Question:

A parallel plate capacitor with plate area \(A\) and separation \(d\) is filled with a dielectric of constant \(K\). The capacitance becomes:

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Dielectric increases capacitance by factor \(K\).
Updated On: Jan 5, 2026
  • \(\varepsilon_0A/d\)
  • \(K\varepsilon_0A/d\)
  • \(\varepsilon_0A/Kd\)
  • \(K^2\varepsilon_0A/d\)
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The Correct Option is B

Solution and Explanation

Insertion of dielectric multiplies capacitance by \(K\): \[ C=K\varepsilon_0\frac{A}{d} \]
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