There are two uncharged identicasl metallic spheres 1 and 2 of radius r separated by a distance `d(dgtgtr)`. A charged metalllic sphere of same radius having charge q is touched with one of the sphere. After some time it is moved away fom the system. Now, the uncharged sphere is earthed. Charge on earthed sphere is

Options

A. `+q/2`
B. `-q/2`
C. `-(qr)/(2d)`
D. `-(qd)/(2r)`

Explanation

To solve the problem, we need to analyze the situation step by step:

### Step 1: Understanding the Initial Condition
We have two identical uncharged metallic spheres (Sphere 1 and Sphere 2) of radius \( r \) separated by a distance \( d \), where \( d \) is much larger than \( r \).

### Step 2: Charging Sphere 1
A charged metallic sphere with charge \( q \) is brought into contact with Sphere 1. When two identical conductive spheres are brought into contact, they will share the charge equally due to their identical nature.

### Step 3: Charge Distribution After Contact
After contact, the total charge \( q \) will be divided equally between Sphere 1 and Sphere 2:
- Charge on Sphere 1 after contact: \( \frac{q}{2} \)
- Charge on Sphere 2 after contact: \( \frac{q}{2} \)

### Step 4: Moving the Charged Sphere Away
Once the charged metallic sphere is moved away, Sphere 1 retains a charge of \( \frac{q}{2} \), while Sphere 2 remains uncharged at this moment.

### Step 5: Earthing Sphere 2
Now, Sphere 2 is earthed. When a conductive object is earthed, it can either gain or lose charge to maintain a potential of zero. The potential of Sphere 2 due to its own charge and the influence of Sphere 1 needs to be calculated.

### Step 6: Calculating the Potential of Sphere 2
The potential \( V \) at Sphere 2 due to Sphere 1 (which has charge \( \frac{q}{2} \)) is given by the formula:
\[ V = \frac{k \cdot \frac{q}{2}}{d} + \frac{k \cdot Q}{r} \]
Where:
- \( k \) is Coulomb's constant,
- \( d \) is the distance between the spheres,
- \( r \) is the radius of Sphere 2.

Since Sphere 2 is earthed, we set the total potential to zero:
\[ \frac{k \cdot \frac{q}{2}}{d} + \frac{k \cdot Q}{r} = 0 \]

### Step 7: Solving for Charge on Sphere 2
Rearranging the equation gives:
\[ \frac{k \cdot Q}{r} = -\frac{k \cdot \frac{q}{2}}{d} \]
This leads to:
\[ Q = -\frac{q \cdot r}{2d} \]

### Final Answer
Thus, the charge on the earthed Sphere 2 is:
\[ Q = -\frac{qr}{2d} \]