Question

Two identical small conducting balls \( B_1 \) and \( B_2 \) are given \(-7\) \(\mu C\) and \(+4\) \(\mu C\) charges respectively. They are brought in contact with a third identical ball \( B_3 \) and then separated. If the final charge on each ball is \(-2\) \(\mu C\), the initial charge on \( B_3 \) was:

• A. \(-2\) pC
• B. \(-3\) pC
• C. \(-5\) pC
• D. \(-15\) pC

Hint:

In charge redistribution problems, when identical conductors are connected, the final charge on each is the arithmetic mean of the total charge.

Answer 1

The Correct Option is B

Charge Redistribution Calculation:
- The total charge on the three balls before contact is:
\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 \] - After bringing them in contact, the charge distributes equally among them:
\[ Q_{\text{final}} = \frac{Q_{\text{total}}}{3} = -2 pC \] - Given, \( Q_1 = -7 pC \), \( Q_2 = +4 pC \), and \( Q_3 \) is unknown:
\[ \frac{-7 + 4 + Q_3}{3} = -2 \] \[ -3 + Q_3 = -6 \] \[ Q_3 = -3 pC \] Thus, the initial charge on \( B_3 \) was \(-3\) pC.