The work done in placing a charge of `8muC` on a condenser of capacity `100muF` is

Options

A. `16xx10^(-5)J`
B. `32xx10^(-6)J`
C. `3.2xx10^(-4)J`
D. `16xx10^(-4)J`

Explanation

To solve the problem of calculating the work done in placing a charge of \(8 \mu C\) on a capacitor of capacity \(100 \mu F\), we can follow these steps:

### Step-by-Step Solution:

1. **Convert the given values to standard units**:
- The charge \(Q\) is given as \(8 \mu C\). To convert microcoulombs to coulombs:
\[ Q = 8 \mu C = 8 \times 10^{-6} C \]
- The capacitance \(C\) is given as \(100 \mu F\). To convert microfarads to farads:
\[ C = 100 \mu F = 100 \times 10^{-6} F \]

2. **Use the formula for work done in charging a capacitor**:
The work done \(W\) in charging a capacitor is given by the formula:
\[ W = \frac{1}{2} \frac{Q^2}{C} \]

3. **Substitute the values into the formula**:
- Substitute \(Q\) and \(C\) into the formula:
\[ W = \frac{1}{2} \frac{(8 \times 10^{-6})^2}{100 \times 10^{-6}} \]

4. **Calculate \(Q^2\)**:
- Calculate \(Q^2\):
\[ Q^2 = (8 \times 10^{-6})^2 = 64 \times 10^{-12} C^2 \]

5. **Substitute \(Q^2\) back into the work formula**:
\[ W = \frac{1}{2} \frac{64 \times 10^{-12}}{100 \times 10^{-6}} \]

6. **Simplify the expression**:
- Divide \(64 \times 10^{-12}\) by \(100 \times 10^{-6}\):
\[ W = \frac{1}{2} \times \frac{64}{100} \times 10^{-12 + 6} = \frac{1}{2} \times \frac{64}{100} \times 10^{-6} \]
- This simplifies to:
\[ W = \frac{32}{100} \times 10^{-6} = 0.32 \times 10^{-6} J \]

7. **Convert to standard scientific notation**:
\[ W = 3.2 \times 10^{-7} J \]

### Final Answer:
The work done in placing a charge of \(8 \mu C\) on a capacitor of capacity \(100 \mu F\) is:
\[ W = 3.2 \times 10^{-7} \text{ Joules} \]