Two concentric hollow conducting spheres of radius a and b `(b gt a)` contains charges `Q_(a)` and `Q_(b)` respectively. If they are connected by a conducting wire then find out following
(i) Final charges on inner and outer spheres.
(ii) Heat produced during the process.

Answer

To solve the problem of two concentric hollow conducting spheres with charges \( Q_a \) and \( Q_b \), we will follow these steps:

### Step 1: Understand the System
We have two concentric hollow conducting spheres:
- Inner sphere with radius \( a \) and charge \( Q_a \)
- Outer sphere with radius \( b \) and charge \( Q_b \)

When connected by a conducting wire, the charges will redistribute until the potential on both spheres is equal.

### Step 2: Use the Concept of Electric Potential
For a conducting sphere, the electric potential \( V \) at the surface is given by:
\[ V = \frac{kQ}{R} \]
where \( k \) is the Coulomb's constant, \( Q \) is the charge, and \( R \) is the radius of the sphere.

- For the inner sphere (radius \( a \)):
\[ V_a = \frac{kQ_a}{a} \]
- For the outer sphere (radius \( b \)):
\[ V_b = \frac{kQ_b}{b} \]

### Step 3: Set the Potentials Equal
Since the spheres are connected by a wire, the potentials must be equal:
\[ \frac{kQ_a'}{a} = \frac{kQ_b'}{b} \]
where \( Q_a' \) and \( Q_b' \) are the final charges on the inner and outer spheres, respectively.

### Step 4: Conservation of Charge
The total charge before and after connecting the spheres must be conserved:
\[ Q_a + Q_b = Q_a' + Q_b' \]

### Step 5: Solve the Equations
From the potential equality, we can express \( Q_b' \) in terms of \( Q_a' \):
\[ Q_b' = \frac{b}{a} Q_a' \]

Substituting this into the conservation of charge equation:
\[ Q_a + Q_b = Q_a' + \frac{b}{a} Q_a' \]
\[ Q_a + Q_b = Q_a' \left(1 + \frac{b}{a}\right) \]
\[ Q_a' = \frac{Q_a + Q_b}{1 + \frac{b}{a}} = \frac{(Q_a + Q_b) a}{a + b} \]

Now substituting \( Q_a' \) back to find \( Q_b' \):
\[ Q_b' = \frac{b}{a} Q_a' = \frac{b}{a} \cdot \frac{(Q_a + Q_b) a}{a + b} = \frac{b(Q_a + Q_b)}{a + b} \]

### Step 6: Final Charges
Thus, the final charges on the inner and outer spheres are:
\[ Q_a' = \frac{(Q_a + Q_b) a}{a + b} \]
\[ Q_b' = \frac{b(Q_a + Q_b)}{a + b} \]

### Step 7: Calculate Heat Produced
The heat produced during the process can be calculated by finding the difference in potential energy before and after the connection.

Initial potential energy \( U_i \):
\[ U_i = \frac{k Q_a^2}{2a} + \frac{k Q_b^2}{2b} \]

Final potential energy \( U_f \):
\[ U_f = \frac{k (Q_a')^2}{2a} + \frac{k (Q_b')^2}{2b} \]

The heat produced \( Q \) is given by:
\[ Q = U_i - U_f \]

### Step 8: Substitute and Simplify
Substituting the expressions for \( U_i \) and \( U_f \) and simplifying will yield the total heat produced.

### Final Answers
(i) Final charges:
\[ Q_a' = \frac{(Q_a + Q_b) a}{a + b} \]
\[ Q_b' = \frac{b(Q_a + Q_b)}{a + b} \]

(ii) Heat produced:
\[ Q = U_i - U_f \]