Question

Consider a uniformly charged ring of radius R. Find the point on the axis where the electric field is a maximum.

Answer 1

Use the formula for electric field on the axis of a charged ring at a distance x from the centre. At a point where the electric field is a maximum, the differentiation of the electric field becomes zero. Therefore, differentiate the expression for the electric field on the axis of the charged ring at a distance x from the centre.

Formula used:
$E = \dfrac{{kxQ}}{{{{\left( {{R^2} + {x^2}} \right)}^{\dfrac{3}{2}}}}}$..........(1)
Here, k is the constant, Q is the charge on the ring, R is the radius of the ring and x is the distance from the centre of the ring to the point where the electric field is the maximum.

Complete step by step solution:
We know that, at a certain point if the electric field is to be the maximum, the differential $\dfrac{{dE}}{{dx}} = 0$.
We differentiate equation (1) with respect to x as follows,
$\dfrac{{dE}}{{dx}} = kQ\dfrac{d}{{dx}}\left( {\left( x \right){{\left( {{R^2} + {x^2}} \right)}^{ - \dfrac{3}{2}}}} \right) = 0$
We can differentiate the above equation using the identity,
$\dfrac{d}{{dx}}\left( {u \cdot v} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$
Therefore, we have,
$\Rightarrow \dfrac{{dE}}{{dx}} = kQ\left( {x\dfrac{d}{{dx}}\left( {{{\left( {{R^2} + {x^2}} \right)}^{ - \dfrac{3}{2}}}} \right) + \left( {{{\left( {{R^2} + {x^2}} \right)}^{ - \dfrac{3}{2}}}} \right)\dfrac{d}{{dx}}\left( x \right)} \right) = 0$
$\Rightarrow kQ\left( {x\left( { - \dfrac{3}{2}{{\left( {{R^2} + {x^2}} \right)}^{ - \dfrac{5}{2}}}\dfrac{d}{{dx}}\left( {{R^2} + {x^2}} \right)} \right) + \left( {{{\left( {{R^2} + {x^2}} \right)}^{ - \dfrac{3}{2}}}} \right)\left( 1 \right)} \right) = 0$
$\Rightarrow kQ\left( {x\left( { - \dfrac{3}{2}{{\left( {{R^2} + {x^2}} \right)}^{ - \dfrac{5}{2}}}\left( {2x} \right)} \right) + {{\left( {{R^2} + {x^2}} \right)}^{ - \dfrac{3}{2}}}} \right) = 0$
$\Rightarrow kQ\left( { - \dfrac{{3{x^2}}}{{{{\left( {{R^2} + {x^2}} \right)}^{\dfrac{5}{2}}}}} + \dfrac{1}{{{{\left( {{R^2} + {x^2}} \right)}^{\dfrac{3}{2}}}}}} \right) = 0$
$\Rightarrow \dfrac{{3{x^2}}}{{{{\left( {{R^2} + {x^2}} \right)}^{\dfrac{5}{2}}}}} = \dfrac{1}{{{{\left( {{R^2} + {x^2}} \right)}^{\dfrac{3}{2}}}}}$
$\Rightarrow 3{x^2} = {R^2} + {x^2}$
$\Rightarrow 2{x^2} = {R^2}$
$\therefore x = \dfrac{R}{{\sqrt 2 }}$

Therefore, the electric field will be the maximum at a distance $\dfrac{R}{{\sqrt 2 }}$ from the centre of the ring.

Note: We have not used the identity, $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$ to solve equation (1) as it involves lot of calculations to do. Students don’t need to derive the expression (1) since it is the proper expression for calculating the electric field along the axis of a charged ring. The distance $\dfrac{R}{{\sqrt 2 }}$ is the distance from the centre of the ring where the electric field is the maximum. This distance is symmetric on both sides of the ring.