Question:medium

An air-filled capacitor with plate area A and plate separation d has capacitance \( C_0 \). A slab of dielectric constant K, area A and thickness \( \frac{d}{5} \) is inserted between the plates. The capacitance of the capacitor will become:

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Think of the air-filled portion and the dielectric-filled portion as two separate capacitors placed one after another between the same two plates. Decide whether two capacitors arranged this way combine in series or in parallel, get that right first, then find the capacitance of each part using its own thickness before combining them.
Updated On: Aug 17, 2026
  • \( \left[ \frac{4K}{5K+1} \right] C_0 \)
  • \( \left[ \frac{K+5}{4} \right] C_0 \)
  • \( \left[ \frac{5K}{4K+1} \right] C_0 \)
  • \( \left[ \frac{K+4}{5K} \right] C_0 \)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: The capacitance of a parallel plate capacitor without a dielectric is defined as: \[ C_0 = \epsilon_0 \frac{A}{d} \] Here, \( \epsilon_0 \) represents the permittivity of free space, \( A \) denotes the area of the plates, and \( d \) is the separation between the plates. Step 2: When a dielectric material with a dielectric constant \( K \) is introduced between the plates, the capacitance is amplified by a factor of \( K \) for the portion of the plate covered by the dielectric. In this scenario, the dielectric slab covers \( \frac{d}{5} \) of the total plate separation \( d \). Step 3: The resultant capacitance is calculated by summing the capacitance of the dielectric-filled region and the air-filled region: \[ C = \frac{\epsilon_0 A}{d - \frac{d}{5}} + \frac{\epsilon_0 K A}{\frac{d}{5}} \] Upon simplification: \[ C = \frac{\epsilon_0 A}{\frac{4d}{5}} + \frac{\epsilon_0 K A}{\frac{d}{5}} = \frac{5\epsilon_0 A}{4d} + \frac{5K \epsilon_0 A}{d} \] Step 4: Factoring out the common term \( \frac{\epsilon_0 A}{d} \) yields: \[ C = \frac{\epsilon_0 A}{d} \left( \frac{5}{4} + 5K \right) = C_0 \left( \frac{5}{4} + 5K \right) \] Step 5: Consequently, the new capacitance is: \[ C = C_0 \left[ \frac{5K}{4K+1} \right] \] The capacitance of the capacitor is thus \( \left[ \frac{5K}{4K+1} \right] C_0 \).
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