Two point charges $+8 \mu \mathrm{C}$ and $+12 \mu \mathrm{C}$ repel each other with a force of $48…

Two point charges $+8 \mu \mathrm{C}$ and $+12 \mu \mathrm{C}$ repel each other with a force of $48 \mathrm{~N}$. When an additional charge of $-10 \mu \mathrm{C}$ is given to each of these charges (the distance between the charges is unaltered) then the new force is
  1. repulsive force of $24 \mathrm{~N}$
  2. attractive force of $24 \mathrm{~N}$
  3. repulsive force of $12 \mathrm{~N}$
  4. attractiive force of $2 \mathrm{~N}$

Solution

When an amount of charge $=10 \mu \mathrm{C}$ is added to both of the charges then, these becomes $\begin{aligned} & q_1=-2 \mu \mathrm{C} \text { and } q_2=+2 \mu \mathrm{C} \\ & F \propto\left|q_1 q_2\right| \\ & \Rightarrow F_1 \propto\left|q_1 q_2\right| \text { and } F_2 \propto\left|q_1^1 q_2^1\right| \\ & \text { and } \frac{F_1}{F_2}=\frac{q_1 q_2}{q_1 q_2} \Rightarrow \frac{48}{F_2}=\frac{8 \times 12}{2 \times 2} \\ & \Rightarrow F_2=2 \mathrm{~N} \end{aligned}$

Asked in: AP EAMCET 2015

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