Two point charges +4 $μ$ C and –6 $μ$ C are separated by a distance of 20 cm in air. At what point on the line joining the two charges from the charge +4 C the electric potential will be zero?

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6 cm

10 cm

40cm

12 cm

answer is B.

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Detailed Solution

When charges are dissimilar, we get null pts outside, and inside nearer to smaller charge $x = \frac{r}{\frac{q_{2}}{q_{1}} - 1} \Rightarrow x = \frac{20}{\frac{6}{4} - 1} = 40 c m$ out side

$x = \frac{r}{\frac{q_{2}}{q_{1}} + 1} \Rightarrow x = \frac{20}{\frac{6}{4} + 1} = 8 c m$ inside

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