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Question

In the above figure, what is the net electric potential at point $$P$$ due to the four particles if $$V=0$$ at infinity, $$q=5.00 \ fC$$ and $$d=4.00 \ cm$$?

Solution
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A charge $$-5q$$ is a distance $$2d$$ from $$P$$, a charge $$-5q$$ is a distance $$d$$ from $$P$$, and two charges $$+5q$$ are each a distance $$d$$ from $$P$$, so the electric potential at $$P$$ is

$$ V = \dfrac{q}{4 \pi \epsilon_0} \bigg[ -\dfrac{1}{2d} - \dfrac{1}{d} + \dfrac{1}{d} + \dfrac{1}{d} \bigg] = \dfrac{q}{8 \pi \epsilon_0 d} = \dfrac{(8.99 \times 10^9 \ N.m^2/C^2)(5.00 \times 10^{-15} \ C)}{2(4.00 \times 10^{-2} \ m)}$$

The zero of the electric potential was taken to be at infinity.

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