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Unformatted text preview: 975 Chapter 24 1. If the electric potential is zero at infinity then at the surface of a uniformly charged sphere it is V = q /4 πε R , where q is the charge on the sphere and R is the sphere radius. Thus q = 4 πε RV and the number of electrons is ( 29 ( 29 ( 29( 29 6 5 9 2 2 19 1.0 10 m 400V 4 2.8 10 . 8.99 10 N m C 1.60 10 C q R V n e e ε × π = = = = × × ⋅ × 2. The magnitude is ∆ U = e ∆ V = 1.2 × 10 9 eV = 1.2 GeV. 3. (a) An Ampere is a Coulomb per second, so 84 84 3600 30 10 5 A h C h s s h C ⋅ = ⋅ F H G I K J F H G I K J = × . . (b) The change in potential energy is ∆ U = q ∆ V = (3.0 × 10 5 C)(12 V) = 3.6 × 10 6 J. 4. (a) V B – V A = ∆ U / q = – W /(– e ) = – (3.94 × 10 –19 J)/(–1.60 × 10 –19 C) = 2.46 V. (b) V C – V A = V B – V A = 2.46 V. (c) V C – V B = 0 (Since C and B are on the same equipotential line). 5. The electric field produced by an infinite sheet of charge has magnitude E = σ /2 ε , where σ is the surface charge density. The field is normal to the sheet and is uniform. Place the origin of a coordinate system at the sheet and take the x axis to be parallel to the field and positive in the direction of the field. Then the electric potential is V V E dx V Ex s x s = = z , where V s is the potential at the sheet. The equipotential surfaces are surfaces of constant x ; that is, they are planes that are parallel to the plane of charge. If two surfaces are separated by ∆ x then their potentials differ in magnitude by ∆ V = E ∆ x = ( σ /2 ε ) ∆ x . Thus, ∆ ∆ x V = = × ⋅ × = × 2 2 885 10 50 010 10 88 10 12 2 6 3 ε σ . . . . C N m V C m m 2 2 c hb g CHAPTER 24 976 6. (a) ( 29 ( 29 15 19 4 4 3.9 10 N 1.60 10 C 2.4 10 N C 2.4 10 V/m. E F e = = × × = × = × (b) ∆ ∆ V E s = = × = × 2 4 10 012 2 9 10 4 3 . . . . N C m V c hb g 7. (a) The work done by the electric field is 19 12 2 12 2 2 21 (1.60 10 C)(5.80 10 C/m )(0.0356 m) 2 2 2(8.85 10 C /N m ) 1.87 10 J. f d i q q d W q E ds dz σ σ ε ε × × = ⋅ = = = × ⋅ = × ∫ ∫ r r (b) Since V – V = – W / q = – σ z /2 ε , with V set to be zero on the sheet, the electric potential at P is 12 2 2 12 2 2 (5.80 10 C/m )(0.0356 m) 1.17 10 V. 2 2(8.85 10 C /N m ) z V σ ε × =  =  =  × × ⋅ 8. (a) By Eq. 2418, the change in potential is the negative of the “area” under the curve. Thus, using the areaofatriangle formula, we have V E ds x =  ⋅ = = z 10 1 2 2 20 2 r r bgbg which yields V = 30 V. (b) For any region within 0 3 < < ⋅ z x E ds m, r r is positive, but for any region for which x > 3 m it is negative. Therefore, V = V max occurs at x = 3 m. V E ds x =  ⋅ = = z 10 1 2 3 20 3 r r bgb g which yields V max = 40 V....
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 Summer '10
 Prof.Yang
 Physics, Charge, Electric Potential

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