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

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

In this one should notice that there symmetry in this situation. Every element dq can be paired with a similar

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}}
$
 

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 - 

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