Question

Two plates of a capacitor of the area $100{\text{cm}}^2$ are separated by a distance of $8\text{cm}$, and charged by a battery of emf $56\text{V}$. If the separation is increased by $8.5\text{mm}$, then the new capacitance will be

• A. ${10}^{-13}\text{F}$
• B. ${10}^{-11}\text{F}$
• C. ${10}^{-14}\text{F}$
• D. ${10}^{-12}\text{F}$

Answer 1

The correct option is D ${10}^{-12}\text{F}$

Given:
$A=100{\text{cm}}^2;d=8\text{cm};V=56\text{V}$

Let $l$ be the increased distance between plates.

$l=8.5\text{mm}$

It is pretty clear that the new capacitance of capacitor will be

$C_{new}=\frac{A{ϵ}_0}{d+l}$

Change in voltage or charge will not influence the capacitance .

Let's substitute the values

$A=100{\text{cm}}^2={10}^{-2}{\text{m}}^2$

$d=8\text{cm}=8\times {10}^{-2}\text{m}$

$l=8.5\text{mm}=0.85\times {10}^{-2}\text{m}$

$\therefore d+l=8.85\times {10}^{-2}\text{m}$

$C_{new}=\frac{{10}^{-2}\times 8.85\times {10}^{-12}}{8.85\times {10}^{-2}}$

$\Rightarrow C_{new}={10}^{-12}\text{F}$

Key concept- Capacitance does not depend on its charge or voltage across it.