Chapter-24 - 975 Chapter 24 1. If the electric potential is...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. 24-18, the change in potential is the negative of the “area” under the curve. Thus, using the area-of-a-triangle 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....
View Full Document

Page1 / 44

Chapter-24 - 975 Chapter 24 1. If the electric potential is...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online