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

Options

A. `16 xx 10^(-32) J`
B. `31 xx 10^(-26) J`
C. `4 xx 10^(-10) J`
D. `32 xx 10^(-32)J`

Answer

To find the work done in placing a charge of \(8 \times 10^{-18}\) coulombs on a capacitor with a capacitance of \(100 \, \mu F\), we can follow these steps:

### Step 1: Convert Capacitance to Farads
The capacitance given is \(100 \, \mu F\). We need to convert this to farads:
\[ C = 100 \, \mu F = 100 \times 10^{-6} \, F = 1 \times 10^{-4} \, F \]

### Step 2: Use the Formula for Work Done
The work done (or energy stored) in charging a capacitor can be calculated using the formula:
\[ U = \frac{1}{2} C V^2 \]
However, we do not have the voltage \(V\) directly. We can relate charge \(Q\), capacitance \(C\), and voltage \(V\) using the formula:
\[ Q = C V \implies V = \frac{Q}{C} \]

### Step 3: Substitute for Voltage in the Energy Formula
Substituting the expression for \(V\) into the energy formula:
\[ U = \frac{1}{2} C \left(\frac{Q}{C}\right)^2 \]
This simplifies to:
\[ U = \frac{1}{2} C \cdot \frac{Q^2}{C^2} = \frac{Q^2}{2C} \]

### Step 4: Substitute the Values
Now, substitute the values of \(Q\) and \(C\):
\[ Q = 8 \times 10^{-18} \, C \]
\[ C = 1 \times 10^{-4} \, F \]
Substituting these into the equation:
\[ U = \frac{(8 \times 10^{-18})^2}{2 \times (1 \times 10^{-4})} \]

### Step 5: Calculate \(Q^2\)
Calculating \(Q^2\):
\[ Q^2 = (8 \times 10^{-18})^2 = 64 \times 10^{-36} \, C^2 \]

### Step 6: Calculate Work Done
Now substitute \(Q^2\) into the equation for \(U\):
\[ U = \frac{64 \times 10^{-36}}{2 \times 1 \times 10^{-4}} = \frac{64 \times 10^{-36}}{2 \times 10^{-4}} = \frac{64 \times 10^{-36}}{2} \times 10^{4} = 32 \times 10^{-32} \, J \]

### Final Result
Thus, the work done in placing the charge on the capacitor is:
\[ U = 32 \times 10^{-32} \, J \]