A charge q is to be distributed on two conducting spheres. What should be the value of the charges on the spheres so that the repulsive force between them is maximum when they are placed at a fixed distance from each other in air ?

Options

A. `(q)/(2)" and "(q)/(2)`
B. `(q)/(4)` and `(3q)/(4)`
C. `(q)/(3)` and `(2q)/(3)`
D. `(q)/(5)` and `(4q)/(5)`

Answer

To solve the problem of distributing a charge \( q \) on two conducting spheres such that the repulsive force between them is maximized, we can follow these steps:

### Step 1: Define the Charges on the Spheres
Let the charge on the first sphere be \( x \) and the charge on the second sphere be \( q - x \). Since the spheres are conducting, the charge will distribute uniformly on each sphere.

### Step 2: Write the Expression for the Force
According to Coulomb's law, the force \( F \) between two point charges is given by:
\[ F = k \frac{|x \cdot (q - x)|}{d^2} \]
where \( k \) is Coulomb's constant, and \( d \) is the distance between the centers of the spheres.

### Step 3: Simplify the Force Equation
We can express the force as:
\[ F = \frac{k}{d^2} (x(q - x)) = \frac{k}{d^2} (qx - x^2) \]

### Step 4: Differentiate the Force with Respect to \( x \)
To find the value of \( x \) that maximizes the force, we differentiate \( F \) with respect to \( x \):
\[ \frac{dF}{dx} = \frac{k}{d^2} (q - 2x) \]

### Step 5: Set the Derivative to Zero
To find the maximum force, we set the derivative equal to zero:
\[ q - 2x = 0 \]
Solving for \( x \), we get:
\[ 2x = q \implies x = \frac{q}{2} \]

### Step 6: Determine the Charges on Each Sphere
Since \( x = \frac{q}{2} \), the charge on the first sphere is \( \frac{q}{2} \) and the charge on the second sphere is:
\[ q - x = q - \frac{q}{2} = \frac{q}{2} \]

### Conclusion
Thus, the charges on the two spheres should be:
- Charge on Sphere 1: \( \frac{q}{2} \)
- Charge on Sphere 2: \( \frac{q}{2} \)

This distribution maximizes the repulsive force between the two spheres when they are placed at a fixed distance from each other in air.