Question

Two spheres A & B of radii 4 cm & 6 cm are given charges of 80 µC & 40 µC respectively. If they are connected by a fine wire, the amount of charge flowing from one to the other is:

• A. \( 32 \) µC from A to B
• B. \( 32 \) µC from B to A
• C. \( 20 \) µC from A to B
• D. \( 16 \) µC from B to A

Hint:

When two conductors are connected, charge redistributes to equalize potentials. Use the potential formula \( V = \frac{Q}{R} \) and the conservation of charge to determine the charge flow direction and amount.

Answer 1

The Correct Option is B

Step 1: Understanding Potential on a Conducting Sphere The potential \( V \) of a charged conducting sphere of radius \( R \) carrying charge \( Q \) is given by: \[ V = \frac{1}{4\pi \varepsilon_0} \frac{Q}{R} \] When two conductors are connected by a wire, charge flows until their potentials become equal. 
Step 2: Finding Initial Potentials of Spheres A and B For sphere A: \[ V_A = \frac{Q_A}{R_A} = \frac{80}{4} = 20 \text{ V} \] For sphere B: \[ V_B = \frac{Q_B}{R_B} = \frac{40}{6} = 6.67 \text{ V} \] Since \( V_A>V_B \), charge will flow from sphere A to sphere B until their potentials equalize. 
Step 3: Finding Common Potential After connection, the total charge is conserved, and the final common potential \( V_f \) is given by: \[ V_f = \frac{Q_A + Q_B}{R_A + R_B} \] \[ V_f = \frac{80 + 40}{4 + 6} = \frac{120}{10} = 12 \text{ V} \] 
Step 4: Charge Redistribution Using \( Q = V_f R \), the final charges on A and B will be: \[ Q_A' = V_f \times R_A = 12 \times 4 = 48 \text{ µC} \] \[ Q_B' = V_f \times R_B = 12 \times 6 = 72 \text{ µC} \] The charge transfer is: \[ \Delta Q = Q_B' - Q_B = 72 - 40 = 32 \text{ µC} \] Since charge increases on B, the charge flows from B to A. 
Final Answer: \( 32 \) µC flows from B to A.