Question

A positively charged particle having some mass is resting in equilibrium at a height \[H\] above the centre of a fixed, uniformly and positively charged ring of radius \[R\]. The force of gravity (\[mg\]) acts downwards. The equilibrium of the particle at the given position (\[H\]) for small vertical displacement will be: (Assuming \[g\] is uniform)
A. Stable if \[H < \dfrac{R}{2}\]
B. Stable if \[H = \dfrac{R}{{\sqrt 2 }}\]
C. Unstable if \[H < \dfrac{R}{{\sqrt 2 }}\]
D. stable if \[H = \dfrac{R}{2}\]

Answer 1

Hint:Use the formula for the electric field at a point on the axis of the ring. Also use the formula for the electrostatic force acting on a charged particle. Hence, derive the expression for the electrostatic force acting on a charged particle on the axis of the ring. Determine the position of the particle from the ring and then determine at which position the particle will be stable.

Formulae used:
The electric field \[E\] at a point on the axis of the ring is
\[E = \dfrac{{KQx}}{{{{\left( {{R^2} + {x^2}} \right)}^{3/2}}}}\] …… (1)
Here, \[K\] is the constant, \[Q\] is the charge on the ring, \[x\] is the distance of the point on the axis from the centre of the ring and \[R\] is the radius of the ring.
The electrostatic force \[F\] acting on a charge is
\[F = qE\] …… (2)
Here, \[q\] is the charge and \[E\] is the electric field.

Complete step by step answer:
We have given that the position of the positively charged particle from the ring is \[H\].The electric field due at this positively charged particle is
\[E = \dfrac{{KQH}}{{{{\left( {{R^2} + {H^2}} \right)}^{3/2}}}}\]
The electrostatic force acting on this charged particle is
\[F = \dfrac{{KqQH}}{{{{\left( {{R^2} + {H^2}} \right)}^{3/2}}}}\]
At the equilibrium position, the weight of the charged particle is balanced by the electrostatic force on the particle. Also
\[\dfrac{{dF}}{{dH}} = 0\]
\[ \Rightarrow \dfrac{{d\left[ {\dfrac{{KqQH}}{{{{\left( {{R^2} + {H^2}} \right)}^{3/2}}}}} \right]}}{{dH}} = 0\]
\[ \Rightarrow KqQ\left[ {\dfrac{{{{\left( {{R^2} + {H^2}} \right)}^{3/2}}\dfrac{{dH}}{{dH}} - H\dfrac{{d{{\left( {{R^2} + {H^2}} \right)}^{3/2}}}}{{dH}}}}{{{{\left( {{R^2} + {H^2}} \right)}^3}}}} \right] = 0\]
\[ \Rightarrow \dfrac{{{{\left( {{R^2} + {H^2}} \right)}^{3/2}}\left( 1 \right) - H\dfrac{3}{2}{{\left( {{R^2} + {H^2}} \right)}^{1/2}}\dfrac{{d{H^2}}}{{dH}}}}{{{{\left( {{R^2} + {H^2}} \right)}^3}}} = 0\]
\[ \Rightarrow \dfrac{{{{\left( {{R^2} + {H^2}} \right)}^{3/2}} - H\dfrac{3}{2}{{\left( {{R^2} + {H^2}} \right)}^{1/2}}\left( {2H} \right)}}{{{{\left( {{R^2} + {H^2}} \right)}^3}}} = 0\]
\[ \Rightarrow \dfrac{{{{\left( {{R^2} + {H^2}} \right)}^{3/2}} - 3{H^2}{{\left( {{R^2} + {H^2}} \right)}^{1/2}}}}{{{{\left( {{R^2} + {H^2}} \right)}^3}}} = 0\]
\[ \Rightarrow {\left( {{R^2} + {H^2}} \right)^{3/2}} - 3{H^2}{\left( {{R^2} + {H^2}} \right)^{1/2}} = 0\]
\[ \Rightarrow {\left( {{R^2} + {H^2}} \right)^{3/2}} = 3{H^2}{\left( {{R^2} + {H^2}} \right)^{1/2}}\]
\[ \Rightarrow {R^2} + {H^2} = 3{H^2}\]
\[ \Rightarrow {R^2} = 2{H^2}\]
\[ \Rightarrow H = \dfrac{R}{{\sqrt 2 }}\]
Hence, in the stable position, the distance of the charged particle from the ring is \[\dfrac{R}{{\sqrt 2 }}\].
Hence, the options A and D are incorrect.
If the position of the particle is \[H < \dfrac{R}{{\sqrt 2 }}\] then the electric field at this charge will decreases and hence the electrostatic force on this particle will also decreases and the particle will move downwards. Hence, the particle will not be in equilibrium at position \[H < \dfrac{R}{{\sqrt 2 }}\].
Hence, the option C is incorrect.
When the particle is at the position \[H = \dfrac{R}{{\sqrt 2 }}\], the weight of the charged particle is balanced by the electrostatic position and the particle will be in stable equilibrium.

Hence, the correct option is B.

Note:The students should keep in mind that the charged particle will be in more equilibrium if its position is away from its previous position. At this position, the particle will displace slightly upwards because of which the electric field and hence the electrostatic force decreases. Hence, the particle will move downwards at its original equilibrium position.