Question

A capacitor of capacitance 2\(\mu\)F is charged to 50 V and then disconnected from the source. Later the gap between the plates of the capacitor is filled with a dielectric material. If the energy stored in the capacitor is decreased by 25% of its initial value, then the dielectric constant of the dielectric material is

• A. \(\frac{2}{3}\)
• B. \(\frac{4}{3}\)
• C. \(\frac{3}{4}\)
• D. \(\frac{3}{2}\)

Hint:

When a charged capacitor is disconnected and dielectric inserted, energy changes inversely with dielectric constant.

Answer 1

The Correct Option is B

Step 1: Initial energy stored
Energy stored in capacitor before dielectric is: \[ E_0 = \frac{1}{2} C V^2 \] Step 2: After dielectric is inserted and capacitor disconnected
Since capacitor is disconnected, charge \( Q \) remains constant.
Energy stored with dielectric \( E = \frac{Q^2}{2 C'} \), where \( C' = K C \), \(K\) is dielectric constant.
Step 3: Relation of energies
Given \( E = 0.75 E_0 \) (energy decreased by 25%).
Express \(E\) in terms of \(E_0\): \[ E = \frac{Q^2}{2 K C} = \frac{1}{K} \times \frac{Q^2}{2 C} = \frac{E_0}{K} \] So, \[ \frac{E_0}{K} = 0.75 E_0 \implies K = \frac{1}{0.75} = \frac{4}{3} \] Step 4: Conclusion
The dielectric constant is \(\frac{4}{3}\).