Question

When a potential difference of $150V$ is applied to the plates of a parallel plate capacitor, the plates carry a surface charge density $30nC/c{m}^2$. Find the distance between the plates.

• A. $1.23\mu m$
• B. $2.21\mu m$
• C. $2.40\mu m$
• D. $4.42\mu m$

Answer 1

The correct option is D $4.42\mu m$

capacitance in terms of charge and applied voltage.

Let $\sigma$ be the surface charge density of plates.
Capacitance is given as $\left(\frac{{\epsilon }_{0}A}{d}\right)$
Where $A$ and $d$ are area of plates and distance between the plates.

$\text{Charge}Q=CV$

$(V=\text{potential difference}$ )

$\sigma A=\left(\frac{{\epsilon }_{0}A}{d}\right)\left(V\right)$

$d=\frac{{\epsilon }_{0}V}{\sigma }$

$=\frac{(8.84\times {10}^{-12})\left(150V\right)}{3\times {10}^{-4}C/m^2}$ = $4.2\mu F$
Hence, option (d) is correct.