Two identical metal spheres having charges +q and +q respectively. When they are separated by distance r, exerts force of repulsion F on each other. The spheres are allowed to touch and then moved back to same separation. The new force of repulsion will be

Options

A. `F/2`
B. F
C. `F/4`
D. `F/10`

Answer

To solve the problem step by step, let's analyze the situation with the two identical metal spheres.

### Step 1: Initial Force Calculation
Initially, we have two identical metal spheres, each with a charge of +q. According to Coulomb's law, the force of repulsion (F) between two charges is given by:

\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q \cdot q}{r^2} = \frac{k \cdot q^2}{r^2} \]

where \( k = \frac{1}{4 \pi \epsilon_0} \) is the Coulomb's constant.

### Step 2: Touching the Spheres
When the two spheres touch, they share their charges equally because they are identical. The total charge before they touch is:

\[ Q_{total} = q + q = 2q \]

After touching, the charge on each sphere becomes:

\[ Q = \frac{Q_{total}}{2} = \frac{2q}{2} = q \]

### Step 3: New Force Calculation
After the spheres are separated again to the same distance \( r \), we need to calculate the new force of repulsion (F') between them. Now, each sphere has a charge of +q. The new force of repulsion is given by:

\[ F' = \frac{1}{4 \pi \epsilon_0} \frac{q \cdot q}{r^2} = \frac{k \cdot q^2}{r^2} \]

### Step 4: Comparing Forces
Notice that the expression for the new force \( F' \) is the same as the initial force \( F \):

\[ F' = F \]

### Conclusion
Thus, the new force of repulsion when the spheres are moved back to the same separation is equal to the original force \( F \).

\[ \text{New Force} = F \]

### Final Answer
The new force of repulsion will be \( F \).

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