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

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

A
$20 μC$ from A to B
B
$16 μC$ from A to B
C
$32 μC$ from B to A
D
$32 μC$ from A to B

Solution:

Total charge $Q = 80 + 40 = 120 μC$ .

By using the formula $Q_{1}' = Q_{1} [\frac{r_{1}}{r_{1} + r_{2}}]$ .

New charge on sphere A is $Q_{A}^{'} = Q [\frac{r_{A}}{r_{A} + r_{B}}] = 120 [\frac{4}{4 + 6}] = 48 μC$ .

Initially it was $80 μC$ i.e., $32 μC$ charge flows from A to B.

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