College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 17 - Problems - Page 653: 22


At the point $(4.0,~0)$, the electric potential is $-2.7\times 10^6~V$

Work Step by Step

We can find the electric potential at the point $(4.0, ~0)$: $V = \frac{k~q_1}{r_1}+ \frac{k~q_2}{r_2}$ $V = k~(\frac{q_1}{r_1}+ \frac{q_2}{r_2})$ $V = (9.0\times 10^9~N~m^2/C^2) (\frac{2.0\times 10^{-3}~C}{4.0~m} + \frac{-4.0\times 10^{-3}~C}{5.0~m})$ $V = -2.7\times 10^6~V$ At the point $(4.0,~0)$, the electric potential is $-2.7\times 10^6~V$.
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