Question

A dielectric is inserted into a capacitor while keeping the charge constant. Identify what happens to the potential difference and the stored energy?

• A. The potential difference decreases and the stored energy increases
• B. Both the potential difference and the stored energy increases
• C. The potential difference increases and the stored energy decreases
• D. Both the potential difference and the stored energy decrease
• E. Both the potential difference and the stored energy remain the same

Answer 1

The correct option is D Both the potential difference and the stored energy decrease

The capacitance of the capacitor increases due to the insertion of the dielectric, $C^{\prime }=kC$ ; where, $k>1$

Initial potential difference on the capacitor $V=\frac{Q}{C}$

New potential difference $V^{\prime }=\frac{Q}{C^{\prime }}=\frac{Q}{kC}=\frac{V}{k}$

Thus potential difference decreases.

Also initial energy stored in the capacitor $E=\frac{Q^2}{2C}$

New energy stored $E^{\prime }=\frac{Q^2}{2C^{\prime }}=\frac{Q^2}{2\left(kC\right)}=\frac{E}{k}$

Thus energy stored in the capacitor also decreases.