If two identical spheres having charge `16 mu C` and `-8 mu C` are kept at certain distance apart , then the force if F. They are touched and again kept at the same distance , the force becomes.

Options

A. F
B. `(F)/( 18)`
C. `(F)/( 8)`
D. `(F)/(2)`

Answer

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the initial charges and force
We have two identical spheres with charges:
- Sphere 1: \( Q_1 = 16 \, \mu C \)
- Sphere 2: \( Q_2 = -8 \, \mu C \)

The force \( F \) between them can be calculated using Coulomb's law:
\[ F = k \frac{|Q_1 \cdot Q_2|}{R^2} \]
where \( k \) is Coulomb's constant and \( R \) is the distance between the charges.

### Step 2: Calculate the initial force
Substituting the values into the formula:
\[ F = k \frac{|16 \cdot (-8)|}{R^2} = k \frac{128}{R^2} \]
Since the charges are opposite, the force is attractive.

### Step 3: Touch the spheres and redistribute charges
When the two spheres are touched, the total charge is shared equally between them. The total charge is:
\[ Q_{total} = Q_1 + Q_2 = 16 \, \mu C - 8 \, \mu C = 8 \, \mu C \]
Since they are identical spheres, the charge on each sphere after touching will be:
\[ Q' = \frac{Q_{total}}{2} = \frac{8 \, \mu C}{2} = 4 \, \mu C \]

### Step 4: Calculate the new force after touching
Now, both spheres have the same charge:
- Sphere 1: \( Q'_1 = 4 \, \mu C \)
- Sphere 2: \( Q'_2 = 4 \, \mu C \)

The new force \( F' \) between them is:
\[ F' = k \frac{|Q'_1 \cdot Q'_2|}{R^2} = k \frac{|4 \cdot 4|}{R^2} = k \frac{16}{R^2} \]
Since both charges are now positive, the force is repulsive.

### Step 5: Compare the two forces
Now we compare the initial force \( F \) and the new force \( F' \):
\[ \frac{F}{F'} = \frac{k \frac{128}{R^2}}{k \frac{16}{R^2}} = \frac{128}{16} = 8 \]
Thus, we find:
\[ F' = \frac{F}{8} \]

### Conclusion
After touching the spheres, the new force \( F' \) becomes:
\[ F' = \frac{F}{8} \]