Two point charges +q_1 and +q_2 repel each ot | vedclass

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(D) The initial electrostatic force between the two charges is given by Coulomb's Law: $F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} = 100 \

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decreased by $1 \ N$

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(D) The initial electrostatic force between the two charges is given by Coulomb's Law: $F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} = 100 \ N$. After the changes, the new charges are $q_1' = q_1 + 0.1q_1 = 1.1q_1$ and $q_2' = q_2 - 0.1q_2 = 0.9q_2$. The new force $F'$ is: $F' = \frac{1}{4 \pi \varepsilon_0} \frac{(1.1q_1)(0.9q_2)}{r^2}$. $F' = (1.1 	imes 0.9) 	imes \left( \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} ight) = 0.99 	imes F$. $F' = 0.99 	imes 100 \ N = 99 \ N$. The change in force is $\Delta F = F - F' = 100 \ N - 99 \ N = 1 \ N$. Since the new force is less than the initial force, the force decreases by $1 \ N$.

Two point charges $+q_1$ and $+q_2$ repel each other with a force of $100 \ N$. $q_1$ is increased by $10 \%$ and $q_2$ is decreased by $10 \%$. If they are kept at their original positions, the change in the force of repulsion between them is:

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