Chapter 20: Electric Potential and Electric Potential Energy Selected Solutions

Selected Solutions

11. (a) No work would be done by the electric fence on a test charge that moves from A to B because the force and the displacement would be perpendicular to each other, therefore, .

(b) Recall that . Notice that E points in the negative x-direction and the displacement Ds, from C to B, is also in the negative x-direction. We, therefore, have

(c) Only the component of the displacement parallel to the electric fence matters; any displacement perpendicular to E doesn't contribute to the change in the electric potential. Notice that the x-component of the displacement is in the positive x-direction. Therefore,

(d) No, it is not possible to determine V_A. We only know the potential differences. We need to know either V_B or V_C to determine V_A.

19. (a) The particle has a positive charge, so it will move in the direction of the electric field, which is the negative x-direction.

(b) From the conservation of energy we know that DK = -DU. The initial kinetic energy is zero and the change in potential energy is given by DU = qDV. Therefore, we have

(c) Its increase in speed will be less than its increase in speed in the first 5.0 cm because v is proportional to the square root of the distance traveled. To see it another way, since E is constant then F is constant which means that the acceleration a is constant. For constant a we know that . Because of the acceleration, the particle will be moving faster during the second 5.0 cm than it was during the first. Therefore, the second 5.0 cm will require less time than the first and it will therefore experience a smaller increase in speed.

31. (a) The work required by an external force to move the +2.7mC charge to infinity equals the work done by the electric force in bringing this charge in from infinity. By definition, this work equals the negative of the change in the electric potential energy when brought in from infinity. Let's call the charges

q₁ = -6.1 mC, q₂ = +2.7 mC, q₃ = -3.3 mC.

Therefore,

(b) Again this work equals -DU:

43. (a) From the definition of capacitance we have that V = Q/C. Plugging in the expression for C for a parallel-plate capacitor gives

(b) The answer to part (a) will decrease because V is inversely proportional to .

(c) Using the same expression as above gives

57.

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