Q. Two spheres $A$ and $B$ of radius $4cm$ 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) $20\mu C$ from $A\ to\ B$
B) $16\mu C$ from $A\ to\ B$
C) $32\mu C$ from $B\ to\ A$
D) $32\mu C$ from $A\ to\ B$
Explanation
(d) 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 $80\mu C\ i.e.,\ 32\mu C$ charge flows from $A$ to $B$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.