Question

A parallel plate capacitor with the plate area $100{\text{cm}}^2$ and the separation between the plates $1.0\text{cm}$ is connected across a battery of emf $24\text{V}$. Find the force of attraction between the plates.

• A. $2.55\times {10}^{-7}\text{N}$
• B. $5\times {10}^{-7}\text{N}$
• C. $7.25\times {10}^{-7}\text{N}$
• D. $9.11\times {10}^{-7}\text{N}$

Answer 1

The correct option is A $2.55\times {10}^{-7}\text{N}$

Given,

Area of a parallel plate capacitor $A=100{\text{cm}}^2$

Separation between the plates $d=1.0\text{cm}$

Applied voltage $V=24\text{volts}$

Capacitance of the parallel plate capacitor $C=\frac{{\epsilon }_{0}A}{d}=\frac{8.85\times {10}^{-12}\times 100\times {10}^{-4}}{{10}^{-2}}=8.85\text{pF}$

Energy stored in the capacitor,

$E=\frac{1}{2}C{V}^2=\frac{1}{2}\times 8.85\times (24{)}^2\times {10}^{-12}=2.55\times {10}^{-9}\text{J}$

Force between the parallel plates $F=\frac{E}{d}=2.55\times {10}^{-7}\text{N}$

Hence, option (a) is the correct answer.