Suppose a uniformly charged wall provides a uniform electric field of 2 × 10 4 N C - 1 normally. A charged…

As we can see from geometry, $\sin \theta =\frac{10}{20}=\frac{1}{2}$

$\Rightarrow \theta =30°$

Vertical component of tension will balance the weight and the horizontal component will balance the electric force. Therefore,

$\tan \theta =\frac{qE}{mg}$

$\Rightarrow \tan 30°=\frac{q\times 2\times {10}^4}{2\times {10}^{-3}\times 10}$

$\Rightarrow \frac{1}{\sqrt{3}}=q\times {10}^6$

$\Rightarrow q=\frac{1}{\sqrt{3}}\times {10}^{-6} \mathrm{C}=\frac{1}{\sqrt{3}} \mathrm{\mu C}$

Hence, $x=3$.