Two ball with charges `5muC and 10muC` are at a distance of 1 m from each other. In order to reduce the distance between them to 0.5 m , the amount of work to be performed is

Options

A. 45 J
B. `0.45xx10^(-6)J`
C. `1.2xx10^(-4)J`
D. 0.45 J

Answer

To solve the problem of calculating the work done to reduce the distance between two charges from 1 meter to 0.5 meters, we can follow these steps:

### Step 1: Identify the charges and their initial distance
We have two charges:
- \( Q_1 = 5 \, \mu C = 5 \times 10^{-6} \, C \)
- \( Q_2 = 10 \, \mu C = 10 \times 10^{-6} \, C \)

The initial distance between the charges is:
- \( R_i = 1 \, m \)

### Step 2: Calculate the initial potential energy
The formula for the electrostatic potential energy \( U \) between two point charges is given by:
\[ U = \frac{k \cdot Q_1 \cdot Q_2}{R} \]
where \( k \) is the Coulomb's constant (\( k \approx 9 \times 10^9 \, N \cdot m^2/C^2 \)).

Substituting the values for initial potential energy \( U_i \):
\[ U_i = \frac{9 \times 10^9 \cdot (5 \times 10^{-6}) \cdot (10 \times 10^{-6})}{1} \]
Calculating this gives:
\[ U_i = 9 \times 10^9 \cdot 50 \times 10^{-12} = 450 \times 10^{-3} \, J = 0.450 \, J \]

### Step 3: Calculate the final potential energy
Now, we need to calculate the final potential energy when the distance is reduced to \( R_f = 0.5 \, m \):
\[ U_f = \frac{k \cdot Q_1 \cdot Q_2}{R_f} \]
Substituting the values for final potential energy \( U_f \):
\[ U_f = \frac{9 \times 10^9 \cdot (5 \times 10^{-6}) \cdot (10 \times 10^{-6})}{0.5} \]
Calculating this gives:
\[ U_f = \frac{9 \times 10^9 \cdot 50 \times 10^{-12}}{0.5} = 9 \times 10^9 \cdot 100 \times 10^{-12} = 900 \times 10^{-3} \, J = 0.900 \, J \]

### Step 4: Calculate the work done
The work done \( W \) to reduce the distance is equal to the change in potential energy:
\[ W = U_f - U_i \]
Substituting the values:
\[ W = 0.900 \, J - 0.450 \, J = 0.450 \, J \]

### Final Answer
The amount of work to be performed is:
\[ \boxed{0.450 \, J} \]