The work done in placing a charge of `8xx10^-18` coulomb on a condenser of capacity 100 micro-farad is

Options

A. (a) `16xx10^-32` joule
B. (b) `3.1xx10^-26` joule
C. (c) `4xx10^-10` joule
D. (d) `32xx10^-32` joule

Explanation

To find the work done in placing a charge on a capacitor, we can use the formula:

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

where:
- \( W \) is the work done,
- \( C \) is the capacitance of the capacitor,
- \( V \) is the voltage across the capacitor.

However, we can also express the voltage in terms of charge and capacitance using the formula:

\[ V = \frac{Q}{C} \]

where:
- \( Q \) is the charge placed on the capacitor.

### Step-by-Step Solution:

1. **Identify the given values:**
- Charge, \( Q = 8 \times 10^{-18} \) coulombs
- Capacitance, \( C = 100 \) microfarads = \( 100 \times 10^{-6} \) farads

2. **Calculate the voltage \( V \) using the formula:**
\[ V = \frac{Q}{C} = \frac{8 \times 10^{-18}}{100 \times 10^{-6}} \]
\[ V = \frac{8 \times 10^{-18}}{1 \times 10^{-4}} = 8 \times 10^{-14} \text{ volts} \]

3. **Substitute \( V \) into the work done formula:**
\[ W = \frac{1}{2} C V^2 \]
First, calculate \( V^2 \):
\[ V^2 = (8 \times 10^{-14})^2 = 64 \times 10^{-28} = 6.4 \times 10^{-27} \text{ volts}^2 \]

4. **Now substitute \( C \) and \( V^2 \) into the work done formula:**
\[ W = \frac{1}{2} \times (100 \times 10^{-6}) \times (6.4 \times 10^{-27}) \]
\[ W = \frac{1}{2} \times 100 \times 10^{-6} \times 6.4 \times 10^{-27} \]
\[ W = 50 \times 10^{-6} \times 6.4 \times 10^{-27} \]
\[ W = 320 \times 10^{-33} = 3.2 \times 10^{-31} \text{ joules} \]

### Final Answer:
The work done in placing the charge on the capacitor is \( 3.2 \times 10^{-31} \) joules.