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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2018
  • Medium
  • MCQ

A parallel plate capacitor of capacitance \(90\ pF\) is connected to a battery of \(emf\) \(20\ V\). If a dielectric material of dielectric constant \(K = \frac{5}{3}\) is inserted between the plates, the magnitude of the induced charge will be.......\(n \) \(C\)

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Answer & Solution

Correct Answer

(D) \(1.2\) 

Step-by-step Solution

Detailed explanation

Charge on Capacitor, \(\mathrm{Q}_{i}=\mathrm{CV}\)

After inserting dielectric of dielectric constant = \(\mathrm{K}\)

\(\mathrm{Q}_{\mathrm{f}}=(\mathrm{k} \mathrm{C}) \mathrm{V}\)

Induced charges on dielectric

\( \mathrm{Q}_{\mathrm{ind}} =\mathrm{Q}_{\mathrm{f}}-\mathrm{Q}_{\mathrm{i}}=\mathrm{KCV}-\mathrm{CV} \)

\(=(\mathrm{K}-1) \mathrm{CV}=\left(\frac{5}{3}-1\right) \times 90 \,\mathrm{pF} \times 2\, \mathrm{V}=1.2\, \mathrm{nc} \)

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