## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 17 - Problems - Page 652: 14

#### Answer

The work done by the electric field is zero.

#### Work Step by Step

Assuming that the electric potential at infinity is zero, we can find the electric potential at the point $b$: $V = k~(\frac{q_1}{r_1}+\frac{q_2}{r_2})$ $V = (9.0\times 10^9~N~m^2/C^2)~(\frac{8.00\times 10^{-9}~C}{0.040~m}+\frac{-8.00\times 10^{-9}~C}{0.04~m})$ $V = 0$ We can find the change in potential energy when $q_3$ is moved to point $b$: $\Delta U = q_3~V$ $\Delta U = (2.00\times 10^{-9}~C)(0)$ $\Delta U = 0$ Since the work done by the electric field is $-\Delta U$, the work done by the electric field is zero.

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