Three identical metallic uncharged spheres A, B and C of radius a are kept on the corners of an equilateral triangle of side `d (d gt gt a)`. A fourth sphere (radius a) which has charge Q touches A and is then removed to a position for away. B is earthed and then the earth connection is removed. c is then earthed. The charge on C is

Options

A. `(Qa)/(2d) ((2d-a)/(2d))`
B. `(Qa)/(2d) ((2d-a)/(d))`
C. `(Qa)/(2d) ((a-d)/d)`
D. `(2Qa)/(d) ((d-a)/(2d))`

Answer

To solve the problem step by step, we will analyze the situation involving the metallic spheres and the charges they acquire through touching and earthing.

### Step 1: Initial Setup
We have three identical metallic uncharged spheres A, B, and C at the corners of an equilateral triangle with side length \( d \) (where \( d \gg a \)). A fourth sphere (let's call it D) with charge \( Q \) touches sphere A.

### Step 2: Charge Distribution After Touching
When sphere D (with charge \( Q \)) touches sphere A, charge will redistribute between them because they are identical spheres. The total charge \( Q \) will be equally divided between A and D.

- Charge on A after touching: \( Q_A = \frac{Q}{2} \)
- Charge on D after touching: \( Q_D = \frac{Q}{2} \)

### Step 3: Earthing Sphere B
Next, sphere B is earthed. When a sphere is earthed, its potential becomes zero. The potential \( V_B \) of sphere B due to its own charge and the influence of sphere A can be expressed as:

\[ V_B = k \frac{Q_A}{d} + k \frac{Q_B}{d} = 0 \]

Where \( k \) is Coulomb's constant, and \( Q_B \) is the charge on sphere B after earthing. Since sphere B is initially uncharged, we can set \( Q_B \) to be the charge that B acquires after earthing.

Substituting \( Q_A = \frac{Q}{2} \):

\[ k \frac{Q}{2d} + k \frac{Q_B}{d} = 0 \]

### Step 4: Solving for Charge on Sphere B
Rearranging the equation gives:

\[ \frac{Q_B}{d} = -\frac{Q}{2d} \]

Thus,

\[ Q_B = -\frac{Q}{2} \]

### Step 5: Removing the Earth Connection from Sphere B
After earthing, when the connection is removed, sphere B retains the charge:

\[ Q_B = -\frac{Q}{2} \]

### Step 6: Earthing Sphere C
Now, we earth sphere C. The potential of sphere C can be expressed as:

\[ V_C = k \frac{Q_A}{d} + k \frac{Q_B}{d} + k \frac{Q_C}{a} = 0 \]

Where \( Q_C \) is the charge on sphere C after earthing.

Substituting the known charges:

\[ k \frac{Q/2}{d} + k \left(-\frac{Q}{2}\right) \frac{1}{d} + k \frac{Q_C}{a} = 0 \]

### Step 7: Simplifying the Equation
This simplifies to:

\[ k \left(\frac{Q/2 - Q/2}{d} + \frac{Q_C}{a}\right) = 0 \]

Thus, we have:

\[ \frac{Q_C}{a} = 0 \]

This implies:

\[ Q_C = 0 \]

### Conclusion
The charge on sphere C after the entire process is \( Q_C = 0 \).