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MHT CET · Physics · Capacitance

If the distance between the plates of a parallel plate capacitor of capacity \(10 \mu \mathrm{F}\) is doubled, then new capacity will be

  1. A \(5 \mu \mathrm{F}\)
  2. B \(20 \mu \mathrm{F}\)
  3. C \(10 \mu \mathrm{F}\)
  4. D \(15 \mu \mathrm{F}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(5 \mu \mathrm{F}\)

Step-by-step Solution

Detailed explanation

In parallel plate capacitor
\(
C=\varepsilon_{0} \frac{A}{d}
\)
Now
\(
d^{\prime}=2 d
\)
Hence,
\(
C^{\prime}=\frac{\varepsilon_{0} A}{d^{\prime}}=\frac{\varepsilon_{0} A}{2 d}
\)
From Eqs. (i) and (ii)
\(
\begin{array}{l}
\frac{C^{\prime}}{C}=\frac{1}{2} \\
C^{\prime}=\frac{C}{2}
\end{array}
\)
Putting \(C=10 \mu \mathrm{F}\)
\(
C^{\prime}=\frac{10}{2}=5 \mu \mathrm{F}
\)
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