Question

For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its centre. Then value of h is:
A. $\dfrac{R}{{\sqrt 2 }}$
B. $R\sqrt 2 $
C. $\dfrac{R}{{\sqrt 5 }}$
D. $R$

Answer 1

Hint: The concept of electric field is very important to study electrostatic phenomena. We know that electric force is developed between two charges when it is placed at a certain distance. This action-at-distance can be explained by using the concept of electric field.

Complete answer:.
Consider a circular ring of radius ‘a’ having a charge ‘q’ uniformly distributed over it. Let o be the center of the ring then electric field at a distance x meters from the center along the axis is given by,
$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{qx}}{{{{\left( {{a^2} + {x^2}} \right)}^{\dfrac{3}{2}}}}}$
Now we will be using this same equation for our further calculations.
Given, radius a=r
Distance from the centre of the ring, x=h

Now above equation becomes,
$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{qh}}{{{{\left( {{R^2} + {h^2}} \right)}^{\dfrac{3}{2}}}}}$…………………………… (1)
In question it is given that the electric field on its axis has the largest magnitude at a distance h from its centre o.
Therefore, we can write, when e is maximum then $\dfrac{{dE}}{{dh}} = 0$
Now differentiate equation (1) with respect to h and equate to zero. We get,
$\dfrac{{dE}}{{dh}} = \dfrac{d}{{dh}}\left( {\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{qh}}{{{{\left( {{R^2} + {h^2}} \right)}^{^{\dfrac{3}{2}}}}}}} \right)$
Use quotient rules for this differentiation. That is,
$\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\left( {\dfrac{{du}}{{dx}}} \right) - u\left( {\dfrac{{dv}}{{dx}}} \right)}}{{{v^2}}}$
We get, $\dfrac{{dE}}{{dh}} = \dfrac{q}{{4\pi {\varepsilon _0}}}\dfrac{d}{{dh}}\left( {\dfrac{h}{{{{\left( {{R^2} + {h^2}} \right)}^{\dfrac{3}{2}}}}}} \right)$
$\dfrac{{dE}}{{dh}} = \dfrac{q}{{4\pi {\varepsilon _0}}}\left( {\dfrac{{{{\left( {{R^2} + {h^2}} \right)}^{\dfrac{3}{2}}}.1 - \dfrac{3}{2}{{\left( {{R^2} + {h^2}} \right)}^{\dfrac{1}{2}}}.2{h^2}}}{{{{\left( {{R^2} + {h^2}} \right)}^3}}}} \right)$
We have $\dfrac{{dE}}{{dh}} = 0$
Therefore, $0 = \dfrac{q}{{4\pi {\varepsilon _0}}}\left( {\dfrac{{{{\left( {{R^2} + {h^2}} \right)}^{\dfrac{3}{2}}}.1 - \dfrac{3}{2}{{\left( {{R^2} + {h^2}} \right)}^{\dfrac{3}{2}}}.2{h^2}}}{{{{\left( {{R^2} + {h^2}} \right)}^3}}}} \right)$
Let us find the value of ‘h’.
but $\dfrac{q}{{4\pi {\varepsilon _0}}} \ne 0$
Then, \[\dfrac{3}{2}{\left( {{R^2} + {h^2}} \right)^{\dfrac{1}{2}}}.2{h^2} = \left( {{{\left( {{R^2} + {h^2}} \right)}^{\dfrac{3}{2}}}} \right)\]
Simplifying above equation,
$\dfrac{3}{2}.2{h^2} = \left( {{{\left( {{R^2} + {h^2}} \right)}^{\dfrac{3}{2} - \dfrac{1}{2}}}} \right)$
$3{h^2} = \left( {{R^2} + {h^2}} \right)$
$3{h^2} - {h^2} = {R^2}$
$3{h^2} = {R^2}$
Value of h will be,$h = \dfrac{R}{{\sqrt 2 }}$

Therefore, the correct option is (a).

Additional information:
A uniform electric field is that in which at every point, the intensity of the electric field is the same both in magnitude and direction. E.g.: electric field between the plates parallel plate condenser.
A non uniform electric field is that point in which the intensity of electric field E changes from point to point in magnitude or in direction or in both. E.g.: Electric field due to point charge.

Note:
> An electric field is a vector quantity that has both magnitude as well as direction.
> Its direction is the direction along the direction of electrostatic force.