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A point charge $Q$ is placed at the center of the line joining two equal point charges $+q$ and $+q$. The value of $Q$ if the system of the charges is equilibrium, is
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Verified Answer
The correct answer is:
$-\frac{\mathrm{q}}{4}$
The forces on charge $Q$ due to the other two charges will be equal and opposite and hence it will be in equilibrium. Force on $+\mathrm{q}$ due to the other two charges should also be equal and opposite. Their magnitudes will be equal if
$$
\begin{aligned}
& \frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}^2}{\mathrm{r}^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{qQ}}{\left(\frac{\mathrm{r}}{2}\right)^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{4 \mathrm{qQ}}{\mathrm{r}^2} \\
& \therefore \mathrm{q}=4 \mathrm{Q} \text { or } \mathrm{Q}=\frac{\mathrm{q}}{4}
\end{aligned}
$$
$\mathrm{Q}$ must be negative, hence $\mathrm{Q}=-\frac{\mathrm{q}}{4}$
$$
\begin{aligned}
& \frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}^2}{\mathrm{r}^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{qQ}}{\left(\frac{\mathrm{r}}{2}\right)^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{4 \mathrm{qQ}}{\mathrm{r}^2} \\
& \therefore \mathrm{q}=4 \mathrm{Q} \text { or } \mathrm{Q}=\frac{\mathrm{q}}{4}
\end{aligned}
$$
$\mathrm{Q}$ must be negative, hence $\mathrm{Q}=-\frac{\mathrm{q}}{4}$
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