Question:
A parallel plate air capacitor has capacity ' \(C\) ' and distance of separation between plates is ' \(d\) '. If a conducting sheet of thickness \(\frac{2 d}{3}\) is inserted in between the plates, the capacitance becomes \(C_1\). The ratio of \(\frac{C_1}{C}\) is
A parallel plate air capacitor has capacity ' \(C\) ' and distance of separation between plates is ' \(d\) '. If a conducting sheet of thickness \(\frac{2 d}{3}\) is inserted in between the plates, the capacitance becomes \(C_1\). The ratio of \(\frac{C_1}{C}\) is
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A conducting sheet of thickness t effectively reduces the plate separation to d minus t.
Updated On: Nov 13, 2026
- \(2:1\)
- \(4:1\)
- \(3:1\)
- \(5:1\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Without the sheet, \(C=\dfrac{\varepsilon_0A}{d}\).
Step 2: With the sheet
A conductor of thickness \(t=\dfrac{2d}{3}\) leaves air gaps of total thickness \(d-t=\dfrac d3\).
Step 3: New capacitance
\(C_1=\dfrac{\varepsilon_0A}{d-t}=\dfrac{\varepsilon_0A}{d/3}=3C\).
Step 4: Ratio
\(\dfrac{C_1}{C}=3:1\)
Final Answer:
The ratio of the capacitances is 3 : 1. \[ \boxed{\text{(C) }3:1} \]
Without the sheet, \(C=\dfrac{\varepsilon_0A}{d}\).
Step 2: With the sheet
A conductor of thickness \(t=\dfrac{2d}{3}\) leaves air gaps of total thickness \(d-t=\dfrac d3\).
Step 3: New capacitance
\(C_1=\dfrac{\varepsilon_0A}{d-t}=\dfrac{\varepsilon_0A}{d/3}=3C\).
Step 4: Ratio
\(\dfrac{C_1}{C}=3:1\)
Final Answer:
The ratio of the capacitances is 3 : 1. \[ \boxed{\text{(C) }3:1} \]
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