First slide

Electrostatic force

Question

Two point charges $3 \times 10^{- 6} C$ and $8 \times 10^{- 6} C$ repel each other by a force of $6 \times 10^{- 3} N$ . If each of them is given an additional charge $- 6 \times 10^{6} C$ , the force between them will be

Easy

A

$2 .4 \times 10^{- 3} N (attractive)$
B

$2 .4 \times 10^{- 9} N (attractive)$
C

$1 .5 \times 10^{- 3} N (repulsive)$
D

$1 .5 \times 10^{- 3} N (attractive)$

Solution

$F \propto Q_{1} Q_{2} \Rightarrow \frac{F_{1}}{F_{2}} = \frac{Q_{1} Q_{2}}{Q_{1}' Q_{2}'}$
$= \frac{3 \times 10^{- 6} \times 8 \times 10^{- 6}}{(3 \times 10^{- 6} - 6 \times 10^{- 6}) (8 \times 10^{- 6} - 6 \times 10^{- 6})} = \frac{3 \times 8}{- 3 \times 2} = - \frac{4}{1}$
$\Rightarrow F_{2} = - \frac{F_{1}}{4} = - \frac{6 \times 10^{- 3}}{4} = - 1 .5 \times 10^{- 3} N (Attractive)$

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