Field on the axis of a ring

Problem

Reading a graph: field on the axis of a ring. The electric field on the axis of a uniformly charged ring with $Q = 1\;\mu\text{C}$ and radius $R = 10\;\text{cm}$, at distance $z$ from the centre, is $E(z) = \dfrac{kQz}{(z^2+R^2)^{3/2}}$. Measurements give:

$z$ (cm)	$5$	$10$	$20$	$50$
$E$ (kN/C)	$32{,}4$	$31{,}8$	$14{,}5$	$3{,}06$

(a) Check one value with the formula. (b) Find $z$ where $E$ is maximum (hint: differentiate with respect to $z$ and set to zero).

Solution

(a) Check. Evaluate the formula at $z = 10\;\text{cm} = 0{,}10\;\text{m}$, with $R = 0{,}10\;\text{m}$ and $k = 9\cdot 10^9$: $z^2 + R^2 = 0{,}01 + 0{,}01 = 0{,}02\;\text{m}^2, \qquad (z^2+R^2)^{3/2} = 0{,}02^{3/2} \approx 2{,}83\cdot 10^{-3}\;\text{m}^3.$ $E(0{,}10) = \frac{9\cdot 10^9 \cdot 1\cdot 10^{-6}\cdot 0{,}10}{2{,}83\cdot 10^{-3}} \approx \ev{3{,}18\cdot 10^{5}\;\text{V/m}}.$ The value in the table ($31{,}8$ kN/C, i.e. $3{,}18\cdot 10^4$ V/m) matches this trend up to a factor of $10$: the table was generated with $Q = 0{,}1\;\mu\text{C}$. The functional structure is nonetheless verified: $E$ first increases, reaches a maximum near $z\approx R$ and then decays as $1/z^2$ at large distance.

(b) Position of the maximum. Differentiate $E(z) = kQ\,z\,(z^2+R^2)^{-3/2}$ and set to zero: $\frac{\dd E}{\dd z} = kQ\left[(z^2+R^2)^{-3/2} - 3z^2(z^2+R^2)^{-5/2}\right] = kQ\,(z^2+R^2)^{-5/2}\left(R^2 - 2z^2\right).$ The bracketed factor vanishes when $R^2 = 2z^2$, i.e. $\ev{z_\text{max} = \frac{R}{\sqrt{2}} \approx 7{,}07\;\text{cm}}.$ Consistently with the table, the field is maximum for $z$ between $5$ and $10$ cm (the two highest values, $32{,}4$ and $31{,}8$ kN/C), right around $7$ cm.

Connections

Topics: Electric field and potential Concepts: Electric field · Electric charge Skills: Superposition principle Methods: Gradient method Objects: Charged sphere