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.......\(n \) \(C\)
Pick an option to check your answer.
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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