A `6muF` capacitor charged from 10 volts to 20 volts. Increase in energy will be -

Options

A. `18 xx 10^(-4)" Joule"`
B. `9xx 10^(-4)" Joule"`
C. `4.5 xx 10^(-4)" Joule"`
D. `9 xx 10^(-9)" Joule"`

Answer

To find the increase in energy of a capacitor when it is charged from 10 volts to 20 volts, we can follow these steps:

### Step 1: Understand the Formula for Energy Stored in a Capacitor
The energy (U) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
where:
- \( U \) is the energy in joules,
- \( C \) is the capacitance in farads,
- \( V \) is the voltage in volts.

### Step 2: Calculate the Initial Energy at 10 Volts
Given:
- Capacitance \( C = 6 \mu F = 6 \times 10^{-6} F \)
- Initial Voltage \( V_1 = 10 V \)

Using the formula:
\[ U_1 = \frac{1}{2} C V_1^2 = \frac{1}{2} \times (6 \times 10^{-6}) \times (10^2) \]
Calculating:
\[ U_1 = \frac{1}{2} \times (6 \times 10^{-6}) \times 100 = 3 \times 10^{-4} \text{ joules} \]

### Step 3: Calculate the Final Energy at 20 Volts
Given:
- Final Voltage \( V_2 = 20 V \)

Using the formula:
\[ U_2 = \frac{1}{2} C V_2^2 = \frac{1}{2} \times (6 \times 10^{-6}) \times (20^2) \]
Calculating:
\[ U_2 = \frac{1}{2} \times (6 \times 10^{-6}) \times 400 = 12 \times 10^{-4} \text{ joules} = 1.2 \times 10^{-3} \text{ joules} \]

### Step 4: Calculate the Increase in Energy
The increase in energy (\( \Delta U \)) is given by:
\[ \Delta U = U_2 - U_1 \]
Substituting the values:
\[ \Delta U = (1.2 \times 10^{-3}) - (3 \times 10^{-4}) \]
Calculating:
\[ \Delta U = 1.2 \times 10^{-3} - 0.3 \times 10^{-3} = 0.9 \times 10^{-3} \text{ joules} = 9 \times 10^{-4} \text{ joules} \]

### Final Answer
The increase in energy when the capacitor is charged from 10 volts to 20 volts is:
\[ \Delta U = 9 \times 10^{-4} \text{ joules} \]

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