The electrical force between two point charge | vedclass

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Question

(C) Let the two point charges be $q_1$ and $q_2$,separated by a distance $r$. The initial force is given by Coulomb's Law: $F = \frac{k q_1 q_2}{r^2}

Vedclass · Accepted Answer

$198$

Vedclass · Answer

(C) Let the two point charges be $q_1$ and $q_2$,separated by a distance $r$. The initial force is given by Coulomb's Law: $F = \frac{k q_1 q_2}{r^2} = 200 \ N$.When the charge $q_1$ increases by $10 \ \%$,the new charge is $q_1' = q_1 + 0.1 q_1 = 1.1 q_1$.When the charge $q_2$ decreases by $10 \ \%$,the new charge is $q_2' = q_2 - 0.1 q_2 = 0.9 q_2$.The new electrical force $F'$ at the same distance $r$ is: $F' = \frac{k q_1' q_2'}{r^2} = \frac{k (1.1 q_1) (0.9 q_2)}{r^2}$.Substituting the initial force value: $F' = (1.1 	imes 0.9) 	imes \frac{k q_1 q_2}{r^2} = 0.99 	imes F$.Therefore,$F' = 0.99 	imes 200 \ N = 198 \ N$.

The electrical force between two point charges is $200 \ N$. If we increase $10 \ \%$ charge on one of the charges and decrease $10 \ \%$ charge on the other,then the electrical force between them for the same distance becomes . . . . . . . . . . (in $N$)

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