Two similar spheres having `+Q` and `-Q` charge are kept at a certain distance. F force acts between the two. If in the middle of two spheres, another similar sphere having `+Q` charge is kept, then it experiences 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, let's analyze the situation step by step.

### Step 1: Understand the Configuration
We have two spheres, one with charge `+Q` and the other with charge `-Q`, separated by a distance `D`. We place a third sphere with charge `+Q` exactly in the middle of the two spheres.

### Step 2: Determine the Forces Acting on the Middle Sphere
The middle sphere experiences forces due to both the `+Q` sphere and the `-Q` sphere.

1. **Force due to the `+Q` sphere**:
- The distance from the middle sphere to the `+Q` sphere is `D/2`.
- The force \( F_1 \) between the two `+Q` charges can be calculated using Coulomb's Law:
\[ F_1 = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q \cdot Q}{(D/2)^2} = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q^2}{D^2/4} = \frac{4Q^2}{4\pi \epsilon_0 D^2} = \frac{4Q^2}{4\pi \epsilon_0 D^2} \]
- The direction of this force is **away** from the `+Q` sphere (repulsive force).

2. **Force due to the `-Q` sphere**:
- The distance from the middle sphere to the `-Q` sphere is also `D/2`.
- The force \( F_2 \) between the `+Q` sphere and the `-Q` sphere is:
\[ F_2 = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q \cdot (-Q)}{(D/2)^2} = \frac{1}{4\pi \epsilon_0} \cdot \frac{-Q^2}{D^2/4} = -\frac{4Q^2}{4\pi \epsilon_0 D^2} \]
- The direction of this force is **towards** the `-Q` sphere (attractive force).

### Step 3: Calculate the Net Force on the Middle Sphere
Now we can find the net force acting on the middle sphere by adding the two forces \( F_1 \) and \( F_2 \):
- Since \( F_1 \) is directed away from the `+Q` sphere and \( F_2 \) is directed towards the `-Q` sphere, we can express the net force \( F_{net} \) as:
\[ F_{net} = F_1 - |F_2| = \frac{4Q^2}{4\pi \epsilon_0 D^2} + \frac{4Q^2}{4\pi \epsilon_0 D^2} = \frac{8Q^2}{4\pi \epsilon_0 D^2} \]
- The net force \( F_{net} \) is directed towards the `-Q` sphere.

### Conclusion
The middle sphere with charge `+Q` experiences a net force of magnitude \( \frac{8Q^2}{4\pi \epsilon_0 D^2} \) directed towards the `-Q` sphere.