The distance between the two charges $25 μ C$ and $36 μ C$ is $11 c m$ At what point on the line joining the two, the intensity will be zero

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At a distance of $5 c m$ from $36 μ C$

At a distance of $55 c m$ from $25 μ C$

At a distance of $5 c m$ from $25 μ C$

At a distance of $11 c m$ from $36 μ C$

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 $x_{1} = \frac{x}{\sqrt{\frac{Q_{2}}{Q_{1}}} + 1} = \frac{11}{\sqrt{\frac{36}{25}} + 1} = 5 c m$

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