A charge q is placed at the centre of the lin | vedclass

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Question

(b) Suppose in the following figure, equilibrium of charge $B$ is considered. Hence for it's equilibrium $|{F_A}|\, = \,|{F_C}|$
$==>$ $\frac{

Vedclass · Accepted Answer

$ - \frac{Q}{4}$

Vedclass · Answer

(b) Suppose in the following figure, equilibrium of charge $B$ is considered. Hence for it's equilibrium $|{F_A}|\, = \,|{F_C}|$
$==>$ $\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{4{x^2}}} = $$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{qQ}}{{{x^2}}}$ $==>$ $q = \frac{{ - \,Q}}{4}$

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.
If charge $A$ is in equilibrium then $q = - {Q_B}\,{\left( {\frac{{{x_1}}}{x}} \right)^2}$
If charge $B$ is in equilibrium then $q = - {Q_A}\,\,{\left( {\frac{{{x_2}}}{x}} \right)^2}$
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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