First slide

Potential energy of system of charge in external field

Question

Two identical rings P and Q of radius 0.1 m are mounted coaxially at a distance 0.5 m apart. The charges on the two rings are 2 $μ$ C and 4 $μ$ C, respectively. The work done in transferring a charge of 5 $μ$ C from the center of P to that of Q is

Easy

A

1.28 J
B

0.72 J
C

0.144 J
D

2.24 J

Solution

$V_{1} = \frac{k \times 2 \times 10^{- 6}}{0.1} + \frac{k \times 4 \times 10^{- 6}}{\sqrt{{(0.1)}^{2} + {(0.5)}^{2}}}$
$V_{2} = \frac{k \times 4 \times 10^{- 6}}{0.1} + \frac{k \times 2 \times 10^{- 6}}{\sqrt{{(0.1)}^{2} + {(0.5)}^{2}}}$
Work done is $q (V_{2} - V_{1}) = 0.72 J$

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