Chapter 17, Problem 2Q

If a negative charge is initially at rest in an electric field, will it move toward a region of higher potential or lower potential? What about a positive charge? How does the potential energy of the charge change in each instance? Explain

To determine

How does the potential energy of the charge change in each instance?

Answer to Problem 2Q

Solution:

Negative charge moves to a region of higher potential, positive charge moves to a region of lower potential and in both cases potential energy decreases.

Explanation of Solution

We have potential difference between two points, ${\text{V}}_{ba}=V_b-V_a=\frac{P{E}_b-P{E}_a}{q}$, where $PE$ is the potential energy and q is the charge.

Now when a negative charge is released, it will move towards positive plate. Because of the acceleration of the charge its kinetic energy increases. When kinetic energy increases potential energy decreases. i.e. $P{E}_{b}P{E}_a$ and $V_{ba}=\frac{P{E}_b-P{E}_a}{q}=\frac{-ve}{-ve}=+ve$, so potential difference between 2 points is positive. Which means negative charge moves to a region of higher potential.

When a positive charge is released, it will move towards negative plate. Because of the acceleration of the charge its kinetic energy increases. When kinetic energy increases potential energy decreases. i.e. $P{E}_{b}P{E}_a$ and ${\text{V}}_{ba}=\frac{P{E}_b-P{E}_a}{q}=\frac{-ve}{+ve}=-ve$, so potential difference between 2 points is negative. Which means positive charge moves to a region of lower potential.