Question

A parallel plate capacitor of capacitance 2 F is charged to a potential V. The energy stored in the capacitor is E. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is E₅. The ratio \( E_5 / E_1 \) is:

• A. 2 : 3
• B. 1 : 2
• C. 1 : 4
• D. 2 : 1

Hint:

When capacitors are connected in parallel, the total capacitance is the sum of individual capacitances. The energy stored in the combination increases due to the increased capacitance.

Answer 1

The Correct Option is D

The energy stored in a capacitor is given by the formula: \[ E = \frac{1}{2} C V^2 \] For a single capacitor with capacitance \( C = 2 \, \text{F} \), the energy is: \[ E_1 = \frac{1}{2} \times 2 \times V^2 = V^2 \] Now, the two capacitors are connected in parallel, and the total capacitance becomes: \[ C_{\text{total}} = C + C = 2C = 4 \, \text{F} \] The energy in the parallel combination is: \[ E_5 = \frac{1}{2} \times 4 \times V^2 = 2V^2 \] Now, the ratio of the energies is: \[ \frac{E_5}{E_1} = \frac{2V^2}{V^2} = 2 \] Thus, the ratio of the energies is \( 2 : 1 \).