ExamBro
ExamBro
MHT CET · Physics · Capacitance

Two circular metal plates each of radius ' \(r\) ' are kept parallel to each other distance 'd' apart. The capacitance of the capacitor formed is ' \(\mathrm{C}_1\) '. If the radius of each of the plates is increased to \(\sqrt{2}\) times the earlier radius and their distance of separation decreased to half the initial value, the capacitance now becomes ' \(\mathrm{C}_2\) '. The ratio of the capacitances \(\mathrm{C}_1: \mathrm{C}_2\). is

  1. A \(1: 1\)
  2. B \(1: 2\)
  3. C \(1: 4\)
  4. D \(4: 1\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(1: 4\)

Step-by-step Solution

Detailed explanation

For the \(1^{\text {st }}\) capacitor,
\(\mathrm{C}_1=\frac{\varepsilon_0 \mathrm{~A}_1}{\mathrm{~d}}=\frac{\varepsilon_0 \pi \mathrm{r}^2}{\mathrm{~d}}\)
For the \(2^{\text {nd }}\) capacitor,
\(\begin{aligned}
& \mathrm{C}_2=\frac{\varepsilon_0 \mathrm{~A}_2}{\mathrm{~d}}=\frac{\varepsilon_0 \pi 2 \mathrm{r}^2}{\frac{d}{2}}=\frac{\varepsilon_0 4 \pi \mathrm{r}^2}{\mathrm{~d}}=4 \mathrm{C}_1 \\
\therefore \quad & \frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{\mathrm{C}_1}{4 \mathrm{C}_1}=\frac{1}{4}
\end{aligned}\)
Same subject
Explore more questions on app
From MHT CET
Explore more questions on app