Electric Field Due To A Uniformly Charged Ring

Introduction

The electric field produced by a uniformly charged ring is an important concept in electrostatics. A uniformly charged ring has equal charge distribution along its circumference, creating a symmetrical electric field. The behavior of the electric field varies with the position of the observation point. Along the axis of the ring, the electric field is directed either towards or away from the ring, depending on the charge's nature (positive or negative). This field is zero at the ring's center and increases along the axis up to a maximum before decreasing as the distance from the ring increases. Understanding this concept is crucial for analyzing charge distributions and electric potentials in more complex geometries.

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

I

In the summation process, all the perpendicular components $d \vec{E}$ add to zero. Thus we only add the $d E_x$ components, which all lie along the $+X$ direction, and this is a simple scalar integral. From Coulomb's Law in vector form,

$
\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_{\text {net }}=\left(\frac{1}{4 \pi \varepsilon_0}\right) \frac{x Q}{\left(x^2+R^2\right)^{3 / 2}}
$

The graph between E and X -

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

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Some Solved Examples

Example 1: Positive charge Q is distributed over a circular Ring of radius a. A point particle of mass m & negative charge -q is placed on its Axis at a distance y from the centre. Find the force on the particle. ()

1) $\frac{k Q}{a^2}$
2) $\frac{k q y Q}{a^3}$
3)0
4) $\frac{k Q y}{a^3}$

Solution

$E$ at a point $P$ that lies on the axis of the ring -

$
E_x=\frac{k Q x}{\left(x^2+R^2\right)^{\frac{3}{2}}}
$

for $x<<R$

$
E_x=\frac{k Q x}{R^3}
$

putting $x=y$ and $R=a$

$
\begin{aligned}
& E=\frac{k Q y}{a^3} \\
& \text { and } \Rightarrow F=q E
\end{aligned}
$
Putting values we get

$
F=\frac{k Q q y}{a^3}
$

Hence, the answer is the option (2).

Example 2: For a uniformly charged ring of radius $R$, the electric field on its axis has the largest magnitude at a distance $h$ from its centre. Then value of $h$ is:
1) $\frac{R}{\sqrt{5}}$
2) $\frac{R}{\sqrt{2}}$
3) $R$
4) $R \sqrt{2}$

Solution
The electric field due to the ring on its axis

$
E_x=\frac{k Q x}{\left(x^2+R^2\right)^{\frac{3}{2}}}
$

here $x=h$

$
E_h=\frac{k Q h}{\left(h^2+R^2\right)^{\frac{3}{2}}}
$

$\underline{d E}$
For finding maximum find $\overline{d h}$ and equate to zero

$
\Rightarrow k Q\left(h^2+R^2\right)^{\frac{3}{2}}=k Q h \cdot \frac{3}{2}\left(h^2+R^2\right)^{\frac{1}{2}} \cdot 2 h
$
$
\begin{aligned}
& h^2+R^2=3 h^2 \\
& 2 h^2=R^2 \\
& h=\frac{R}{\sqrt{2}}
\end{aligned}
$
Hence, the answer is the option (2).
Example 2: At the centre of the uniformly charged ring ( + Q) electric field will be. (Centre at O )
1) $\frac{k Q}{R^2}$
2) $\frac{k Q}{\left(x^2+y^2\right)^{\frac{3}{2}}}$
3) 0
4) None

Solution

At centre $\mathrm{x}=0$

$
\begin{aligned}
& E_c=0 \quad V_c=\frac{k Q}{R} \\
& W_{\text {ext }}=\Delta K=W_{\text {ext }}=\frac{1}{2} m v^2-0 \\
& W=q \cdot V=q_0 \times \frac{k Q}{R} \Rightarrow \frac{k q_0 Q}{R}=\frac{1}{2} m v^2 \\
& \Rightarrow v=\sqrt{\frac{2 k q_0 Q}{m R}}
\end{aligned}
$
Hence, the answer is the option (3).

Summary

The electric field due to a uniformly charged ring at any point along its axis depends on the distance from the center of the ring. At the center, the electric field is zero due to symmetry. As you move along the axis, the electric field increases, reaches a maximum, and then gradually decreases. The formula for the electric field at a point on the axis involves the total charge, the ring's radius, and the distance from the center. This analysis simplifies calculations in many electrostatics problems and is used in various applications like particle accelerators and electric sensors, where uniform charge distributions are common.