Question

If the shown system is in equilibrium, then the surface charge density on the large sheet is
[Take ${ϵ}_0=9\times {10}^{-12}{\text{C}}^2{\text{/Nm}}^2$ and $g=10{\text{m/s}}^2$]

• A. $7.8\times {10}^{-7}{\text{C/m}}^2$
• B. $4.8\times {10}^{-7}{\text{C/m}}^2$
• C. $6.8\times {10}^{-7}{\text{C/m}}^2$
• D. $5.8\times {10}^{-7}{\text{C/m}}^2$

Answer 1

The correct option is A $7.8\times {10}^{-7}{\text{C/m}}^2$

Free body diagram of the system is given below:


From equilibrium of forces along $\text{x}-$axis,
$T\sin {60}^{\circ }=qE.....\left(1\right)$
From equilibrium of forces along $\text{y}-$axis,
$T\cos {60}^{\circ }=mg.....\left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$,
$\tan {60}^{\circ }=\frac{qE}{mg}.....\left(3\right)$

Electric field due to a plate having surface charge density $\sigma$ is
$E=\frac{\sigma }{2{ϵ}_0}$

Substituting this in $\left(3\right)$, we get
$\tan {60}^{\circ }=\frac{q\sigma }{2{ϵ}_{0}mg}$

$\Rightarrow \sigma =\frac{2{ϵ}_{0}mg\tan {60}^{\circ }}{q}$

$\Rightarrow \sigma =\frac{2\times 9\times {10}^{-12}\times 10\times {10}^{-3}\times 10\times \tan {60}^{\circ }}{4\times {10}^{-6}}$

$\therefore \sigma =7.8\times {10}^{-7}{\text{C/m}}^2$

Hence, option (d) is the correct answer.