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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.), , Note3: 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, Note1: 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, along, any path. “, 𝐖, V= 𝐪, From figure 1,, 𝐏, ⃗ . ⃗⃗⃗⃗, 𝐕𝐏 -𝐕𝐑 =, = − ∫𝐑 𝐄, 𝐝𝐫, 𝐪, 𝐏, 𝐖∞𝐏, ⃗⃗⃗⃗, ⃗ . 𝐝𝐫, 𝐕𝐏 -𝐕∞ = 𝐪 = − ∫∞ 𝐄, 𝐖𝐑𝐏, , 𝐅𝐄, , (∵ =E), 𝐪, , 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, IT, , =𝐌𝐋𝟐 𝐓 −𝟑 𝐈 −𝟏, , 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, ⃗⃗⃗⃗, ⃗ . 𝐝𝐫, =− ∫𝐑 𝐄, , The work done is independent of the path taken., q1, R, P, q2, q3, It depends only on the initial and final positions., Unit: The SI unit of potential is J/C called Volt

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8, Electrostatic potential and capacitance, , ELECTROSTAICS OF CONDUCTORS, 10. Important results regarding electrostatics of conductors, Behavior of a conductor in an electric field., ⃗=, 𝐄, , 𝛔, 𝛜𝟎, , ̂, 𝐧, , (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, ⃗E ( θ=0), By Gauss’s law,, ⃗⃗⃗⃗, ds, Pill box, ⃗⃗⃗⃗ = 𝟏 q, ⃗ . ds, ∮E, 𝛜𝟎, , Edscos θ =, 𝟏, , 𝟏, 𝛜𝟎, , (𝛔ds), 𝟏, , E= 𝛜 (𝛔), 𝟎, , E= 𝛜 (𝛔) (∵θ=0, cos 0=1), Surface of conductor, 𝟎, The pill box is partly inside and partly outside, 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, 𝟎, , 1. For σ > 0, electric field is normal to the surface outward, 2. For σ < 0, electric field is normal to the surface inward.

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9, Electrostatic potential and capacitance, , 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., Conductors, Dielectrics, In a dielectric, the free movement of charges is, 1. Free charges move 1. Free movement is, not possible. the external field induces dipole, against the external, not possible., moment by stretching or re-orienting molecules of, field., External, field, the dielectric., 2. Induced electric field, induces, dipole, Dielectrics are of two types, cancels the external, moment., 1. Non-polar dielectrics, field., 2. Induced, electric, 2. Polar dielectrics, 3. Inside the conductor,, field reduces the, EExt +Ein =0, external field., 1. Non-polar dielectrics: If the centre of gravity of, 3. EExt +Ein ≠0, inside, a the +ve nuclei and the electron clouds, the dielectric., coincide, the molecule is called non-polar, dielectric., , H2 , N2 , O2 , CO2 etc., H2, , CO2, , Note: Dipole moment of non-polar molecules, is zero., −σp, +σp, 2. Polar dielectrics: If the centre of gravity of a the, no volume charge density, +ve nuclei and the electron clouds donot, Polarisation, coincide, the molecule is called non-polar, When a dielectric is placed in an, dielectric., external electric field, its molecules gain electric, H2 O, NH3 , HCl, etc p, ⃗, ⃗, p, dipole moment and the dielectric is said to be, polarized. The electric dipole moment induced per, unit volume of the dielectric material is called the, HCl, H2 O, electric polarization of the dielectric. P=𝛘𝐞 E, Note: polar molecules have permanent Where, 𝛘𝐞 is electric susceptibility, dipole moment., , 13. Electric Polarisation by external field, 1. Non- polar molecules, , 1., Charges, are, In the absence of displaced in external, external electric field. field., 1. No polarization., 2. Polarised- An, 2.No dipole moment., induced dipole, moment is produced., , 2. Polar molecules, , In the absence of 1. Dipoles tend to align, external electric field., with the external field., 1. Dipoles are oriented 2. Polarized-there is a, randomly., net dipole moment in, 2. Total dipole moment the direction of field., is zero.

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10, Electrostatic potential and capacitance, , 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., The −ve charge on ‘B’ decreases the potential of ‘A’., Capacitance C =, A single, Conductor, , Another conductor ’B’, is placed near ‘A’, , Q, V, , Smaller the potential larger, the capacitance., ∴Capacitance is increased.

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13, Electrostatic potential and capacitance, , ⃗ and ⃗D are parallel., vectors 𝐏, 𝐄, , ⃗⃗⃗D . 𝐧, ̂=𝛔, 21. Combination of capacitors, 1. Capacitors in series, Consider 3 capcitors of capacitances C1 ,C2 and C3, connected in series as shown in the figure., V1, Q, C1, , V2, Q, C2, , V3, Q, C3, , Substitute (2) in (1), Q, V=C +, 1, , Q, C2, , Q, , + C --------(3), 3, , Now replace the three capacitors by a single, capacitor of capacitance ‘C’,, Q, C, Q, V= C, V, we get, , ,, , Q, C, , Q, , =C +, 1, , Q, C2, , +, , Q, --------(4), C3, , V, OR,, Since the capacitors are connected in series,, 1, 1, 1, 1, =, +, +, charges on them are the same but potential, C, C, C, C, 1, 2, 3, difference across the capacitors are different., Note 1: When there are ‘n’ capacitors,, V = V 1 + V2 + V3 --------(1), 1, 1, 1, 1, But,, = C + C +……….+ C, C, 1, 2, n, Q, V1 = C, 1, Q, Note 2: In series, effective capacitance is less than, V2 =, -------(2), C2, the lowest value of capacitance in that, Q, combination., V3 = }, C3, , Substitute (2) in (1), , 2. Capacitors in parallel, , Q = C1 V + C2 V + C3 V --------(3), , Consider 3 capcitors of capacitances C1 ,C2 and C3, connected in parallel as shown in the figure., C3, , Q3, , C2 Q 2, C1, , Q1, , Now replace the three capacitors by a single, capacitor of capacitance ‘C’,, Q, C, Then,, we get, , V, Q=CV, CV = C1 V + C2 V + C3 V --------(4), , OR,, V, C = C1 + C2 + C3, Since the capacitors are connected in parallel,, potential difference across the capacitors are the, Note 1: When there are ‘n’ capacitors,, same but charges on them are different ., , Q = Q 1 + Q2 + Q 3 --------(1), But,, Q 1 = C1 V, Q 2 = C2 V }-------(2), Q 3 = C3 V, , C = C1 + C2 +………+ Cn, Note 2: In parallel, effective capacitance is larger, than the highest value of capacitance in that, combination.

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14, Electrostatic potential and capacitance, , 22. Energy stored in a capacitor, 𝐐, , W = ∫𝟎 𝐝𝐰, , Energy stored in a capacitor is the work done, in charging it., , 𝐐q, , =∫𝟎, =, , Consider a capacitor of capacitance ‘C’, charged, to a potential difference ‘V’ by giving a charge ‘q’., , =, , C, , dq, , 𝐐, q, ∫, 𝟎, 𝑪, 𝟏, , dq, , 𝐐, 𝟏 𝐪𝟐, , 𝑪, , [𝟐], , 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,, , 𝐐𝟐, , U = 𝟐𝐂, , q, , dw= C dq, , 𝐐𝟐, , The total work done to increase the charge, from 0 to Q,, , (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, , 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, will be transferred to the outer surface of Working, the hollow sphere., Brush B1 is positively charged. The positive, charge on brush B1 is sprayed off to the belt(due, Description of the apparatus, to corona discharge). Belt carries these charges, upwards. Brush B2 collects these charges and, It consists of a large metallic sphere, supported 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

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15, Electrostatic potential and capacitance, , 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.), [Ans: 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?, [Ans: 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?, [Ans: 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?, [Ans: 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, , ;, , C1 =, , t, (d−t+ ), K, , 𝛆𝟎 𝐀, 3, 3d, (d− d+ ), 4, 4K, , =, , 𝛆𝟎 𝐀, , 𝐂, , = 𝟒−𝟑 𝟎 3 =𝟏, , 3 3, d(1− + ), 4 4K, , 𝟒, , +, , 𝐂𝟎, , 3, +, 𝟒 4K, , 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., 𝟏, 𝟏, 𝟏, 𝟏, = + +, 𝐂 𝟏𝟎 𝟏𝟎 𝟏𝟎, , [ (a) ≡, , 𝟑, , 𝟏𝟎, , = 𝟏𝟎; C= 𝟑 =3.33µF;, , 𝐂𝐞𝐟𝐟=c+𝐜𝟒 = 3.33 + 10 = 13.33µF;, (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., Q’=CV’;, , V, , ∴ V’=C ;, , 𝟏, , Total energy E=2 x 𝟐Q’V’ =𝟐 𝒙, , 𝟏𝐐V, 𝟐𝟐 2, , 1, , 1, , =4 QV=4 x9x10−8x100=2.25 x𝟏𝟎−𝟔 𝐉]

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16, Electrostatic potential and capacitance, Problem 5: NCERT e.g. 2.3, charge respectively., , Figures (a), , 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, , [Ans: (a) As V∝ r , 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.], Problem 6: NCERT e.g. 2.6, A molecule of a substance has a permanent electric dipole moment of, magnitude 10–29 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., [Ans: 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]

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18, Electrostatic potential and capacitance, NEET 2016: An electric dipole is placed at an angle of 30° with an electric field intensity 2 × 105 N/C. It, experiences a torque equal to 4 Nm. The charge on the dipole, if the dipole length is 2 cm, is, (1) 8 mC, (2) 2 mC, (3) 5 mC, (4) 7 µC, τ, , 4, , [Ans: (2)][𝛕=pE𝐬𝐢𝐧 𝛉= q(2l) Esin θ; q=(2l) E sin θ=(2x 10−2 )x 2 × 105, , sin 30, , =2x 𝟏𝟎−𝟑 C], , NEET 2016: A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four, dielectric materials having dielectric constants k1, k2, k3 and k4 as shown in the figure below. If a single, dielectric material is used to have the same capacitance C in this capacitor, then its dielectric constant k, is given by, , 𝟏, , 𝟏, , 𝟏, , 𝐏, , 𝐒, , [Ans: (3)][ 𝐂 =𝐂 + 𝐂 ; ϵ, 𝐞𝐟𝐟, , d, , d⁄2, , =ϵ, , +ϵ, , 0 (K1 +K2 +K3 )A/3, , 0K A, , d⁄2, , ;, , 0K 4A, , d, d⁄2, 1, 1 2, 𝟑, 𝟏, =, {, + K };K = (𝐊 +𝐊 +𝐊 ) + 𝐊 ], ϵ0 K A ϵ0 A (K1 +K2 +K3 )/3, 4, 𝟏, 𝟐, 𝟑, 𝟒, , NEET 2017: A capacitor is charged by a battery. The battery is removed and another identical uncharged, capacitor is connected in parallel. The total electrostatic energy of resulting system is, , 1, , 1, , V, , 1 C V2, 2, , [Ans:(1)][ Ui =2 C V 2; Uf =2 2C ( 2 )2 = 2, , ;, , Uf =, , 𝐔𝐢, 𝟐, , Ans: Decreases by a factor of 2 ], , NEET 2018: The electrostatic force between the metal plates of an isolated parallel plate capacitor C, having a charge Q and area A, is, (a) Independent of the distance between the plates (b) Linearly proportional to the distance between, the plates (c) Proportional to the square root of the distance between the plates (d) Inversely, proportional to the distance between the plates., 𝑞, , [Ans:(a)]Electrostatic force Fplate=qE=q 2𝜀, , 0, , 𝑞2, , =, ; F is independent of the distance between the plates.], 𝐴 2𝜀 𝐴, 0, , NEET 2020: In a certain region of space with volume 0.2 m3, the electric potential is found to be 5 V, throughout. The magnitude of electric field in this region is, (a) Zero, , (b) 0.5 N/C, , (c) 1 N/C, , [Ans:(a)][ Electric field in a region, E = −𝑑𝑟𝑑𝑉;, , (d) 5 N/C, , But here electric potential is constant., Therefore electric field will be zero. ], XXXXXXXXXXX