First slide

Pontial due to

Question

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

Moderate

A

20 $μ$ C from A to B
B

16 $μ$ C from A to B
C

32 $μ$ C from B to A
D

32 $μ$ C from A to B.

Solution

Total charge Q = 80 + 40 = 120 $μ$ C. By using the formula $Q_{1}^{'} = Q [\frac{r_{1}}{r_{1} + r_{2}}]$ 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 $μ$ C i.e., 32 $μ$ C charge flows from A to B.

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