules of, against the external, not possible., the dielectric., field., External, field, Dielectrics are of two types, 2. Induced electric field, induces, dipole, 1. Non-polar dielectrics, cancels the external, moment., 2. Polar dielectrics, field., 2. Induced, electric, 3. Inside the conductor,, field reduces the, 1. Non-polar dielectrics: If the centre of gravity of, EExt +Ein =0, external field., a the +ve nuclei and the electron clouds, 3. EExt +Ein ≠0, inside, coincide, the molecule is called non-polar, the dielectric., dielectric., , H2 , N2 , O2 , CO2 etc., H2, , CO2, , Note: Dipole moment of non-polar molecules, is zero., 2. Polar dielectrics: If the centre of gravity of a the, −σp, +σp, +ve nuclei and the electron clouds donot, no volume charge density, coincide, the molecule is called non-polar, Polarisation, dielectric., When a dielectric is placed in an, H2 O, NH3 , HCl, etc p, ⃗, ⃗, p, external electric field, its molecules gain electric, dipole moment and the dielectric is said to be, polarized. The electric dipole moment induced per, HCl, H2 O, unit volume of the dielectric material is called the, Note: polar molecules have permanent electric polarization of the dielectric. P=𝛘𝐞 E, dipole moment., Where, 𝛘𝐞 is electric susceptibility, , 13. Electric Polarization by external field

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10, Electrostatic potential and capacitance, 1.No, external, electric field., 2. No polarization., 3.No dipole moment., , 1.Charges, are, displaced in external, field., 2.PolarizedAn, induced, dipole, moment is produced., , 1. Non- polar molecules, , 1.No external electric, field, 2. Dipoles are oriented, randomly., 3. Total dipole moment, is zero., , 1. Dipoles tends to align, with the external field., 2. Polarized-there is a, net dipole moment in, the direction of field., , 2., Polar molecules, CAPACITORS AND CAPACITANCE, , 14. Capacitor, A capacitor is a device used for storing charges., , If charges are given to a conductor, its potential, rises. The rise in potential is directly proportional, It consists of two conductors separated by a to the charge given to it., dielectric medium or vacuum., i.e., Q∝V, Q=CV, Q, Uses:, C=, 1. In radio and television tuners, V, Where C is capacitance (capacity), 2. To produce and transmit EM waves, Note 1: A capacitor is a system of two conductors separated by an insulator (Fig. 1)., The two conductors have charges Q and – Q, with potential difference V = V1 – V2 between them., Note 2: Even a single conductor can be used as a capacitor by assuming the other at infinity., Note 3: The conductors may be so charged by connecting them to the two terminals of a battery., Note 4: Q is called the charge of the capacitor, though this, in fact, is the charge on one of the, conductors – the total charge of the capacitor is zero, Note 5: C is independent of Q or V, Note 6: The capacitance C depends only on the, 1. Geometrical configuration (shape, size, separation) of the system of two conductors., 2. the nature of the insulator (dielectric), , 15. Capacitance, Q, Define 1Farad, Capacitance, C= V, ”The capacitance of a capacitor is said to be one, “The ratio of the charge given to a conductor to, Farad, if its potential is raised by 1V, when 1 C of, the rise in potential is called capacitance.”, charge is given to it.”, , Unit: Farad (C V–1), Q, It, Dimension: C = V = V ;, IT, , [C] = 𝐌𝐋𝟐𝐓 −𝟑 𝐈−𝟏 = 𝐌 −𝟏 𝐋−𝟐 𝐓 𝟒 𝐈𝟐, , Note:, , 1 milli Farad (1mF)= 10−3F, 1 micro Farad (1μF)= 10−6F, 1 pico Farad (1pF)= 10−12F, , 16. Principle of a Capacitor, A single conductor can store charge., , When ‘B’ is placed near ‘A’, −ve charge is, induced on the near face of ‘B’ and +ve charge on, The capacity of a conductor can be increased by the farther face., placing another earth connected conductor near it., When ‘B’ is earthed, +ve charge flows to the earth.

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14, Electrostatic potential and capacitance, we get, , CV = C1 V + C2 V + C3 V --------(4), , V, Since the capacitors are connected in parallel, OR,, potential difference across the capacitors are the, C = C1 + C2 + C3, same but charges on them are different ., Note 1: When there are ‘n’ capacitors,, Q = Q 1 + Q2 + Q 3 --------(1), But,, C = C1 + C2 +………+ Cn, Q 1 = C1 V, Q 2 = C2 V }-------(2), Note 2: In parallel, effective capacitance is larger, Q 3 = C3 V, than the highest value of capacitance in that, combination., , 22. Energy stored in a capacitor, 𝐐, , Energy stored in a capacitor is the work done, in charging it., Consider a capacitor of capacitance ‘C’, charged, to a potential difference ‘V’ by giving a charge ‘q’., , W = ∫𝟎 𝐝𝐰, 𝐐q, , =∫𝟎, =, =, , C, , dq, , 𝐐, q, ∫, 𝟎, 𝑪, 𝟏, , 𝟏 𝐪𝟐, , [ ], 𝑪 𝟐, , dq, 𝐐, 𝟎, , Then,, q = CV, q, 𝟏 𝐐𝟐, Or,, V = C --------(1), = 𝑪 [ 𝟐 − 𝟎], The work done to increase the charge by a, Work done is stored in the capacitor as its, small amount dq is given by,, potential energy., dw= V dq -----------(2), , Or,, , dw=, , q, C, , 𝐐𝟐, , U = 𝟐𝐂, , dq, , The total work done to increase the charge, from 0 to Q,, , 23. Van de graaff generator, It is a device used to produce charges on large, scale. It can produce very high voltage of about, 15 x 106 V., Use: for accelerating atomic particles in atomsmashing experiments., Principle: It is based on two principles,, 1. The phenomenon of corona discharge., The process, by which the charge at sharp, pointed ends of a conductor gets discharge is, called corona discharge., , 2. If a charged conductor is brought in internal, contact with a hollow sphere, all the charges, , 𝐐𝟐, , (𝐂𝐕)𝟐, , Note 1: Q=CV, U = 𝟐𝐂=, Q, , 𝟏, , 𝟐𝐂, 𝟏Q, , Note 2: C = V, U=𝟐 C𝐕 𝟐 ==𝟐 V 𝐕𝟐, , 𝟏, , U=𝟐 C𝐕 𝟐, 𝟏, , U=𝟐 QV

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15, Electrostatic potential and capacitance, will be transferred to the outer surface of, the hollow sphere., Working, Brush B1 is positively charged. The positive, Description of the apparatus, charge on brush B1 is sprayed off to the belt(due, to corona discharge). Belt carries these charges, It consists of a large metallic sphere, supported upwards. Brush B2 collects these charges and, transfer them to the metallic sphere. These, on insulated pillars T1 and T2 . A non-conducting, positive charges move on to the outer surface of, belt is wound around two pulleys P1 and P2 . A, the hollow sphere. This process continues and, motor rotates this belt., very high potential is produced., , XXXXXXXXXX, 24. Problems, Problem 1:NCERT e.g. 2.7 (a) A comb run through one’s dry hair attracts small bits of paper. Why?, What happens if the hair is wet or if it is a rainy day? (Remember, a paper does not conduct electricity.), [When a comb run through one’s dry hair, it gets charged due to friction. The molecules in the paper gets, polarised by the charged comb. Hence it attracts small bits of paper. If hair is wet, the comb does not get, charged due to the absence of friction.], (b) Ordinary rubber is an insulator. But special rubber tyres of aircraft are made slightly conducting., Why is this necessary?, [To enable them to conduct charge (produced by friction) to the ground], (c) Vehicles carrying inflammable materials usually have metallic ropes touching the ground during, motion. Why?, [To conduct charge (produced by friction) to the ground; as too much of frictional electricity, accumulated may result in spark and result in fire.], (d) A bird perches on a bare high power line, and nothing happens to the bird. A man standing on the, ground touches the same line and gets a fatal shock. Why?, [Current passes only when there is a potential difference.], Problem 2: NCERT e.g. 2.8 A slab of material of dielectric constant K has the same area as the plates of a, parallel-plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the, capacitance changed when the slab is inserted between the plates?, [C=, , 𝛆𝟎 𝐀, t, K, , ;, , (d−t+ ), , C1 =, , 𝛆𝟎 𝐀, 3, 4, , 3d, 4K, , (d− d+ ), , =, , 𝛆𝟎 𝐀, 3 3, 4 4K, , 𝐂, , = 𝟒−𝟑 𝟎 3 =𝟏, , d(1− + ), , 𝟒, , +, , 4K, , 𝐂𝟎, , 3, +, 𝟒 4K, , 𝐂, , = K+3𝟎, , =, , 4K, , 𝟒𝐊 𝐂𝟎, 𝐊+𝟑, , ], , Problem 3: NCERT e.g. 2.9 A network of four 10 μF capacitors is connected to a 500 V supply, as shown, in Fig. 2.29. Determine (a) the equivalent capacitance of the network and (b) the charge on each, capacitor., [ (a) ≡, , 𝟏, 𝟏, 𝟏, 𝟏, = + +, 𝐂 𝟏𝟎 𝟏𝟎 𝟏𝟎, , 𝟑, , 𝟏𝟎, , = 𝟏𝟎; C= 𝟑 =3.33µF;, , 𝐂𝐞𝐟𝐟=c+𝐜𝟒 = 3.33 + 10 = 13.33µF;

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16, Electrostatic potential and capacitance, (b)Q 4 =C4 V=10x10−6x500=5x𝟏𝟎−𝟑 C ;, , Q=CV=3.33x10−6x500=166510−6C=1.7𝟏𝟎−𝟑 C ], , Problem 4: NCERT e.g. 2.10 (a) A 900 pF capacitor is charged by 100 V battery[fig.1] . How much, electrostatic energy is stored by the capacitor? (b) The capacitor is disconnected from the battery and, connected to another 900 pF capacitor [Fig. 2]. What is the electrostatic energy stored by the system?, 𝟏, , 1, , [Q=CV=900x10−12 x100=9x10−8C; E=𝟐QV=2x9x10−8x100= 4.5x𝟏𝟎−𝟔 𝐉], [(b) In the steady situation, the common potential difference be V’, Q’ = Q/2., V, , Q’=CV’; ∴ V’=C ;, , Problem 5: NCERT e.g. 2.3, charge respectively., , Figures (a), , 𝟏, , Total energy E=2 x 𝟐Q’V’, , = E=𝟐 𝒙, , 𝟏𝐐 V, 𝟐𝟐 2, , 1, , 1, , =4 QV=4 x9x10−8x100=2.25 x𝟏𝟎−𝟔 𝐉], , Figures (a) and (b) show the field lines of a positive and negative point, , Figures (b), , (a) Give the signs of the potential difference VP – VQ; VB – VA., (b) Give the sign of the potential energy difference of a small negative charge between the points Q and, P; A and B., (c) Give the sign of the work done by the field in moving a small positive charge from Q to P., (d) Give the sign of the work done by the external agency in moving a small negative charge from B to A., (e) Does the kinetic energy of a small negative charge increase or decrease in going from B to A?, 1, r, , [(a) As V∝ , VP > VQ. Thus, (VP – VQ) is positive. Also VB is less negative than VA. Thus, VB> VAor (VB– VA), is positive., (b) A small negative charge will be attracted towards positive charge.The negative charge moves from, higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a, small negative charge between Q and P is positive., Similarly, (P.E.)A > (P.E.)B and hence sign of potential energy differences is positive., (c) In moving a small positive charge from Q to P, work has to be done by an external agency against the, electric field. Therefore, work done by the field is negative., (d) In moving a small negative charge from B to A work has to be done by the external agency. It is, positive., (e) Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy, decreases in going from B to A.]

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17, Electrostatic potential and capacitance, Problem 6: NCERT e.g. 2.6, A molecule of a substance has a permanent electric dipole moment of, –29, magnitude 10 C m. A mole of this substance is polarised (at low temperature) by applying a strong, electrostatic field of magnitude 106 V m–1. The direction of the field is suddenly changed by an angle of, 60°. Estimate the heat released by the substance in aligning its dipoles along the new direction of the, field. For simplicity, assume 100% polarisation of the sample., [dipole moment of each molecules =, For 1 mole, total dipole moment, p = 6 × 1023 × 10–29 C m= 6 × 10–6 C m:, , 10–29Cm;, , Initial potential energy,, Ui =–pE cos θ =–6×10–6×106 cos 0° =–6J:, Final potential energy (when θ = 60°),, Uf = –6 × 10–6 × 106 cos 60°, =–3J:, Change in potential energy, = –3 J – (–6J) = 3 J:, This must be the energy released], Problem 7: NCERT e.g. 2.5 (a) Determine the electrostatic potential energy of a system consisting, of two charges 7 μC and –2 μC (and with no external field) placed at (–9 cm, 0, 0) and (9 cm, 0, 0), respectively., (b) How much work is required to separate the two charges infinitely away from each other?, 1, r, , (c) Suppose that the same system of charges is now placed in an external electric field E = A ( 2);, A = 9 × 105 C m–2. What would the electrostatic energy of the configuration be?, 𝟏, , [[a) U = 0+𝟒𝛑𝛜, , 𝐪𝟏 𝐪𝟐, ;U, 𝟎 𝐫𝟏𝟐, , =𝟗 𝐱 𝟏𝟎𝟗 𝐱, , 𝟕 𝐱 𝟏𝟎−𝟔 (−𝟐 𝐱 𝟏𝟎−𝟔 ), =−𝟕, 𝟏𝟖 𝐱𝟏𝟎−𝟐, , 𝐱 𝟏𝟎−𝟏 =, , −𝟎. 𝟕J], , b) W = U2 – U1 = 0 – U = 0 – (–0.7) =, , 0.7 J], , c) mutual interaction energy + energy of interaction of the two charges with the external electric field., −dV, ;, dr, , E=, , U=𝐪𝟏 V+𝐪𝟐 V-0.7=, , r−1, , 1, , 𝐀, , 𝐄 𝐝𝐫=-∫ A (r2 ) dr=-𝐴 ∫ r −2 dr=-A[ −1 ]= 𝐫 :, , dv=-E dr; v=-∫, , 𝐀, ( 𝟕 𝐱 𝟏𝟎−𝟔 +−𝟐 𝐱 𝟏𝟎−𝟔 )-0.7, 𝐫, , =, , 9 × 105, 0.09, , 𝐱𝟓 𝐱 𝟏𝟎−𝟔 - 0.7 =50-0.7, , =49.3J], , Problem 8: NCERT EXE 2.2 A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices., Calculate the potential at the centre of the hexagon., [, , 𝟏, , 𝐪, , 𝟓 𝐱 𝟏𝟎−𝟔, , VP =𝟒𝛑𝛜 𝐫 ; VP = 𝟔 𝐱 𝟗𝐱 𝟏𝟎𝟗 𝐱 𝟏𝟎 𝐱 𝟏𝟎−𝟐= 27 x 𝟏𝟎𝟓V, 𝟎, , =2.7 x 𝟏𝟎𝟔 V], , Problem 9: NCERT EXE 2.3 Two charges 2 μC and –2 μC are placed at points A and B 6 cm apart., (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every, point on this surface?, [(a) The plane normal to AB and passing through its mid-point has zero potential everywhere., (b) Normal to the plane in the direction AB.], Problem 10: NCERT EXE 2.5 A parallel plate capacitor with air between the plates has a capacitance of, 8 pF (1pF = 10–12 F). What will be the capacitance if the distance between the plates is reduced by half,, and the space between them is filled with a substance of dielectric constant 6?

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19, Electrostatic potential and capacitance, , XXXXXXXXXXX