{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# CH-24 - Chapter 24 1(a An Ampere is a Coulomb per second so...

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

Chapter 24 1. (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. 2. The magnitude is U = e V = 1.2 10 9 eV = 1.2 GeV. 3. The electric field produced by an infinite sheet of charge has magnitude E = /2 0 , where 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 0 , 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 = ( /2 0 ) x . Thus, x V 2 2 885 10 50 010 10 88 10 0 12 2 6 3 . . . . C N m V C m m 2 2 c hb g 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. (a) E F e 39 10 160 10 2 4 10 15 19 4 . . . . N C N C c hc h (b) V E s 2 4 10 012 2 9 10 4 3 . . . . N C m V c hb g 6. (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 995

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

View Full Document
CHAPTER 24 V E ds x  z 10 1 2 2 20 0 2 bgbg which yields V = 30 V. (b) For any region within 0 3 z x E ds m, 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 0 3 bgbg which yields V max = 40 V. (c) In view of our result in part (b), we see that now (to find V = 0) we are looking for some X > 3 m such that the “area” from x = 3 m to x = X is 40 V. Using the formula for a triangle (3 < x < 4) and a rectangle (4 < x < X ), we require 1 2 1 20 4 20 40 bgbgb gbg X . Therefore, X = 5.5 m. 7. (a) The work done by the electric field is (in SI units) 19 12 21 0 0 0 12 0 0 0 (1.60 10 )(5.80 10 )(0.0356) 1.87 10 J. 2 2 2(8.85 10 ) f d i q q d W q E ds dz (b) Since V – V 0 = – W / q 0 = – z /2 0 , with V 0 set to be zero on the sheet, the electric potential at P is (in SI units) 12 2 12 0 (5.80 10 )(0.0356) 1.17 10 V. 2 2(8.85 10 ) z V    8. We connect A to the origin with a line along the y axis, along which there is no change of potential (Eq. 24-18: E ds z 0 ). Then, we connect the origin to B with a line along the x axis, along which the change in potential is V E ds xdx x    F H G I K J z z 4 00 4 00 4 2 2 0 4 0 4 . .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}