Two similar spheres having `+q` and `-q` charges are kept at a certain distance. The force acts between the two is `F`. If in the middle of two spheres, another similar sphere having `+q` charge is kept,then it experience a force in magnitude and direction as

Options

A. zero having no direction
B. 8 F towards `+q` charge
C. 8 F towards `- q` charge
D. 4 F towards `+q` charge

Answer

To solve the problem, we need to analyze the forces acting on the charge placed in the middle of the two spheres. Here is the step-by-step solution:

### Step 1: Identify the Charges and Their Positions
We have two spheres:
- Sphere A with charge \( +q \)
- Sphere B with charge \( -q \)

They are separated by a distance \( d \). A third sphere, Sphere C, with charge \( +q \) is placed exactly in the middle of the two spheres.

### Step 2: Calculate the Distance from Sphere C to A and B
Since Sphere C is in the middle, the distance from Sphere C to Sphere A (and Sphere B) is:
\[ \text{Distance from C to A} = \text{Distance from C to B} = \frac{d}{2} \]

### Step 3: Calculate the Forces Acting on Sphere C
1. **Force due to Sphere A (F1)**:
The force between Sphere C and Sphere A is repulsive (since both are positive charges). The magnitude of this force can be calculated using Coulomb's law:
\[ F_1 = k \frac{q \cdot q}{\left(\frac{d}{2}\right)^2} = k \frac{q^2}{\frac{d^2}{4}} = \frac{4kq^2}{d^2} \]
This force acts to the right (away from Sphere A).

2. **Force due to Sphere B (F2)**:
The force between Sphere C and Sphere B is attractive (since Sphere B is negatively charged). The magnitude of this force is also:
\[ F_2 = k \frac{q \cdot (-q)}{\left(\frac{d}{2}\right)^2} = k \frac{q \cdot (-q)}{\frac{d^2}{4}} = -\frac{4kq^2}{d^2} \]
This force acts to the left (toward Sphere B).

### Step 4: Determine the Net Force on Sphere C
Since the forces \( F_1 \) and \( F_2 \) are equal in magnitude but opposite in direction, we can find the net force \( F_{net} \) acting on Sphere C:
\[ F_{net} = F_1 + F_2 = \frac{4kq^2}{d^2} - \frac{4kq^2}{d^2} = 0 \]
However, we need to consider the net effect of both forces acting on Sphere C. The forces are not equal in terms of direction; thus, we need to add their magnitudes considering the direction:
\[ F_{net} = F_1 + |F_2| = \frac{4kq^2}{d^2} + \frac{4kq^2}{d^2} = \frac{8kq^2}{d^2} \]

### Step 5: Relate to Given Force F
We know from the original problem that the force \( F \) between Sphere A and Sphere B is given by:
\[ F = k \frac{q^2}{d^2} \]
Thus, we can express \( F_{net} \) in terms of \( F \):
\[ F_{net} = 8F \]

### Step 6: Determine the Direction of the Net Force
The net force \( F_{net} \) is directed toward the negative charge (Sphere B) since the attractive force from Sphere B is stronger than the repulsive force from Sphere A.

### Final Answer
The magnitude of the force experienced by Sphere C is \( 8F \), and the direction is toward Sphere B (the negative charge).