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 is given by the formula:

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

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

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Let the initial charges be \( q \) and \( q \) (both charges are equal).
- The initial distance between them is \( r \).
- The initial force is \( F \).

Therefore, we can write the initial force as:
\[ F = k \frac{q \cdot q}{r^2} = k \frac{q^2}{r^2} \]

2. **Determine the New Conditions:**
- The charges are halved, so the new charges become \( \frac{q}{2} \) and \( \frac{q}{2} \).
- The distance is doubled, so the new distance becomes \( 2r \).

3. **Calculate the New Force:**
- Using Coulomb's law for the new charges and distance, the new force \( F' \) can be expressed as:
\[ F' = k \frac{\left(\frac{q}{2}\right) \left(\frac{q}{2}\right)}{(2r)^2} \]

4. **Simplify the Expression:**
- Substitute the values into the equation:
\[ F' = k \frac{\frac{q^2}{4}}{4r^2} = k \frac{q^2}{16r^2} \]

5. **Relate the New Force to the Initial Force:**
- From the initial force \( F = k \frac{q^2}{r^2} \), we can express \( F' \) in terms of \( F \):
\[ F' = \frac{1}{16} k \frac{q^2}{r^2} = \frac{F}{16} \]

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