Two spheres carrying charges `+6muC and +9muC`, separated by a distance d, experiences a force of repulsion F. When a charge of `-3muC` is given to both the spheres and kept at the same distance as before, the new force of repulsion is

Options

A. `(F)/(3)`
B. `F`
C. `(F)/(g)`
D. `3F`

Explanation

To solve the problem, we will use Coulomb's law, which states that the electrostatic force \( F \) between two point charges is given by:

\[ F = k \frac{q_1 q_2}{r^2} \]

where:
- \( F \) is the electrostatic force,
- \( k \) is Coulomb's constant,
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.

### Step 1: Calculate the initial force of repulsion

The initial charges on the two spheres are:
- \( q_1 = +6 \, \mu C = 6 \times 10^{-6} \, C \)
- \( q_2 = +9 \, \mu C = 9 \times 10^{-6} \, C \)

The initial force \( F \) between the two spheres separated by distance \( d \) is:

\[ F = k \frac{(6 \times 10^{-6})(9 \times 10^{-6})}{d^2} \]

### Step 2: Determine the new charges after adding \(-3 \, \mu C\)

When a charge of \(-3 \, \mu C\) is added to both spheres, the new charges become:
- New charge on sphere 1: \( q_1' = 6 \, \mu C - 3 \, \mu C = 3 \, \mu C = 3 \times 10^{-6} \, C \)
- New charge on sphere 2: \( q_2' = 9 \, \mu C - 3 \, \mu C = 6 \, \mu C = 6 \times 10^{-6} \, C \)

### Step 3: Calculate the new force of repulsion

Now, we calculate the new force \( F' \) with the new charges:

\[ F' = k \frac{(3 \times 10^{-6})(6 \times 10^{-6})}{d^2} \]

### Step 4: Relate the initial force \( F \) to the new force \( F' \)

Now, we can express \( F' \) in terms of \( F \):

\[ F' = k \frac{(3 \times 10^{-6})(6 \times 10^{-6})}{d^2} = k \frac{(6 \times 10^{-6})(9 \times 10^{-6})}{d^2} \cdot \frac{3}{9} \]

This means:

\[ F' = F \cdot \frac{3}{9} = \frac{1}{3} F \]

### Step 5: Conclusion

Thus, the new force of repulsion \( F' \) is:

\[ F' = \frac{F}{3} \]

### Final Answer

The new force of repulsion is \( \frac{F}{3} \).

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