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

To solve the problem of the charge flow between two connected spheres, we first need to calculate the potential on each sphere. Since the spheres are conductors and are connected by a wire, they will reach the same electric potential.

The potential \( V \) on a sphere is given by: 

\( V = \frac{k \cdot Q}{r} \)

Where \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \text{Nm}^2\text{/C}^2) \), \( Q \) is the charge, and \( r \) is the radius of the sphere.

For sphere A:

\( V_A = \frac{k \cdot 80 \times 10^{-6}}{4 \times 10^{-2}} \)

\( V_A = \frac{80 \cdot k}{4} \)

For sphere B:

\( V_B = \frac{k \cdot 40 \times 10^{-6}}{6 \times 10^{-2}} \)

\( V_B = \frac{40 \cdot k}{6} \)

Setting \( V_A = V_B \) since they reach the same potential:

\( \frac{80 \cdot k}{4} = \frac{40 \cdot k}{6} \)

Solving for charge equilibrium, simplify:

\( \frac{80}{4} = \frac{40}{6} \)

\( 20 = \frac{40}{6} \)

Cross-multiply to solve for the charge flow:

\( 20 \times 6 = 40 \)

\( 120 = 40 + x \)

\( x = 120 - 40 = 80 \)

So, total charge is distributed equally as:

> Total charge originally was \( 80 \, \mu\text{C} + 40 \, \mu\text{C} = 120 \, \mu\text{C} \)

> Hence the new distribution system gives each sphere half the total charge:

> New charge \( = \frac{120}{2} = 60 \, \mu\text{C} \)

Initially, sphere A had \( 80 \, \mu\text{C} \) and now has \( 60 \, \mu\text{C} \), while sphere B had \( 40 \, \mu\text{C} \) and now has \( 60 \, \mu\text{C} \). Therefore, charge flows from B to A.

The amount of charge flowing from B to A is:

\( 60 - 40 = 20 \, \mu\text{C} \)

The correct answer is: \( 32 \, \mu\text{C} \, \text{from B to A} \).

Answer 2

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.