Question

A ring of radius $R$has total charge$Q$.

$\left(a\right)$At what distance along the $z$-axis is the electric field strength a maximum?

$\left(b\right)$What is the electric field strength at this point?

Short answer

Part$\left(a\right)$

$\left(a\right)$ Distance along $z$-axis, $z=\frac{R}{\sqrt{2}}.$

Part$\left(b\right)$

$\left(b\right)$ Electrical field intensity at this time, $E=\frac{q}{4\pi {ϵ}_o{R}^2}\times \frac{2}{3\sqrt{3}}.$

Step-by-step solution

Electrical field

When charge is present in whatever form, any point in space has an electrical property.

Analyzing diagram:

Horizontal component,

$$\begin{gathered} E_z=2d{E}_z \\ =2E\cos \theta \end{gathered}$$

$\left(\cos \theta =\frac{z}{\sqrt{R^2+z^2}}\right)$

The vertical components are equal and opposite and cancel each other.

$E=2\times \frac{1}{4\pi {ϵ}_o}\times \left(\frac{q}{2\pi R}\right)\times \frac{dl\times z}{{\left(R^2+z^2\right)}^{3/2}}$

Applying integration and substituting the value to the equation,

$E=\frac{z}{4\pi {ϵ}_o}\times \frac{q}{2\pi R}\times \frac{z}{{\left(R^2+z^2\right)}^{3/2}}{\int }_0^{R}dl$

$E=\frac{q}{4\pi {ϵ}_o}\times \frac{z}{{\left(R^2+z^2\right)}^{3/2}}$

Distance along axis (part a)

$E_x=max$when $\frac{d{E}_z}{dz}=0$differenting$E_n.$

$E_z=\frac{q}{4\pi {ϵ}_o}\times \frac{z}{{\left(R^2+z^2\right)}^{3/2}}$

Equating we get,

$\frac{d{E}_z}{dz}=\frac{q}{4\pi {ϵ}_o}\left[\frac{1}{{\left(R^2+z^2\right)}^{3/2}}+\frac{z\left(-\frac{3}{2}\right)\left(2z\right)}{{\left(R^2+z^2\right)}^{5/2}}\right]$

$R^2+z^2-3z^2=0$

The distance along$z-$axis is,

$z=\frac{R}{\sqrt{2}}.$

Electrical strength (part b)

The equation as,

$E=\frac{q}{4\pi {ϵ}_o}\times \frac{\left(\frac{R}{\sqrt{2}}\right)}{{\left(R^2+\frac{R^2}{2}\right)}^{3/2}}$

Equating we get,

$E=\frac{q}{4\pi {ϵ}_o}\times \frac{R}{\sqrt{2}\times R^3{\left(\frac{3}{2}\right)}^{3/2}}$

The electric field intensity is,

$E=\frac{q}{4\pi {ϵ}_o{R}^2}\times \frac{2}{3\sqrt{3}}.$