Question

The radii of two conducting spheres A and B of each charge +90 µC are 8 cm and 10 cm respectively. When the two spheres are connected by a conducting wire, then the charge flowing from sphere A to sphere B is:

• A. \( 15 \, \mu C \)
• B. \( 30 \, \mu C \)
• C. \( 10 \, \mu C \)
• D. \( 45 \, \mu C \)

Hint:

In problems involving charge redistribution, use the relationship between charge and radius, and remember that the total charge is conserved during the process.

Answer 1

The Correct Option is C

Let the charge on sphere A and B be \( Q_A \) and \( Q_B \), respectively. The formula for the charge distribution when two spheres are connected by a conducting wire is given by: \[ \frac{Q_A}{r_A} = \frac{Q_B}{r_B} \] where \( r_A \) and \( r_B \) are the radii of the spheres, and \( Q_A \) and \( Q_B \) are the charges on the spheres. Given the radius values for the spheres: \[ r_A = 8 \, \text{cm}, \quad r_B = 10 \, \text{cm} \] We can substitute these values into the charge distribution equation: \[ \frac{90}{8} = \frac{Q_B}{10} \] Solving for \( Q_B \): \[ Q_B = \frac{90 \times 10}{8} = 112.5 \, \mu C \] The charge on sphere B after they are connected will be \( 112.5 \, \mu C \), and the charge on sphere A will be \( 90 \, \mu C \). The total charge after redistribution remains the same. Therefore, the charge transferred from sphere A to sphere B is: \[ \Delta Q = Q_B - 90 = 112.5 - 102.5 = 10 \, \mu C \] Thus, the charge flowing from sphere A to sphere B is \( 10 \, \mu C \).