Charges $3 \mathrm{Q}, \mathrm{q}$ and Q are placed along x -axis at positions $\mathrm{x}=0,…

Charges $3 \mathrm{Q}, \mathrm{q}$ and Q are placed along x -axis at positions $\mathrm{x}=0, \mathrm{x}=\frac{1}{3}$ and $\mathrm{x}=1$ respectively. When the force on charge Q is zero, the value of $q$ is
  1. $\frac{\mathrm{Q}}{3}$
  2. $-\frac{\mathrm{Q}}{3}$.
  3. $\frac{4}{3} \mathrm{Q}$
  4. $-\frac{4}{3} \mathrm{Q}$

Solution


Force $F_1$ due to charge $3 Q$ on $Q$ will be, $\mathrm{F}_1=\frac{1}{4 \pi \varepsilon_0} \frac{3 \mathrm{Q}^2}{l^2}$
Force $F_2$ due to charge $q$ on charge $Q$ will be, $\begin{aligned} & \mathrm{F}_2=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\left(\frac{2 l}{3}\right)^2} \\ \therefore \quad & \mathrm{~F}_2=\frac{1}{4 \pi \varepsilon_0} \frac{9 \mathrm{Qq}}{4 l^2} \end{aligned}$ For the force on charge Q to be zero, $\begin{array}{ll} & \mathrm{F}_1+\mathrm{F}_2=0 \text { or } \mathrm{F}_1=-\mathrm{F}_2 \\ \therefore & \cdot \frac{3 \mathrm{Q}^2}{l^2}=-\frac{9 \mathrm{Qq}}{4 l^2} \\ \therefore & \mathrm{Q}=-\frac{3 \mathrm{q}}{4} \\ \therefore \quad & \mathrm{q}=-\frac{4}{3} \mathrm{Q} \end{array}$

Asked in: MHT CET 2024 (11 May Shift 1)

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