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Two charges $+80 \mu \mathrm{C}$ and $+20 \mu \mathrm{C}$ are separated by a distance in air. An unknown third charge $q$ is placed at the centre of the line joining the two charges. If the charges are in equilibrium, then $q$.
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Verified Answer
The correct answer is:
$-4 \mu \mathrm{C}$
Given, $Q_1=+80 \mu \mathrm{C}, Q_2=+20 \mu \mathrm{C}$
Let distance between charges be r.
According to question,
Since, charges are in equilibrium,
$\begin{aligned} & \frac{k Q_1 q}{r^2 / 4}+\frac{k Q_1 Q_2}{r^2}+\frac{k Q_2 q}{r^2 / 4}=0 \\ & \Rightarrow \frac{4 k \times 80 \times q}{r^2}+\frac{k \times 80 \times 20}{r^2}+\frac{4 k \times 20 \times q}{r^2}=0 \\ & \Rightarrow 4 \times 80 \times q+4 \times 20 \times q=-80 \times 20 \\ & \Rightarrow \quad 4 q+q=-20 \\ & \Rightarrow \quad 5 q=-20 \\ & \Rightarrow \quad q=-4 \mu \mathrm{C} \\ & \end{aligned}$
Let distance between charges be r.
According to question,
Since, charges are in equilibrium,
$\begin{aligned} & \frac{k Q_1 q}{r^2 / 4}+\frac{k Q_1 Q_2}{r^2}+\frac{k Q_2 q}{r^2 / 4}=0 \\ & \Rightarrow \frac{4 k \times 80 \times q}{r^2}+\frac{k \times 80 \times 20}{r^2}+\frac{4 k \times 20 \times q}{r^2}=0 \\ & \Rightarrow 4 \times 80 \times q+4 \times 20 \times q=-80 \times 20 \\ & \Rightarrow \quad 4 q+q=-20 \\ & \Rightarrow \quad 5 q=-20 \\ & \Rightarrow \quad q=-4 \mu \mathrm{C} \\ & \end{aligned}$
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