Two point charges 3 times 10^{-6} , C and 8 t | vedclass

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(D) According to Coulomb's Law,the force between two point charges is $F = k \frac{Q_1 Q_2}{r^2}$. Since the distance $r$ remains constant,$F \propto

Vedclass · Accepted Answer

$1.5 	imes 10^{-3} \, N$ (attractive)

Vedclass · Answer

(D) According to Coulomb's Law,the force between two point charges is $F = k \frac{Q_1 Q_2}{r^2}$. Since the distance $r$ remains constant,$F \propto Q_1 Q_2$.Initial charges: $Q_1 = 3 	imes 10^{-6} \, C$,$Q_2 = 8 	imes 10^{-6} \, C$.Initial force: $F_1 = 6 	imes 10^{-3} \, N$.New charges after adding $-6 	imes 10^{-6} \, C$ to each:$Q_1' = 3 	imes 10^{-6} - 6 	imes 10^{-6} = -3 	imes 10^{-6} \, C$$Q_2' = 8 	imes 10^{-6} - 6 	imes 10^{-6} = 2 	imes 10^{-6} \, C$Using the ratio: $\frac{F_2}{F_1} = \frac{Q_1' Q_2'}{Q_1 Q_2}$$\frac{F_2}{6 	imes 10^{-3}} = \frac{(-3 	imes 10^{-6}) 	imes (2 	imes 10^{-6})}{(3 	imes 10^{-6}) 	imes (8 	imes 10^{-6})}$$\frac{F_2}{6 	imes 10^{-3}} = \frac{-6}{24} = -\frac{1}{4}$$F_2 = -\frac{6 	imes 10^{-3}}{4} = -1.5 	imes 10^{-3} \, N$.The negative sign indicates that the force is attractive.

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

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