Two charges of equal magnitudes and at a distance r exert a force F on each other. If the charges are halved and distance between them is doubled, then the new force acting on each charge is

Options

A. `F//8`
B. `F//4`
C. `4F`
D. `F//16`

Answer

To solve the problem, we will use Coulomb's Law, which states that the force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by:

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

where \( k \) is Coulomb's constant.

### Step-by-Step Solution:

1. **Initial Setup**:
Let the initial charges be \( q_1 \) and \( q_2 \) such that \( q_1 = q_2 = q \). The initial distance between them is \( r \). The initial force \( F \) can be expressed as:
\[ F = k \frac{q \cdot q}{r^2} = k \frac{q^2}{r^2} \]

2. **Halving the Charges**:
According to the problem, the charges are halved. Thus, the new charges become:
\[ q_1' = \frac{q}{2}, \quad q_2' = \frac{q}{2} \]

3. **Doubling the Distance**:
The distance between the charges is doubled, so the new distance is:
\[ r' = 2r \]

4. **Calculating the New Force**:
We will now calculate the new force \( F' \) using the new charges and the new distance:
\[ F' = k \frac{q_1' \cdot q_2'}{(r')^2} = k \frac{\left(\frac{q}{2}\right) \cdot \left(\frac{q}{2}\right)}{(2r)^2} \]

5. **Substituting Values**:
Substitute the values into the equation:
\[ F' = k \frac{\left(\frac{q^2}{4}\right)}{(4r^2)} = k \frac{q^2}{4 \cdot 4r^2} = k \frac{q^2}{16r^2} \]

6. **Relating to Initial Force**:
We know from the initial force \( F \):
\[ F = k \frac{q^2}{r^2} \]
Therefore, we can express \( F' \) in terms of \( F \):
\[ F' = \frac{1}{16} \cdot k \frac{q^2}{r^2} = \frac{F}{16} \]

### Conclusion:
The new force acting on each charge is:
\[ F' = \frac{F}{16} \]

### Final Answer:
The new force acting on each charge is \( \frac{F}{16} \).

---