F is the force and r is the distance between two charges q. If charges are halved and distance is doubled, then the new force will

Options

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

Explanation

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 the formula:

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

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

### Step 1: Write the initial force equation
Let the initial charges be \( q_1 \) and \( q_2 \), and the initial distance be \( r \). The initial force \( F \) can be expressed as:

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

### Step 2: Halve the charges
According to the problem, both charges are halved. Therefore, the new charges will be:

\[ q_1' = \frac{q_1}{2}, \quad q_2' = \frac{q_2}{2} \]

### Step 3: Double the distance
The distance is doubled, so the new distance will be:

\[ r' = 2r \]

### Step 4: Write the new force equation
Now we can express the new force \( F' \) using the new charges and new distance:

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

Substituting the values of \( q_1' \), \( q_2' \), and \( r' \):

\[ F' = k \frac{\left(\frac{q_1}{2}\right) \left(\frac{q_2}{2}\right)}{(2r)^2} \]

### Step 5: Simplify the new force equation
Now, simplify the equation:

\[ F' = k \frac{\frac{q_1 q_2}{4}}{4r^2} \]

This simplifies to:

\[ F' = k \frac{q_1 q_2}{16r^2} \]

### Step 6: Relate the new force to the initial force
We know that the initial force \( F \) is:

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

Now, we can relate \( F' \) to \( F \):

\[ F' = \frac{F}{16} \]

### Conclusion
Thus, the new force \( F' \) when the charges are halved and the distance is doubled is:

\[ F' = \frac{F}{16} \]