Question

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

Answer 1

Given that mass of particle $m=10mg$

The Charge of the particle is $1=-0.01\mu C$

The area of the plates of capacitor is $A=100c{m}^2$

Let potential $=V$

The equation capacitance

$C=\frac{Q}{V}$

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

At equilibrium, the weight of the particle is balanced by the force of electric field.

$\therefore qE=Mg$

Electric force $=qE=q\frac{V}{d}$ where V - Potential , d - separation of both the plates

$=q\frac{VC}{{\epsilon }_{0}A}$

Because $C=\frac{{\epsilon }_{0}A}{d}⟹d=\frac{{\epsilon }_{0}A}{C}$

$qE=mg$

$\Rightarrow \frac{QVC}{{\epsilon }_{0}A}=mg$

$=\frac{0.01\times 0.02\times V}{8.85\times {10}^{-12}\times 100}=0.1\times 980$

$\Rightarrow V=\frac{0.1\times 980\times 8.85\times {10}^{-10}}{0.0002}=43mV$