Two parallel air capacitors have their plate areas $100 \mathrm{~cm}^2$ and $500 \mathrm{~cm}^2$,…
Two parallel air capacitors have their plate areas $100 \mathrm{~cm}^2$ and $500 \mathrm{~cm}^2$, respectively. They have the same charge and potential. If the distance between the plates of the first capacitor is $0.5 \mathrm{~mm}$, the distance between the plates of the second capacitor is
$1 \mathrm{~cm}$
$0.75 \mathrm{~cm}$
$0.25 \mathrm{~cm}$
$0.52 \mathrm{~cm}$
Solution
Here, two parallel plate capacitors have same charge q and same potential V, so they have equal capacitances as
$\begin{aligned} & \mathrm{C}=\frac{\mathrm{q}}{\mathrm{V}} \\ & \mathrm{C}_1=\mathrm{C}_2 \\ & \frac{\varepsilon_0 \mathrm{~A}_1}{\mathrm{~d}_1}=\frac{\varepsilon_0 \mathrm{~A}_2}{\mathrm{~d}_2} \\ & \text { Or } \mathrm{d}_2=\frac{\mathrm{A}_2}{\mathrm{~A}_1} \mathrm{~d}_1 \\ & \text { Now, } \mathrm{A}_1=100 \mathrm{~cm}^2, \mathrm{~A}_2=500 \mathrm{~cm}^2 \\ & \mathrm{~d}_1=0.5 \mathrm{~mm}=0.05 \mathrm{~cm} \\ & \therefore \mathrm{d}_2=\frac{500 \times 0.05}{100}=0.25 \mathrm{~cm}=2.5 \mathrm{~mm}\end{aligned}$