Question

The distance between the plates of a reactangular parallel plate capacitor (PPC) of plate area $3140{\text{mm}}^2$, whose capacitance is equal to that of an isolated metallic sphere of radius $2\text{m}$ will be

• A. $1.25\text{mm}$
• B. $0.125\text{mm}$
• C. $2.5\text{mm}$
• D. $0.25\text{mm}$

Answer 1

The correct option is B $0.125\text{mm}$

Given,

Radius of the isolated spherical capacitor, $R=2\text{m}$

Capacitance of isolated spherical capacitor is given by,
$$C_{sphere}=4\pi {\epsilon }_{0}R=4\pi {\epsilon }_0\times 2...\left(1\right)$$Capacitance of parallel plate capacitor is given by, $$C_{ppc}=\frac{{\epsilon }_{0}A}{d}=\frac{3140\times {10}^{-6}{\epsilon }_0}{d}...\left(2\right)$$ According to the problem statement,

$C_{sphere}=C_{ppc}$

$\Rightarrow 4\pi {\epsilon }_0\times 2=\frac{3140\times {10}^{-6}{\epsilon }_0}{d}...\left(2\right)$

$\Rightarrow 4\times 3.14\times {\epsilon }_0\times 2=\frac{3140\times {10}^{-6}{\epsilon }_0}{d}$

$\Rightarrow d=1.25\times {10}^{-4}\text{m}=0.125\text{mm}$

Hence, option $\left(b\right)$ is the correct answer.

Key Concept- Capacitance of parallel plate capacitor and capacitance of an isolated metallic sphere.