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The distance between the two charges $25 μC$ and $36 μC$ is 11 cm. At what point on the line joining the two, the intensity will be zero

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At a distance of 5 cm from $25 μC$

At a distance of 5 cm from $36 μC$

At a distance of 10 cm from $25 μC$

At a distance of 11 cm from $36 μC$

answer is A.

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Detailed Solution

Suppose electric field is zero at point N in the figure then

At $N |E_{1}| = |E_{2}|$

which gives $x_{1} = \frac{x}{\sqrt{\frac{Q_{2}}{Q_{1}}} + 1} = \frac{11}{\sqrt{\frac{36}{25}} + 1} = 5 cm$

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