The two parallel plates of a condenser have been connected to a battery of 300 V and the charge collected at each plate is `1muC.` The energy supplied by battery is :

Options

A. `6 xx 10^(-4)J`
B. `3 xx 10^(-4)J`
C. `1.5 xx 10^(-4) J`
D. `4.5 xx 10^(-4)J`

Answer

To solve the problem, we need to calculate the energy supplied by the battery to the parallel plate capacitor. We can use the formula for energy stored in a capacitor, which is given by:

\[ U = \frac{1}{2} Q V \]

Where:
- \( U \) is the energy stored in the capacitor,
- \( Q \) is the charge on the capacitor,
- \( V \) is the voltage across the capacitor.

### Step-by-Step Solution:

1. **Identify the given values:**
- Voltage \( V = 300 \, \text{V} \)
- Charge \( Q = 1 \, \mu\text{C} = 1 \times 10^{-6} \, \text{C} \)

2. **Substitute the values into the energy formula:**
\[ U = \frac{1}{2} Q V \]
\[ U = \frac{1}{2} \times (1 \times 10^{-6} \, \text{C}) \times (300 \, \text{V}) \]

3. **Calculate the energy:**
\[ U = \frac{1}{2} \times 1 \times 10^{-6} \times 300 \]
\[ U = \frac{300 \times 10^{-6}}{2} \]
\[ U = 150 \times 10^{-6} \, \text{J} \]
\[ U = 1.5 \times 10^{-4} \, \text{J} \]

4. **Final Result:**
The energy supplied by the battery is:
\[ U = 1.5 \times 10^{-4} \, \text{J} = 0.00015 \, \text{J} \]