Question 799: Electrostats & Capacitance

Question

Two point charges $+q_1$ and $q_2$ repel each other with a force of $100 \mathrm{~N} . q_1$ 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

Accepted Answer

To calculate how the force changes when the charges are altered, consider the following:
The original force between the charges is given by:
$ F_{	ext{net}}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}  $
The new conditions change the charges: $q_1$ is increased by 10%, and $q_2$ is decreased by 10%. This changes the charges to $1.1q_1$ and $0.9q_2$, respectively.
The new force is then calculated as:
$ \begin{aligned} F_{	ext{net}'} &= \frac{1}{4 \pi \varepsilon_0} \frac{(1.1q_1)(0.9q_2)}{r^2} \ &= \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} 	imes \left(\frac{110}{100}ight) 	imes \left(\frac{90}{100}ight)\ &= \frac{99}{100} 	imes F_{	ext{net}} \end{aligned} $
Since the original force is 100 N, the new force becomes:
$ F_{	ext{net}'} = \frac{99}{100} 	imes 100 = 99 \, 	ext{N} $
Thus, the force decreases by:
$ 100 \, 	ext{N} - 99 \, 	ext{N} = 1 \, 	ext{N} $
Therefore, the net force decreases by 1 N.

Answer

decreased by 10 N

Answer

increased by 10 N

Answer

increased by 1 N

Answer

decreased by 1 N

Electrostats & Capacitance

Two point charges +q_1 and q_2 repel each other with a force of 100 \mathrm{~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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