`-10mu C, 40 mu C` and q are the charges on three identical conductors P,Q and R respectively, Now P and Q attract each other with a force F when they are separated by a distance d. Now P and Q are made in contact with each other and then separated . Again Q and R are touched and they are separated by a distance 'd' . The repulsive force between Q and R is 4F . Then the charge q is:

Options

A. `10 mu C`
B. `30 mu C`
C. `40 mu C`
D. `65 mu C`

Answer

To solve the problem step by step, we will analyze the forces acting on the charges and apply the principles of electrostatics.

### Step 1: Understand the Initial Attraction between P and Q
Given:
- Charge on P, \( q_P = -10 \, \mu C \)
- Charge on Q, \( q_Q = 40 \, \mu C \)

The force \( F \) between P and Q when they are separated by a distance \( d \) can be expressed using Coulomb's Law:
\[ F = k \frac{|q_P \cdot q_Q|}{d^2} \]
Substituting the values:
\[ F = k \frac{|-10 \cdot 40|}{d^2} = k \frac{400}{d^2} \]

### Step 2: Contact and Separation of P and Q
When P and Q are brought into contact, the total charge is shared equally because they are identical conductors. The total charge before contact is:
\[ q_{total} = q_P + q_Q = -10 \, \mu C + 40 \, \mu C = 30 \, \mu C \]
After contact, each conductor will have:
\[ q_{P,Q} = \frac{q_{total}}{2} = \frac{30 \, \mu C}{2} = 15 \, \mu C \]
So, after separation:
- Charge on P, \( q_P' = 15 \, \mu C \)
- Charge on Q, \( q_Q' = 15 \, \mu C \)

### Step 3: Analyze the Repulsion between Q and R
Now, Q (with charge \( 15 \, \mu C \)) is brought into contact with R (with charge \( q \)). After contact, the total charge on Q and R will be:
\[ q_{total, QR} = q_Q' + q = 15 \, \mu C + q \]
After contact, the charge on both Q and R will be:
\[ q_{Q,R} = \frac{q_{total, QR}}{2} = \frac{15 + q}{2} \]

### Step 4: Calculate the Repulsive Force between Q and R
The repulsive force between Q and R when separated by distance \( d \) is given as \( 4F \):
\[ 4F = k \frac{(q_{Q,R})^2}{d^2} \]
Substituting \( q_{Q,R} \):
\[ 4F = k \frac{\left(\frac{15 + q}{2}\right)^2}{d^2} \]
From Step 1, we know:
\[ F = k \frac{400}{d^2} \]
Thus:
\[ 4F = 4 \cdot k \frac{400}{d^2} = k \frac{1600}{d^2} \]
Setting the two expressions for \( 4F \) equal gives:
\[ k \frac{1600}{d^2} = k \frac{\left(\frac{15 + q}{2}\right)^2}{d^2} \]
Canceling \( k \) and \( d^2 \) (since they are non-zero), we have:
\[ 1600 = \left(\frac{15 + q}{2}\right)^2 \]

### Step 5: Solve for q
Taking the square root of both sides:
\[ 40 = \frac{15 + q}{2} \]
Multiplying by 2:
\[ 80 = 15 + q \]
Thus:
\[ q = 80 - 15 = 65 \, \mu C \]

### Final Answer
The charge \( q \) is:
\[ \boxed{65 \, \mu C} \]