ExamBro
ExamBro
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

  1. A \(\frac{v_0}{K}\)
  2. B \({v_0}{K^2}\)
  3. C \(\frac{v_0}{K^2}\)
  4. D \({v_0}^2\)
Verified Solution

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 \(v_0^{\prime}=\frac{\mathrm{Q}^2}{2 \mathrm{C}^{\prime}}\)
\(v_0^1=\frac{Q^2}{2 \mathrm{KC}}=\frac{v_0}{\mathrm{~K}}\)