Question:

An isolated capacitor of capacitance 100 \(\mu\)F is charged to 32 C. If 16 C of charge is discharged from it, then its capacitance value is

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Capacitance \( C = \epsilon_0 A / d \).
Unless you change the area (\( A \)), the distance (\( d \)), or the dielectric (\( \epsilon_r \)), the capacitance will not change.
Updated On: Apr 29, 2026
  • 50 \(\mu\)F
  • 100 \(\mu\)F
  • 5 \(\mu\)F
  • 200 \(\mu\)F
  • 32 \(\mu\)F
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
Capacitance (\( C \)) is an intrinsic property of a capacitor that depends on its physical dimensions (area of plates, distance between them) and the dielectric material between the plates.

Step 3: Detailed Explanation:

The relationship between charge (\( Q \)), capacitance (\( C \)), and potential difference (\( V \)) is given by \( Q = CV \).
However, \( C = Q/V \) does not mean that the capacitance depends on the charge.
Just as the capacity of a bucket to hold water does not change whether it is full or half-empty, the capacitance of a capacitor remains constant regardless of the amount of charge stored in it.
Since the physical structure of the isolated capacitor has not changed, its capacitance remains 100 \(\mu\)F.

Step 4: Final Answer:

The capacitance value remains 100 \(\mu\)F.
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