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Question

Suppose in the following figure. equilibrium of charge $\mathrm{B}$ is considered. Hence for it's

equilibrium $\left|\mathrm{F}_{\mathrm{A}}\ri

Vedclass · Accepted Answer

$ - \frac{Q}{4}$

Vedclass · Answer

Suppose in the following figure. equilibrium of charge $\mathrm{B}$ is considered. Hence for it's

equilibrium $\left|\mathrm{F}_{\mathrm{A}}\right|=\left|\mathrm{F}_{\mathrm{C}}\right|$

$\Rightarrow \frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}^{2}}{4 \mathrm{x}^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{qQ}}{\mathrm{x}^{2}} \Rightarrow \mathrm{q}=\frac{-\mathrm{Q}}{4}$

Short Trick : For such type of problem the magnitude of middle charge can be determined if either of the extreme charge is in equilibrium by using the following formula.

In charge $\mathrm{A}$ is in equilibrium then $\mathrm{q}=$ $-Q_{B}\left(\frac{x_{1}}{x}\right)^{2}$

In charge $\mathrm{B}$ is in equilibrium the $\mathrm{q}=-\mathrm{Q}_{\mathrm{A}}\left(\frac{\mathrm{x}_{2}}{\mathrm{x}}\right)^{2}$

If the whole system is in equilibrium then use either of the above formula.

A charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium, if $q$ is equal to

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