A capcitor of capacitance `(1)/(300) muF` is connected to a battery of 300V and charged. Then the energy stored in the condenser is

Options

A. `3 xx 10^(-4)J`
B. `6 xx 10^(-14)J`
C. `1.5 xx 10^(-4)J`
D. `12 xx 10^(-5)J`

Answer

To find the energy stored in a capacitor, we can use the formula:

\[ U = \frac{1}{2} C V^2 \]

where:
- \( U \) is the energy stored in the capacitor (in joules),
- \( C \) is the capacitance (in farads),
- \( V \) is the voltage (in volts).

### Step 1: Convert Capacitance to Farads
The given capacitance is:

\[ C = \frac{1}{300} \mu F = \frac{1}{300} \times 10^{-6} F = \frac{1}{300000} F \]

### Step 2: Identify Voltage
The voltage provided by the battery is:

\[ V = 300 V \]

### Step 3: Substitute Values into the Energy Formula
Now we can substitute the values of \( C \) and \( V \) into the energy formula:

\[ U = \frac{1}{2} \left( \frac{1}{300000} \right) (300)^2 \]

### Step 4: Calculate \( V^2 \)
Calculating \( V^2 \):

\[ V^2 = 300^2 = 90000 \]

### Step 5: Substitute \( V^2 \) Back into the Formula
Now substitute \( V^2 \) back into the energy formula:

\[ U = \frac{1}{2} \left( \frac{1}{300000} \right) (90000) \]

### Step 6: Simplify the Expression
Now simplify the expression:

\[ U = \frac{1}{2} \times \frac{90000}{300000} \]

\[ U = \frac{1}{2} \times \frac{9}{30} = \frac{1}{2} \times \frac{3}{10} = \frac{3}{20} \]

### Step 7: Convert to Decimal
Now convert \( \frac{3}{20} \) to decimal form:

\[ U = 0.15 \, \text{J} \]

### Step 8: Convert to Scientific Notation
Convert \( 0.15 \) to scientific notation:

\[ U = 1.5 \times 10^{-1} \, \text{J} \]

### Final Answer
Thus, the energy stored in the capacitor is:

\[ U = 1.5 \times 10^{-4} \, \text{J} \]