JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2022
- Medium
- MCQ
A parallel plate capacitor with width \(4\,cm\), length \(8\,cm\) and separation between the plates of \(4\,mm\) is connected to a battery of \(20\,V\). A dielectric slab of dielectric constant \(5\) having length \(1\,cm\), width \(4\,cm\) and thickness \(4\,mm\) is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be......... \(\in_{0}\,J\). (Where \(\epsilon_{0}\) is the permittivity of free space)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(240\)
Step-by-step Solution
Detailed explanation
\(C _{\text {eff }}=\left[\frac{\varepsilon_{0}(7 \times 4)}{4 / 10}+\frac{5 \varepsilon_{0}(1 \times 4)}{4 / 10}\right] \times 10^{-2}\)
\(C _{\text {eff }}=1.2 \varepsilon_{0}\)
Energy \(=\frac{1}{2} C _{\text {eff }} V ^{2}\)
\(=\frac{1}{2}(1.2) \varepsilon_{0}(20)(20)=240 \varepsilon_{0}\)
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