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# ch24 - Chapter 24 1(a An ampere is a coulomb per second so...

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973 Chapter 24 1. (a) An ampere is a coulomb per second, so 84 A h 84 C h s 3600 s h 3.0 10 5 C. (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. If the electric potential is zero at infinity then at the surface of a uniformly charged sphere it is V = q /4 0 R , where q is the charge on the sphere and R is the sphere radius. Thus q = 4 0 RV and the number of electrons is  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 4. (a)   15 19 4 4 3.3 10 N 1.60 10 C 2.1 10 N C 2.1 10 V/m. E F e (b) V E s 2.4 10 4 N C 0.12m 2.9 10 3 V. 5. 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 s E dx 0 x V s Ex , 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 2 0 V 2 8.85 10 12 C 2 N m 2 50V 0.10 10 6 C m 2 8.8 10 3 m.

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CHAPTER 24 974 6. (a) V B – V A = U / q = – W /(– e ) = – (4.78 10 –19 J)/(–1.60 10 –19 C) = 2.99 V. (b) V C – V A = V B – V A = 2.99 V. (c) V C – V B = 0 (since C and B are on the same equipotential line). 7. 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 d r s 0 ). Then, we connect the origin to B with a line along the x axis, along which the change in potential is V   r E d r s   4.00 xdx   4.00 4 2 2 0 4 0 x 4 which yields V B – V A = –32.0 V. 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  2 0 1 10 2 10 2 x V E ds   which yields V = 20 V. (b) For any region within 0 3m, x E ds 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.   3 0 1 10 3 10 2 x V E ds   which yields V max = 25 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 25 V. Using the formula for a triangle (3 < x < 4) and a rectangle (4 < x < X ), we require    1 1 10 4 10 25.
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ch24 - Chapter 24 1(a An ampere is a coulomb per second so...

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