Question
Consider a system of three charges $$\frac{q}{3},\frac{q}{3}$$ and $$ - \frac{{2q}}{3}$$ placed at points $$A, B$$ and $$C,$$ respectively, as shown in the figure. Take $$O$$ to be the centre of the circle of radius $$R$$ and angle $$CAB = {60^ \circ }$$
Consider a system of three charges $$\frac{q}{3},\frac{q}{3}$$ and $$ - \frac{{2q}}{3}$$ placed at points $$A, B$$ and $$C,$$ respectively, as shown in the figure. Take $$O$$ to be the centre of the circle of radius $$R$$ and angle $$CAB = {60^ \circ }$$
A.
The electric field at point $$O$$ is $$\frac{q}{{8\pi {\varepsilon _0}{R^2}}}$$ directed along the negative $$x$$-axis
B.
The potential energy of the system is zero
C.
The magnitude of the force between the charges at $$C$$ and $$B$$ is $$\frac{q^2}{{54\pi {\varepsilon _0}{R^2}}}$$
D.
The potential at point $$O$$ is $$\frac{q}{{12\pi {\varepsilon _0}R}}$$
Answer :
The magnitude of the force between the charges at $$C$$ and $$B$$ is $$\frac{q^2}{{54\pi {\varepsilon _0}{R^2}}}$$
Solution :
The electric field due to $$A$$ and $$B$$ at $$O$$ are equal and opposite producing a resultant which is zero. The electric field at $$O$$ due to $$C$$ is
$$E = \frac{1}{{4\pi { \in _0}}}\frac{{\frac{{2q}}{3}}}{{{R^2}}} = \frac{q}{{6\pi { \in _0}{R^2}}}.$$
$$\therefore $$ Option [A] is not correct. The electric potential at $$O$$ is
$${V_O} = K\left[ {\frac{{\frac{{ + q}}{3}}}{R}} \right] + K\left[ {\frac{{\frac{{ + q}}{3}}}{R}} \right] + K\left[ {\frac{{\frac{{ - 2q}}{3}}}{R}} \right] = 0$$
$$\therefore $$ Option [D] is wrong
In $$\Delta ABC\frac{{AC}}{{AB}} = \sin {30^ \circ } \Rightarrow AC = \frac{{AB}}{2} = R$$
Also $$\frac{{BC}}{{AB}} = \sin {60^ \circ } \Rightarrow BC = \frac{{\sqrt 3 AB}}{2} = \sqrt 3 R$$
Potential energy of the system
$$\eqalign{ & K\left[ {\frac{{\left( {\frac{q}{3}} \right)\left( {\frac{2}{3}} \right)}}{{2R}}} \right] + K\left[ {\frac{{\left( {\frac{q}{3}} \right)\left( {\frac{{ - 2q}}{3}} \right)}}{R}} \right] + K\left[ {\frac{{\left( {\frac{q}{3}} \right)\left( {\frac{{ - 2q}}{3}} \right)}}{{\sqrt 3 R}}} \right] \cr & = \frac{{k{q^2}}}{{9R}}\left[ {\frac{1}{2} - 2 - \frac{2}{{\sqrt 3 }}} \right] \ne 0 \cr} $$
$$\therefore $$ Option [B] is wrong
Magnitude of force between $$B$$ and $$C$$ is
$$F = \frac{1}{{4\pi { \in _0}}}\frac{{\left( {\frac{{2q}}{3}} \right)\left( {\frac{q}{3}} \right)}}{{{{(\sqrt 3 R)}^2}}} = \frac{{{q^2}}}{{54\pi { \in _0}{R^2}}}$$
The electric field due to $$A$$ and $$B$$ at $$O$$ are equal and opposite producing a resultant which is zero. The electric field at $$O$$ due to $$C$$ is
$$E = \frac{1}{{4\pi { \in _0}}}\frac{{\frac{{2q}}{3}}}{{{R^2}}} = \frac{q}{{6\pi { \in _0}{R^2}}}.$$
$$\therefore $$ Option [A] is not correct. The electric potential at $$O$$ is
$${V_O} = K\left[ {\frac{{\frac{{ + q}}{3}}}{R}} \right] + K\left[ {\frac{{\frac{{ + q}}{3}}}{R}} \right] + K\left[ {\frac{{\frac{{ - 2q}}{3}}}{R}} \right] = 0$$
$$\therefore $$ Option [D] is wrong
In $$\Delta ABC\frac{{AC}}{{AB}} = \sin {30^ \circ } \Rightarrow AC = \frac{{AB}}{2} = R$$
Also $$\frac{{BC}}{{AB}} = \sin {60^ \circ } \Rightarrow BC = \frac{{\sqrt 3 AB}}{2} = \sqrt 3 R$$
Potential energy of the system
$$\eqalign{ & K\left[ {\frac{{\left( {\frac{q}{3}} \right)\left( {\frac{2}{3}} \right)}}{{2R}}} \right] + K\left[ {\frac{{\left( {\frac{q}{3}} \right)\left( {\frac{{ - 2q}}{3}} \right)}}{R}} \right] + K\left[ {\frac{{\left( {\frac{q}{3}} \right)\left( {\frac{{ - 2q}}{3}} \right)}}{{\sqrt 3 R}}} \right] \cr & = \frac{{k{q^2}}}{{9R}}\left[ {\frac{1}{2} - 2 - \frac{2}{{\sqrt 3 }}} \right] \ne 0 \cr} $$
$$\therefore $$ Option [B] is wrong
Magnitude of force between $$B$$ and $$C$$ is
$$F = \frac{1}{{4\pi { \in _0}}}\frac{{\left( {\frac{{2q}}{3}} \right)\left( {\frac{q}{3}} \right)}}{{{{(\sqrt 3 R)}^2}}} = \frac{{{q^2}}}{{54\pi { \in _0}{R^2}}}$$
