When identical point charges are placed at the vertices of a cube of edge length `'a'` each of them experiences a net force of magnitude `F`.
Now these charges are placed on the vertices of another cube of edge length `'b'`. What will be magnitude of the net force on any of the charges? These cubes are simply geometrical constructs and not made by any matter.

Options

A. `(a^(2)F)/(b^(2))`
B. `(b^(2)F)/(a^(2))`
C. `(aF)/b`
D. Insufficient information

Explanation

To solve the problem, we need to determine the net force experienced by a charge when it is placed at the vertices of a cube with a different edge length. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have identical point charges placed at the vertices of a cube. The first cube has an edge length of `a`, and the second cube has an edge length of `b`.

2. **Identifying the Forces**:
- Each charge experiences forces due to the other charges. The forces depend on the distances between the charges and the magnitude of the charges.

3. **Calculating the Force in the First Cube**:
- For the cube with edge length `a`, let the force experienced by each charge be `F`. The net force on any charge is the vector sum of the forces due to the other 7 charges.

4. **Force Calculation**:
- The force between two point charges is given by Coulomb's law:
\[ F = k \frac{Q^2}{r^2} \]
where `k` is Coulomb's constant, `Q` is the charge, and `r` is the distance between the charges.

5. **Distances in the Cube**:
- In a cube, the distances between charges can be categorized as:
- **Adjacent charges**: Distance = `a`
- **Face diagonal charges**: Distance = \(\sqrt{2}a\)
- **Body diagonal charges**: Distance = \(\sqrt{3}a\)

6. **Net Force Calculation**:
- The net force on a charge will be a combination of the forces due to the adjacent, face diagonal, and body diagonal charges.
- Using symmetry, we can determine that the net force is proportional to:
\[ F \propto \frac{1}{a^2} \]
- Therefore, the net force can be expressed as:
\[ F = k \frac{Q^2}{a^2} \]

7. **Transitioning to the Second Cube**:
- Now, when the charges are placed at the vertices of a cube with edge length `b`, the distances change:
- The forces will now be:
- Adjacent charges: Distance = `b`
- Face diagonal charges: Distance = \(\sqrt{2}b\)
- Body diagonal charges: Distance = \(\sqrt{3}b\)

8. **New Net Force Calculation**:
- The new net force on a charge in the second cube will be:
\[ F' = k \frac{Q^2}{b^2} \]

9. **Relating the Forces**:
- Since we know the force in the first cube was `F`, we can relate the two forces:
\[ F' = F \cdot \frac{a^2}{b^2} \]

### Final Expression:
Thus, the magnitude of the net force on any of the charges when they are placed on the vertices of the second cube is:
\[ F' = F \cdot \frac{a^2}{b^2} \]