Two spheres carrying charges +6 mu C and +9 m | vedclass

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(C) The initial force between the two charges is given by Coulomb's Law: $F = k \frac{q_1 q_2}{d^2}$.Substituting the initial values: $F = k \frac{(6

Vedclass · Accepted Answer

$F/3$

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(C) The initial force between the two charges is given by Coulomb's Law: $F = k \frac{q_1 q_2}{d^2}$.Substituting the initial values: $F = k \frac{(6 \mu C)(9 \mu C)}{d^2} = k \frac{54 \mu C^2}{d^2} \dots (1)$.When a charge of $-3 \mu C$ is added to both spheres,the new charges are:$q_1' = 6 \mu C - 3 \mu C = 3 \mu C$$q_2' = 9 \mu C - 3 \mu C = 6 \mu C$.The new force $F'$ is: $F' = k \frac{q_1' q_2'}{d^2} = k \frac{(3 \mu C)(6 \mu C)}{d^2} = k \frac{18 \mu C^2}{d^2} \dots (2)$.Dividing equation $(2)$ by equation $(1)$:$\frac{F'}{F} = \frac{18}{54} = \frac{1}{3}$.Therefore,the new force is $F' = F/3$.

Two spheres carrying charges $+6 \mu C$ and $+9 \mu C$ are separated by a distance $d$ and experience a force of repulsion $F$. When a charge of $-3 \mu C$ is added to both spheres and they are kept at the same distance as before,the new force of repulsion is:

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