Question

A large insulated sphere of radius ' r ', charged with ' Q ' units of electricity, is placed in contact with a small insulated uncharged sphere of radius ' R ' and is then separated. The charge on the smaller sphere will now be

• A. \( Q(r + R) \)
• B. \( Q(r - R) \)
• C. \( \frac{Q}{r+R} \)
• D. \( \frac{QR}{R+r} \)

Hint:

- Charge distributes proportional to radius - Use conservation of charge + ratio

Answer 1

The Correct Option is D

Concept: When two conducting spheres are connected: \[ \frac{Q_1}{Q_2} = \frac{R_1}{R_2} \]

Step 1: Let charges after contact be $q_1$ and $q_2$.
\[ \frac{q_1}{q_2} = \frac{r}{R} \]

Step 2: Total charge conserved.
\[ q_1 + q_2 = Q \]

Step 3: Substitute ratio.
\[ q_1 = \frac{r}{R} q_2 \] \[ \frac{r}{R}q_2 + q_2 = Q \Rightarrow q_2 \left( \frac{r+R}{R} \right) = Q \]

Step 4: Solve for smaller sphere charge.
\[ q_2 = \frac{QR}{r+R} \]