Question

Two parallel plate capacitors have their plate areas $100{\text{cm}}^2$ and $500{\text{cm}}^2$ respectively. If they have the same charge and potential difference, and the distance between the plates of first capacitor is $0.5\text{mm},$ then the distance between the plates of second capacitor is

• A. $0.10\text{cm}$
• B. $0.15\text{cm}$
• C. $0.20\text{cm}$
• D. $0.25\text{cm}$

Answer 1

The correct option is D $0.25\text{cm}$

Since the potential difference and charge on capacitor is same for both capacitors.

From the relation of capacitance,

$Q=CV\Rightarrow \frac{Q}{V}=C=\text{Constant}$

$\therefore C_1=C_2$

Using the formula of capacitance of parallel plate capacitor, we get

$\therefore \frac{A_1{ϵ}_0}{d_1}=\frac{A_2{ϵ}_0}{d_2}.........\left(1\right)$


Here $A_1=100{\text{cm}}^2;A_2=500{\text{cm}}^2;d_1=0.5\text{mm};d_2=d$

Substituting in eq. $\left(1\right)$

$\Rightarrow \frac{100{ϵ}_0}{0.5}=\frac{500{ϵ}_0}{d}$

$\Rightarrow d=2.5\text{mm}=0.25\text{cm}$

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

$\begin{array}{c}\text{Why this question?}\\ \text{Concept : The capacitance of a parallel plate}\\ \text{capacitor is the ratio of charge on capacitor}\\ \text{to the potential difference between its plates.}\end{array}$