JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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)
- A \(240\)
- B \(241\)
- C \(242\)
- D \(243\)
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 \(=\frac{1}{2} C _{\text {eff }} V ^{2}\)…
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