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1. Electric Charges and Fields
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The distance between the two charges $25\,\mu C$ and $36\,\mu C$ is $11\,cm$ At what point on the line joining the two, the intensity will be zero
A
At a distance of $5\,cm$ from $25\,\mu C$
B
At a distance of $5\,cm$ from $36\,\mu C$
C
At a distance of $10\,cm$ from $25\,\mu C$
D
At a distance of $11\,cm$ from $36\,\mu C$
Solution
(a) Suppose electric field is zero at point $N$ in the figure then
At $N$ |$E_1$| = |$E_2$|
which gives ${x_1} = \frac{x}{{\sqrt {\frac{{{Q_2}}}{{{Q_1}}}} + 1}} = \frac{{11}}{{\sqrt {\frac{{36}}{{25}}} + 1}} = 5\,cm$
Standard 12
Physics
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