Question

The particle $P$ shown in figure has a mass of $10\text{mg}$ and a charge of $-0.01\mu \text{C}$. Each plate has a surface area $100{\text{cm}}^2$ on one side. What potential differences $V$ should be applied to the combination to hold $P$ in equilibrium ?

• A. $95\text{V}$
• B. $43\text{V}$
• C. $70\text{V}$
• D. $80\text{V}$

Answer 1

The correct option is B $43\text{V}$

Equivalent capacitance of the system,

$C_{eq}=\frac{C}{2}=\frac{0.04}{2}=0.02\mu \text{F}$

Charge on each capacitor will be

$q=C_{eq}\times V=(0.02\times {10}^{-6})V\text{Coulomb}$

Electric field inside capacitor is given by

$E=\frac{\sigma }{{ϵ}_o}=\frac{q}{A{ϵ}_o}[A=100{\text{cm}}^2]$

$\Rightarrow E=\frac{0.02\times {10}^{-6}\times V}{(100\times {10}^{-4})\times (8.85\times {10}^{-12})}$


For equilibrium of particle, net force on particle will be zero.

$qE=mg$

$(0.01\times {10}^{-6})\times \frac{0.02\times {10}^{-6}V}{100\times {10}^{-4}\times 8.85\times {10}^{-12}}=10\times {10}^{-3}\times 9.8$

$\Rightarrow V=\frac{1000\times 8.85\times {10}^{-7}\times 9.8}{0.02\times 0.01}$

$\therefore$ $V=43.36\approx 43\text{V}$

Hence, option $\left(a\right)$ is correct.