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Question

Charges of $$5 \times 10^{-6}$$ and $$-3 \times 10^{-6} C$$ are kept 16 cm apart. Find the position of points on the line joining the charges where the electric potential is zero.


Solution

Potential at P iss zero
$$\dfrac{k(5 \times 10^{-6})}{x} = \dfrac{k(3 \times 10^{-6})}{0.16-x}$$

$$5(0.16 -x) = 3x$$
$$0.8 = 8x$$
$$x = 0.1m$$
$$x = 10cm$$

987386_1088301_ans_afc3f5bf3d6843088c4887bc21674da0.jpeg

Physics

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