Question

A capacitor is charged by a battery to a potential difference \( V \). It is disconnected from the battery and connected across another identical uncharged capacitor. Calculate the ratio of total energy stored in the combination to the initial energy stored in the capacitor.

Hint:

When two capacitors are connected in parallel after one is charged, the total energy stored is reduced because the charge is divided between the two capacitors, resulting in a lower potential difference and, hence, lower stored energy.

Answer 1

Let the capacitance of each capacitor be \( C \). 1. Initial Energy Stored: The energy stored in the initial capacitor is given by: \[ E_{\text{initial}} = \frac{1}{2} C V^2 \] 2. Connection to Identical Capacitor: When the charged capacitor is connected to an identical uncharged capacitor, the total charge is redistributed. The total charge \( Q \) on the initially charged capacitor is: \[ Q = C \cdot V \] After connection, this charge is shared equally. The new voltage across each capacitor is: \[ V_{\text{new}} = \frac{Q}{2C} = \frac{C \cdot V}{2C} = \frac{V}{2} \] 3. Total Energy in Combination: The total energy stored in the two capacitors after connection is the sum of their individual energies: \[ E_{\text{total}} = 2 \cdot \left( \frac{1}{2} C \left( \frac{V}{2} \right)^2 \right) = 2 \cdot \frac{1}{2} C \cdot \frac{V^2}{4} = \frac{C V^2}{4} \] 4. Ratio of Energies: The ratio of the total stored energy in the combination to the initial stored energy is: \[ \frac{E_{\text{total}}}{E_{\text{initial}}} = \frac{\frac{C V^2}{4}}{\frac{1}{2} C V^2} = \frac{1}{2} \] Therefore, the ratio of total energy stored in the combination to the initial energy is \( \frac{1}{2} \).