Question 72 of 311

Suppose a uniformly charged wall provides a uniform electric field of $2\times {10}^{4}N{C}^{-1}$ normally. A charged particle of mass $2g$ being suspended through a silk thread of length $20cm$ and remain stayed at a distance of $10cm$ from the wall. Then the charge on the particle will be $\dfrac{1}{\sqrt{x}}\mu C$ where $x=$ ________ [use $g=10m{s}^{-2}$]

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Answer: 3

As we can see from geometry, $\sin \theta =\dfrac{10}{20}=\dfrac{1}{2}$
$\Rightarrow \theta =30^{\circ}$
Vertical component of tension will balance the weight and the horizontal component will balance the electric force. Therefore,
$\tan \theta =\dfrac{qE}{mg}$
$\Rightarrow \tan 30^{\circ}=\dfrac{q\times 2\times {10}^{4}}{2\times {10}^{-3}\times 10}$
$\Rightarrow \dfrac{1}{\sqrt{3}}=q\times {10}^{6}$
$\Rightarrow q=\dfrac{1}{\sqrt{3}}\times {10}^{-6}C=\dfrac{1}{\sqrt{3}}\mu C$
Hence, $x=3$.