Question:
A parallel plate capacitor having a separation between the plates d, plate area A and material with dielectric constant K has capacitance \(C_0\). Now one-third of the material is replaced by another material with dielectric constant 2 K, so that effectively there are two capacitors one with area \(\frac{1}{3}\)A, dielectric constant 2 K and another with area \(\frac{2}{3}\)A and dielectric constant K. If the capacitance of this new capacitor is C then \(\frac{C}{C_0}\) is
A parallel plate capacitor having a separation between the plates d, plate area A and material with dielectric constant K has capacitance \(C_0\). Now one-third of the material is replaced by another material with dielectric constant 2 K, so that effectively there are two capacitors one with area \(\frac{1}{3}\)A, dielectric constant 2 K and another with area \(\frac{2}{3}\)A and dielectric constant K. If the capacitance of this new capacitor is C then \(\frac{C}{C_0}\) is
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The two portions share the same plate separation, so they are in parallel; add their capacitances.
Updated On: Nov 4, 2026
- 1
- \(\frac{4}{3}\)
- \(\frac{2}{3}\)
- \(\frac{1}{3}\)
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The Correct Option is B
Solution and Explanation
\(C_0 = \dfrac{K\epsilon_0 A}{d}\).
Portion 1: \(C_1 = \dfrac{(2K)\epsilon_0 (A/3)}{d} = \dfrac{2}{3}C_0\).
Portion 2: \(C_2 = \dfrac{K\epsilon_0 (2A/3)}{d} = \dfrac{2}{3}C_0\).
They are in parallel: \(C = C_1 + C_2 = \dfrac{4}{3}C_0\).
Answer: (2) \(\dfrac{4}{3}\)
Portion 1: \(C_1 = \dfrac{(2K)\epsilon_0 (A/3)}{d} = \dfrac{2}{3}C_0\).
Portion 2: \(C_2 = \dfrac{K\epsilon_0 (2A/3)}{d} = \dfrac{2}{3}C_0\).
They are in parallel: \(C = C_1 + C_2 = \dfrac{4}{3}C_0\).
Answer: (2) \(\dfrac{4}{3}\)
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