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

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• 12 Questions around this concept.

## Solve by difficulty

The electric field strength due to a ring of radius $2 R$ at a distance $X$ from its center on the axis of the ring carrying a charge $Q$ is given by:

$\mathrm{E}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qx}}{\left((2 \mathrm{R})^2+\mathrm{x}^2\right)^{3 / 2}}$.

At what distance from the center will the electric field be maximum?

## Concepts Covered - 1

Electric field on the axis of a charged ring

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 element on the opposite side of the ring. Every component $\vec{dE}$ perpendicular to the x-axis is thus cancelled by a component $\vec{dE}$ in the opposite direction.
In the summation process, all the perpendicular components $\vec{dE}$ add to zero. Thus we only add the $dE_x$ components, which all lie along the +X direction, and this is a simple scalar integral. From Coulomb’s Law in vector form,

$\begin{array}{c}{\overline{d E}=\frac{1}{4 \pi \varepsilon_{0}} \frac{d q}{r^{2}} \hat{r}} \\ \\ {\text { whose magnitude is }} \\ \\ {\qquad d E=\frac{1}{4 \pi \varepsilon_{0}} \frac{d q}{\left(R^{2}+x^{2}\right)}}\end{array}$

\begin{array}{l}{\text { The } X \text { -component is }} \\ \\ {\qquad \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}}\end{array}

As we integrate around the ring, all the terms remain constant

Also,          $\int dq = Q$

$\begin{array}{l}{\text { So the total field }\left(E_{x}\right) \text { is }} \\ \\ {\qquad=\frac{1}{4 \pi \varepsilon_{0}} \frac{x}{\left(x^{2}+R^{2}\right)^{3 / 2}} \int d q} \\ \\ {=\left(\frac{1}{4 \pi \varepsilon_{0}}\right) \frac{x Q}{\left(x^{2}+R^{2}\right)^{3 / 2}}}\end{array}$

So, the Net electric field is -

$E_{net} =\left(\frac{1}{4 \pi \varepsilon_{0}}\right) \frac{x Q}{\left(x^{2}+R^{2}\right)^{3 / 2}}$

Graph  between E and X -

If,   $\dpi{100} x= \pm \frac{R}{\sqrt{2}}$

$\dpi{100} E_{max}=\frac{Q}{6\sqrt{3}\pi \varepsilon _{0}R^{2}}$  ,

## Study it with Videos

Electric field on the axis of a charged ring

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