Question

A parallel plate capacitor of $20\mu \text{F}$ is connected to a battery of emf $8.85\text{V}$. If the plate area of each plate is $2{\text{cm}}^2$, with what magnitude of force the positively charged plate attracts the negative one?

• A. $4.425\times {10}^6\text{N}$
• B. $17.7\times {10}^6\text{N}$
• C. $8.85\times {10}^6\text{N}$
• D. $2.21\times {10}^6\text{N}$

Answer 1

The correct option is C $8.85\times {10}^6\text{N}$

Given:
$C=20\mu \text{F};V=8.85\text{V};A=2{\text{cm}}^2$

The force of attraction in such a case in given by,

$F=\frac{Q^2}{2A{ϵ}_0}.......\left(1\right)$

From definition of capacitance,

$Q=CV$

$\Rightarrow Q=20\times 8.85\mu \text{C}$

Substituiting all data in eq. $\left(1\right)$,

$F=\frac{{20}^2\times {8.85}^2\times {10}^{-12}}{2\times 2\times {10}^{-4}\times 8.85\times {10}^{-12}}$

$\Rightarrow F=8.85\times {10}^6\text{N}$

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