Question

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

• A. $1.6\times {10}^{-5}\text{cm}$
• B. $1.0\times {10}^{-7}\text{N}$
• C. $4\times {10}^{-5}\text{N}$
• D. $2.5\times {10}^{-7}\text{N}$

Answer 1

The correct option is D $2.5\times {10}^{-7}\text{N}$

$A=100{\text{cm}}^2=100\times {10}^{-4}{\text{m}}^{2}d=1.0\times {10}^{-2}\text{m}E=24\text{V}$

Force between plates of capacitor, $F=\frac{{\epsilon }_{0}A{E}^2}{2d^2}$

Substituting the data given in the question,

$F=\frac{8.89\times {10}^{-12}\times 100\times {10}^{-4}\times 24\times 24}{2\times ({10}^{-2}{)}^2}$

$=8.89\times 100\times 12\times 24\times {10}^{-10}$

$=2.5\times {10}^{-7}\text{N}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (d) is the correct answer.