Question
Two charge $$q$$ and $$-3q$$ are placed fixed on $$x$$-axis separated by distance $$d.$$ Where should a third charge $$2q$$ be placed such that it will not experience any force?
Two charge $$q$$ and $$-3q$$ are placed fixed on $$x$$-axis separated by distance $$d.$$ Where should a third charge $$2q$$ be placed such that it will not experience any force?
A.
$$\frac{{d - \sqrt 2 d}}{2}$$
B.
$$\frac{{d + \sqrt 3 d}}{2}$$
C.
$$\frac{{d + 3d}}{2}$$
D.
$$\frac{{d - \sqrt 5 d}}{2}$$
Answer :
$$\frac{{d + \sqrt 3 d}}{2}$$
Solution :
Let a charge $$2q$$ be placed at $$P,$$ at a distance $$I$$ from $$A$$ where charge $$q$$ is placed, as shown in figure.
The charge $$2q$$ will not experience any force, when force, when force of repulsion on it due to $$q$$ is balanced by force of attraction on it due to $$-3q$$ at $$B$$ where $$AB = d$$
$$\eqalign{ & {\text{or}}\,\,\frac{{\left( {2q} \right)\left( q \right)}}{{4\pi {\varepsilon _0}{\ell ^2}}} = \frac{{\left( {2q} \right)\left( { - 3q} \right)}}{{4\pi {\varepsilon _0}{{\left( {\ell + d} \right)}^2}}} \cr & {\left( {\ell + d} \right)^2} = 3{\ell ^2}\,\,{\text{or}}\,\,2{\ell ^2} - 2\ell d - {d^2} = 0 \cr & \therefore \ell = \frac{{2d \pm \sqrt {4{d^2} + 2{d^2}} }}{4} = \frac{d}{2} \pm \frac{{\sqrt 3 d}}{2} \cr} $$
Let a charge $$2q$$ be placed at $$P,$$ at a distance $$I$$ from $$A$$ where charge $$q$$ is placed, as shown in figure.
The charge $$2q$$ will not experience any force, when force, when force of repulsion on it due to $$q$$ is balanced by force of attraction on it due to $$-3q$$ at $$B$$ where $$AB = d$$
$$\eqalign{ & {\text{or}}\,\,\frac{{\left( {2q} \right)\left( q \right)}}{{4\pi {\varepsilon _0}{\ell ^2}}} = \frac{{\left( {2q} \right)\left( { - 3q} \right)}}{{4\pi {\varepsilon _0}{{\left( {\ell + d} \right)}^2}}} \cr & {\left( {\ell + d} \right)^2} = 3{\ell ^2}\,\,{\text{or}}\,\,2{\ell ^2} - 2\ell d - {d^2} = 0 \cr & \therefore \ell = \frac{{2d \pm \sqrt {4{d^2} + 2{d^2}} }}{4} = \frac{d}{2} \pm \frac{{\sqrt 3 d}}{2} \cr} $$
