Question

A ring of radius R carries a charge Q, uniformly distributed along its circumference. What is the ratio of the electric field strength at a distance $R$ to that at a distance $\dfrac{R}{{\sqrt 2 }}$ along the axis?
  $\left( A \right)\dfrac{{\sqrt 3 }}{8}$
$\left( B \right)\dfrac{{3\sqrt 2 }}{8}$
$\left( C \right)\dfrac{{3\sqrt 3 }}{{4\sqrt 2 }}$
$\left( D \right)\dfrac{{2\sqrt 2 }}{{3\sqrt 3 }}$

Answer 1

Hint: Electric field strength is a quantitative expression of the intensity of an electric field at a particular location. We need to find Electric Field strength along the x-axis due to the charged ring. By applying this in two different radii we can find the ratio between two different distances.

Complete step by step solution:
As per the question ,
A ring of radius $R$ carries a charge $Q$ .
Where $Q$ is uniformly distributed along the circumference of the ring .
We need to find the Electric Field Strength at a distance $R$ in the axis and also at a distance $\dfrac{R}{{\sqrt 2 }}$ along the axis.
We know,
Electric Field Strength along the axis due to the charge ring is E.
Hence,
$E = kQ\dfrac{x}{{\sqrt {{{\left( {{x^2} + {R^2}} \right)}^3}} }}$
Let it be along the x-axis.
According to the question
Case I:
$x = R$
Now putting case I in electric field strength formula,we get
$E = kQ\dfrac{R}{{\sqrt {{{\left( {{R^2} + {R^2}} \right)}^3}} }}$
$ \Rightarrow E = kQ\dfrac{R}{{\sqrt {{{\left( {2{R^2}} \right)}^3}} }}$
$ \Rightarrow E = kQ\dfrac{R}{{2\sqrt 2 {R^3}}}$
$ \Rightarrow E = kQ\dfrac{1}{{2\sqrt 2 {R^2}}}$
Case II:
$x = \dfrac{R}{{\sqrt 2 }}$
Now putting case II in electric strength formula, we get
$E = kQ\dfrac{{\dfrac{R}{{\sqrt 2 }}}}{{\sqrt {^{{{\left( {{{\left( {\dfrac{R}{{\sqrt 2 }}} \right)}^2} + {R^2}} \right)}^3}}} }}$
$ \Rightarrow E = kQ\dfrac{{\dfrac{R}{{\sqrt 2 }}}}{{\sqrt {{{\left( {\dfrac{{{R^2}}}{2} + {R^2}} \right)}^3}} }}$
$ \Rightarrow E = kQ\dfrac{{\dfrac{R}{{\sqrt 2 }}}}{{\sqrt {{{\left( {\dfrac{{3{R^2}}}{2}} \right)}^3}} }}$
$ \Rightarrow E = kQ\dfrac{{\dfrac{R}{{\sqrt 2 }}}}{{\dfrac{{3\sqrt 3 {R^3}}}{{2\sqrt 2 }}}}$
$ \Rightarrow E = kQ\dfrac{2}{{3\sqrt 3 {R^2}}}$
Now taking the ratio of both the cases ,we get
$Ratio = \dfrac{{{E_R}}}{{{E_{\dfrac{R}{{\sqrt 2 }}}}}}$
Putting the value in the above equation, we get
$Ratio = \dfrac{{kQ\dfrac{1}{{2\sqrt 2 {R^2}}}}}{{kQ\dfrac{2}{{3\sqrt 3 {R^2}}}}}$
$ \Rightarrow Ratio = \dfrac{{3\sqrt 3 }}{{4\sqrt 2 }}$
Therefore, the correct option is $\left( C \right)$ .

Note:
Electric fields calculate the strength of a field at different positions. Different structures have different formulas of electric field strength, so it is very important to learn all the formulas .While solving the problem, go step by step because the solution is a bit complicated.