Question

A ring made of wire with a radius of $10{\text{cm}}$ is charged negatively and carries a charge of $ - 5 \times {10^{ - 9}}{\text{C}}$. Find the distance from the centre to the point on the axis of the ring where the intensity of the electric field is maximum.
A) $0 \cdot 71{\text{m}}$
B) $7 \cdot 1 \times {10^{ - 2}}{\text{m}}$
C) $7 \cdot 1 \times {10^{ - 3}}{\text{m}}$
D) $1 \cdot 7 \times {10^{ - 2}}{\text{m}}$

Answer 1

Hint: The electric field at a point on the axis of a charged ring is known to be dependent on the charge of the ring, the distance between the point and the centre of the ring and on the radius of the ring. The general formula for this is known. At the point on the axis where the electric field is maximum, the derivative of the electric field will be zero.

Formula used:
The electric field at a point on the axis of a charged ring is given by, ${E_x} = \dfrac{{kQx}}{{{{\left( {{x^2} + {R^2}} \right)}^{\dfrac{3}{2}}}}}$ where $k$ is a constant, $Q$ is the charge on the ring, $R$ is the radius of the ring and $x$ is the distance between the centre of the ring and the point on the axis.

Complete step by step answer:
Step 1: Sketch the given ring and list its known parameters.

Here the centre of the ring is marked as O and we take P as the point on the axis where the strength of the electric field is maximum.
The charge on the ring is given to be $Q = - 5 \times {10^{ - 9}}{\text{C}}$ .
The radius of the ring is given to be $R = 10{\text{cm}}$ .
Let $x$ be the distance between the centre of the ring and point P which is to be determined.
Step 2: Express the relation for the electric field at point P on the axis of the ring and obtain its first derivative.
The electric field at point P on the axis of the given charged ring can be expressed as
${E_x} = \dfrac{{kQx}}{{{{\left( {{x^2} + {R^2}} \right)}^{\dfrac{3}{2}}}}}$ --------- (1) ($k$ is a constant)
The derivative of equation (1) with respect to $x$ is expressed as
 $\dfrac{{d{E_x}}}{{dx}} = \dfrac{d}{{dx}}\left[ {\dfrac{{kQx}}{{{{\left( {{x^2} + {R^2}} \right)}^{\dfrac{3}{2}}}}}} \right]$ -------- (2)
Equation (2) then becomes $\dfrac{{d{E_x}}}{{dx}} = kQ\left[ {\dfrac{{{{\left( {{x^2} + {R^2}} \right)}^{\dfrac{3}{2}}} \cdot 1 - \left( {x \cdot \dfrac{3}{2}{{\left( {{x^2} + {R^2}} \right)}^{\dfrac{1}{2}}} \cdot 2x} \right)}}{{{{\left( {{x^2} + {R^2}} \right)}^3}}}} \right]$
On simplifying, we get $\dfrac{{d{E_x}}}{{dx}} = kQ{\left( {{x^2} + {R^2}} \right)^{\dfrac{1}{2}}} \times \dfrac{{\left( {{x^2} + {R^2} - 3{x^2}} \right)}}{{{{\left( {{x^2} + {R^2}} \right)}^3}}}$ --------- (3)
Step 3: Apply the condition for the electric field to be maximum.
For the electric field to be maximum at point P, $\dfrac{{d{E_x}}}{{dx}} = 0$ .
The equation (3) becomes $kQ{\left( {{x^2} + {R^2}} \right)^{\dfrac{1}{2}}} \times \dfrac{{\left( {{x^2} + {R^2} - 3{x^2}} \right)}}{{{{\left( {{x^2} + {R^2}} \right)}^3}}} = 0$
$ \Rightarrow {\left( {{x^2} + {R^2}} \right)^{\dfrac{1}{2}}} \times \dfrac{{\left( {{x^2} + {R^2} - 3{x^2}} \right)}}{{{{\left( {{x^2} + {R^2}} \right)}^3}}} = 0$ ---------- (4)
But we know that ${\left( {{x^2} + {R^2}} \right)^{\dfrac{1}{2}}} > 0$ and so the above equation can be simplified as $\dfrac{{\left( {{x^2} + {R^2} - 3{x^2}} \right)}}{{{{\left( {{x^2} + {R^2}} \right)}^3}}} = 0$
$ \Rightarrow {x^2} - 3{x^2} + {R^2} = 0$ or $ \Rightarrow 2{x^2} = {R^2}$
$ \Rightarrow x = \pm \dfrac{R}{{\sqrt 2 }}$------- (5)
We take $x = \dfrac{R}{{\sqrt 2 }}$ ------- (6)
Substituting for $R = 10{\text{cm}}$ in equation (6) we get, $x = \dfrac{{10}}{{\sqrt 2 }} = 7 \cdot 1{\text{cm}}$
So the field has maximum strength at $x = 7 \cdot 1 \times {10^{ - 2}}{\text{m}}$ .

Thus the correct option is B.

Note:
The derivative of the electric field expression in equation (2) is obtained using the quotient rule given by, $\dfrac{{dy}}{{dx}} = \dfrac{{\left( {v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}} \right)}}{{{v^2}}}$ and in equation (2), $u = x$ and $v = {\left( {{x^2} + {R^2}} \right)^{\dfrac{3}{2}}}$. As both $x$ and $R$ are essentially distances, they cannot be negative and so ${\left( {{x^2} + {R^2}} \right)^{\dfrac{1}{2}}} > 0$ in equation (4). In equation (5) we obtain two values of $x$ i.e. $x = + \dfrac{R}{{\sqrt 2 }}$ and $x = - \dfrac{R}{{\sqrt 2 }}$ as the square of both the negative value and positive value of$x$ has the same value.