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(D) The initial force between two charges $q_1 = 2 C$ and $q_2 = 6 C$ separated by distance $r$ is given by Coulomb's Law: $F_1 = k \frac{q_1 q_2}{r^2

Vedclass · Accepted Answer

$4 	imes 10^3 \,N$ (attraction)

Vedclass · Answer

(D) The initial force between two charges $q_1 = 2 C$ and $q_2 = 6 C$ separated by distance $r$ is given by Coulomb's Law: $F_1 = k \frac{q_1 q_2}{r^2} = k \frac{(2)(6)}{r^2} = \frac{12k}{r^2}$.Given $F_1 = 12 	imes 10^3 \,N$, so $\frac{k}{r^2} = 10^3$.After adding $-4 C$ to each charge, the new charges are $q_1' = 2 - 4 = -2 C$ and $q_2' = 6 - 4 = 2 C$.The new force $F_2$ is $F_2 = k \frac{q_1' q_2'}{r^2} = k \frac{(-2)(2)}{r^2} = -4 \frac{k}{r^2}$.Substituting $\frac{k}{r^2} = 10^3$, we get $F_2 = -4 	imes 10^3 \,N$.The negative sign indicates that the force is attractive. Thus, the force is $4 	imes 10^3 \,N$ (attraction).

Two charges $2 C$ and $6 C$ are separated by a finite distance. If a charge of $-4 C$ is added to each of them, the initial force of $12 \times 10^3 \,N$ will change to

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