A parallel plate capacitor is charged to a potential difference of 50 volts. It is then discharged through a resistance for 2 seconds and its potential drops by 10 volts. Calculate the fraction of energy stored in the capacitance.

Options

A. 0.14
B. 0.25
C. `0.50`
D. 0.64

Answer

To solve the problem, we need to calculate the fraction of energy stored in the capacitor after it has been discharged. Let's go through the steps methodically.

### Step 1: Calculate the initial energy stored in the capacitor (U)
The energy (U) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
Where:
- \( C \) is the capacitance,
- \( V \) is the potential difference.

Given that the initial potential difference \( V = 50 \) volts, we can express the initial energy as:
\[ U = \frac{1}{2} C (50)^2 = \frac{1}{2} C \cdot 2500 = 1250C \]

### Step 2: Calculate the final potential difference (V')
After discharging, the potential difference drops by 10 volts. Therefore, the final potential difference \( V' \) is:
\[ V' = 50 - 10 = 40 \text{ volts} \]

### Step 3: Calculate the final energy stored in the capacitor (U')
Using the same energy formula for the final potential difference:
\[ U' = \frac{1}{2} C (V')^2 = \frac{1}{2} C (40)^2 = \frac{1}{2} C \cdot 1600 = 800C \]

### Step 4: Calculate the fraction of energy stored in the capacitor
We need to find the fraction of the energy stored after discharging compared to the initial energy:
\[ \text{Fraction} = \frac{U'}{U} = \frac{800C}{1250C} \]

The capacitance \( C \) cancels out:
\[ \text{Fraction} = \frac{800}{1250} = \frac{64}{100} = 0.64 \]

### Final Answer
The fraction of energy stored in the capacitor after discharging is **0.64**.

---