Electric Field Due To A Uniformly Charged Ring - Practice Questions & MCQ

In this concept we are going to derive the electric field due to continous charge on a ring -

$
\overline{d E}=\frac{1}{4 \pi \varepsilon_0} \frac{d q}{r^2} \hat{r}
$

whose magnitude is

$
d E=\frac{1}{4 \pi \varepsilon_0} \frac{d q}{\left(R^2+x^2\right)}
$

The $X$-component is

$
\begin{aligned}
d E_x & =\frac{1}{4 \pi \varepsilon_0} \frac{d q}{\left(R^2+x^2\right)}(\cos \theta) \\
& =\frac{1}{4 \pi \varepsilon_0} \frac{d q}{\left(R^2+x^2\right)}\left(\frac{x}{\sqrt{R^2+x^2}}\right) \\
E_x & =\int d E_x=\int \frac{1}{4 \pi \varepsilon_0} \frac{x d q}{\left(x^2+R^2\right)^{3 / 2}}
\end{aligned}
$

As we integrate around the ring, all the terms remain constant
Also, $\quad \int d q=Q$

So the total field $\left(E_x\right)$ is

$
\begin{aligned}
& \quad=\frac{1}{4 \pi \varepsilon_0} \frac{x}{\left(x^2+R^2\right)^{3 / 2}} \int d q \\
& =\left(\frac{1}{4 \pi \varepsilon_0}\right) \frac{x Q}{\left(x^2+R^2\right)^{3 / 2}}
\end{aligned}
$

So, the Net electric field is -

$
E_{n e t}=\left(\frac{1}{4 \pi \varepsilon_0}\right) \frac{x Q}{\left(x^2+R^2\right)^{3 / 2}}
$
 

Graph  between E and X - 

                                                                        $\begin{aligned} x & = \pm \frac{R}{\sqrt{2}} \\ E_{\max } & =\frac{Q}{6 \sqrt{3} \pi \varepsilon_0 R^2}\end{aligned}$