MHT CET · Physics · Capacitance
The potential energy of charged parallel plate capacitor is \(v_0\). If a slab of dielectric constant \(\mathrm{K}\) is inserted between the plates, then the new potential energy will be
- A \(\frac{v_0}{K}\)
- B \({v_0}{K^2}\)
- C \(\frac{v_0}{K^2}\)
- D \({v_0}^2\)
Answer & Solution
Correct Answer
(A) \(\frac{v_0}{K}\)
Step-by-step Solution
Detailed explanation
We know, \(v_0=\frac{Q^2}{2 C}\)
On inserting the slab of dielectric constant \(\mathrm{k}\), the new capacitance \(\mathrm{C}^{\prime}=\mathrm{KC}\)
\(\therefore \quad\) New potential energy \(v_0^{\prime}=\frac{\mathrm{Q}^2}{2 \mathrm{C}^{\prime}}\)
\(v_0^1=\frac{Q^2}{2 \mathrm{KC}}=\frac{v_0}{\mathrm{~K}}\)
On inserting the slab of dielectric constant \(\mathrm{k}\), the new capacitance \(\mathrm{C}^{\prime}=\mathrm{KC}\)
\(\therefore \quad\) New potential energy \(v_0^{\prime}=\frac{\mathrm{Q}^2}{2 \mathrm{C}^{\prime}}\)
\(v_0^1=\frac{Q^2}{2 \mathrm{KC}}=\frac{v_0}{\mathrm{~K}}\)
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