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, $\int dq=Q$
So the total field $\left(E_x\right)$ is

$\begin{array}{rl}& =\frac{1}{4\pi {\epsilon }_0}\frac{x}{{(x^2+R^2)}^{3/2}}\int dq\\ & =\left(\frac{1}{4\pi {\epsilon }_0}\right)\frac{xQ}{{(x^2+R^2)}^{3/2}}\end{array}$

So, the Net electric field is -

$E_{net}=\left(\frac{1}{4\pi {\epsilon }_0}\right)\frac{xQ}{{(x^2+R^2)}^{3/2}}$
 

As we integrate around the ring, all the terms remain constant
Also, $\int dq=Q$
So the total field $\left(E_x\right)$ is

$\begin{array}{rl}& =\frac{1}{4\pi {\epsilon }_0}\frac{x}{{(x^2+R^2)}^{3/2}}\int dq\\ & =\left(\frac{1}{4\pi {\epsilon }_0}\right)\frac{xQ}{{(x^2+R^2)}^{3/2}}\end{array}$

So, the Net electric field is -

$E_{net}=\left(\frac{1}{4\pi {\epsilon }_0}\right)\frac{xQ}{{(x^2+R^2)}^{3/2}}$
 

Graph  between E and X - 

$\begin{array}{rl}& x=\pm \frac{R}{\sqrt{2}}\\ & \text{ If, }\\ & E_{max}=\frac{Q}{6\sqrt{3}\pi {\epsilon }_0{R}^2}\end{array}$