Two spheres of radii R1 and R2 have equal charge are joint together with a copper wire. If the potential on each sphere after they are separated to each other is V, then initial charge on any sphere was `(k = 1/(4piin_(0)))`-

Options

A. `(V)/(k)(R_(1) - R_(2))`
B. `(V)/(2k)(R_(1) + R_(2))`
C. `(V)/(3k)(R_(1) + R_(2))`
D. `(V)/(k) ((R_(1)R_(2)))/((R_(1) + R_(2)))`

Answer

To solve the problem, we need to find the initial charge on either of the spheres after they are connected by a copper wire and then separated. Here’s a step-by-step solution:

### Step 1: Understand the setup
We have two spheres with radii \( R_1 \) and \( R_2 \) that have equal charge \( Q \) initially. When they are connected by a copper wire, charge can flow between them until they reach the same electric potential.

### Step 2: Write the expression for potential
The electric potential \( V \) of a charged sphere is given by the formula:
\[ V = \frac{kQ}{R} \]
where \( k = \frac{1}{4\pi \epsilon_0} \).

### Step 3: Set up the equations for both spheres
When the spheres are connected, let \( q \) be the charge that flows from one sphere to another. After charge redistribution:
- The charge on the first sphere becomes \( Q - q \).
- The charge on the second sphere becomes \( Q + q \).

The potentials of the spheres after they are separated are equal, so we can write:
\[ \frac{k(Q - q)}{R_1} = \frac{k(Q + q)}{R_2} \]

### Step 4: Simplify the equation
Since \( k \) is a constant and appears on both sides, we can cancel it out:
\[ \frac{Q - q}{R_1} = \frac{Q + q}{R_2} \]

### Step 5: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ (Q - q)R_2 = (Q + q)R_1 \]

### Step 6: Expand and rearrange the equation
Expanding both sides:
\[ QR_2 - qR_2 = QR_1 + qR_1 \]
Rearranging gives:
\[ QR_2 - QR_1 = qR_1 + qR_2 \]
Factoring out \( q \):
\[ Q(R_2 - R_1) = q(R_1 + R_2) \]

### Step 7: Solve for \( q \)
Now, we can solve for \( q \):
\[ q = \frac{Q(R_2 - R_1)}{R_1 + R_2} \]

### Step 8: Substitute \( q \) back into the potential equation
We know that the potential \( V \) on each sphere after separation is given by:
\[ V = \frac{k(Q - q)}{R_1} = \frac{k(Q + q)}{R_2} \]

Using either expression, we can substitute \( q \) to find \( Q \) in terms of \( V \).

### Step 9: Final expression for \( Q \)
Using the expression for \( q \):
\[ V = \frac{k(Q - \frac{Q(R_2 - R_1)}{R_1 + R_2})}{R_1} \]
This simplifies to:
\[ V = \frac{kQ(R_1 + R_2 - (R_2 - R_1))}{R_1(R_1 + R_2)} \]
\[ V = \frac{2kQ}{R_1 + R_2} \]
Thus, we can solve for \( Q \):
\[ Q = \frac{V(R_1 + R_2)}{2k} \]

### Final Answer:
The initial charge on either sphere was:
\[ Q = \frac{V(R_1 + R_2)}{2k} \]