Two spheres A and B of radius 4 cm and 6 cm are given charges of `80 muC and 40 muC`, respectively. If they are connected by a fine wire, then the amount of charge flowing from one to the other is

Options

A. `20 muC` from A to B
B. `20 muC` from B to A
C. `32 muC` from B to A
D. `32 muC` from A to B

Answer

To solve the problem of how much charge flows between two spheres A and B when they are connected by a fine wire, we can follow these steps:

### Step 1: Calculate the initial charge densities of both spheres.
The charge density \( \sigma \) on a sphere is given by the formula:
\[ \sigma = \frac{Q}{A} \]
where \( Q \) is the charge and \( A \) is the surface area of the sphere. The surface area \( A \) of a sphere is given by:
\[ A = 4\pi r^2 \]
For sphere A (radius = 4 cm = 0.04 m):
\[ A_A = 4\pi (0.04)^2 = 4\pi (0.0016) = 0.0201 \, \text{m}^2 \]
Charge on sphere A, \( Q_A = 80 \, \mu C = 80 \times 10^{-6} \, C \):
\[ \sigma_A = \frac{80 \times 10^{-6}}{0.0201} \approx 3.975 \times 10^{-3} \, C/m^2 \]

For sphere B (radius = 6 cm = 0.06 m):
\[ A_B = 4\pi (0.06)^2 = 4\pi (0.0036) = 0.0452 \, \text{m}^2 \]
Charge on sphere B, \( Q_B = 40 \, \mu C = 40 \times 10^{-6} \, C \):
\[ \sigma_B = \frac{40 \times 10^{-6}}{0.0452} \approx 8.84 \times 10^{-3} \, C/m^2 \]

### Step 2: Calculate the potential on each sphere.
The electric potential \( V \) on the surface of a sphere is given by:
\[ V = \frac{Q}{4\pi \epsilon_0 r} \]
Where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \).

For sphere A:
\[ V_A = \frac{80 \times 10^{-6}}{4\pi (8.85 \times 10^{-12})(0.04)} \approx 1.79 \times 10^4 \, V \]

For sphere B:
\[ V_B = \frac{40 \times 10^{-6}}{4\pi (8.85 \times 10^{-12})(0.06)} \approx 3.77 \times 10^4 \, V \]

### Step 3: Determine the final charge distribution.
When the two spheres are connected by a wire, charge will flow until the potentials are equal. Let \( Q_A' \) and \( Q_B' \) be the final charges on spheres A and B, respectively. The total charge is conserved:
\[ Q_A + Q_B = Q_A' + Q_B' \]
Substituting the known values:
\[ 80 \times 10^{-6} + 40 \times 10^{-6} = Q_A' + Q_B' \]
\[ 120 \times 10^{-6} = Q_A' + Q_B' \]

### Step 4: Set the potentials equal.
Since the potentials must be equal:
\[ V_A' = V_B' \]
\[ \frac{Q_A'}{4\pi \epsilon_0 (0.04)} = \frac{Q_B'}{4\pi \epsilon_0 (0.06)} \]
This simplifies to:
\[ \frac{Q_A'}{0.04} = \frac{Q_B'}{0.06} \]
Cross-multiplying gives:
\[ 0.06 Q_A' = 0.04 Q_B' \]
Substituting \( Q_B' = 120 \times 10^{-6} - Q_A' \) into the equation:
\[ 0.06 Q_A' = 0.04 (120 \times 10^{-6} - Q_A') \]

### Step 5: Solve for \( Q_A' \).
Expanding and rearranging:
\[ 0.06 Q_A' + 0.04 Q_A' = 0.04 \times 120 \times 10^{-6} \]
\[ 0.10 Q_A' = 4.8 \times 10^{-6} \]
\[ Q_A' = \frac{4.8 \times 10^{-6}}{0.10} = 48 \times 10^{-6} \, C \]

### Step 6: Calculate the charge that flowed.
The charge that flowed from A to B is:
\[ Q_{\text{flow}} = Q_A - Q_A' = 80 \times 10^{-6} - 48 \times 10^{-6} = 32 \times 10^{-6} \, C \]

### Final Answer:
The amount of charge flowing from sphere A to sphere B is \( 32 \, \mu C \).

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