Question

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

• A. $20\mu C$fromA to B
• B. $16\mu C$ from A to B
• C. $32\mu C$fromB to A
• D. $32\mu C$ from A to B

Answer 1

Answer: (D)

$32\mu C$ from A to B

Answer 2

Total charge $Q = 80 + 40 = 120\mu C$. By using the formula $Q_{1} ⥂ ' = Q\left\lbrack \frac{r_{1}}{r_{1} + r_{2}} \right\rbrack$. New charge on sphere A is $Q_{A}^{'} = Q\left\lbrack \frac{r_{A}}{r_{A} + r_{B}} \right\rbrack = 120\left\lbrack \frac{4}{4 + 6} \right\rbrack = 48\mu C$. Initially it was $1.694 \times 10^{9}m^{2}$ i.e., $32\mu C$ charge flows from A to B.