Question:

Two circular plates each of radius '\(r\)' are kept parallel to each other distance '\(d\)' apart. The capacitance of the capacitor formed is '\(C_1\)'. If the radius of each of the plates is increased to \(\sqrt{3}\) times the earlier radius and their distance of separation decreased to half the initial value, the capacitance now becomes '\(C_2\)'. The ratio \(C_1:C_2\) is

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Capacitance is proportional to area and inversely proportional to separation.
Updated On: Oct 1, 2026
  • \(1:2\)
  • \(1:4\)
  • \(1:6\)
  • \(6:1\)
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
For a parallel plate capacitor, \(C = \frac{\varepsilon_0 A}{d}\). For circular plates, \(A = \pi r^2\).

Step 2: Key Formula or Approach:
New radius \(= \sqrt3\,r\), so the new area is \(\pi\cdot 3r^2 = 3A\). New separation \(= \frac d2\).

Step 3: Detailed Explanation:
\[ C_2 = \frac{\varepsilon_0 (3A)}{d/2} = 6\cdot\frac{\varepsilon_0 A}{d} = 6C_1 \]
So \(\frac{C_1}{C_2} = \frac16\), which is \(1 : 6\).
Option D, \(6:1\), is the inverse ratio.

Final Answer:
\(C_1 : C_2 = 1 : 6\), option (C). \[ \boxed{1:6} \]
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