Question

A charge is uniformly distributed over a ring of radius a. Obtain an expression for the electric field at its centre . Hence show that for large distances it behaves like a point charge.​

Answer 1

Consider a ring of radius $\sf{a}$ uniformly distributed by a charge $\sf{q}$ over it. We find the electric field due to the ring at a point P at a distance $\sf{r}$ from the center of the ring along the axial line.

The ring is placed in YZ plane as in fig. 1. We consider an element of length $\sf{dl }$ which subtends an angle $\sf{d\theta}$ at the center of the ring and carries a charge $\sf{dq.}$

Since the charge is uniformly distributed over the ring,

$\sf{\longrightarrow\dfrac{dq}{dl}=\dfrac{q}{2\pi a}}$

$\sf{\longrightarrow\dfrac{dq}{a\,d\theta}=\dfrac{q}{2\pi a}}$

$\sf{\longrightarrow dq=\dfrac{q}{2\pi}\ d\theta\quad\quad\dots(1)}$

Also the radius vector $\vec{\sf{a}}$ is given by,

$\longrightarrow\vec{\sf{a}}=\sf{a\sin\theta\,\hat j-a\cos\theta\,\hat k}$

in fig. 2, the vector $\vec{\sf{r}}$ is along X axis and can be given by,

• $\vec{\sf{r}}=\sf{r\,\hat i}$

The vector $\vec{\sf{s}}$ is the vector drawn from the element to the point P and is related to $\vec{\sf{a}}$ and $\vec{\sf{r}}$ by triangle law of vector addition as,

$\longrightarrow\vec{\sf{r}}=\vec{\sf{a}}+\vec{\sf{s}}$

$\longrightarrow\vec{\sf{s}}=\vec{\sf{r}}-\vec{\sf{a}}$

$\longrightarrow\vec{\sf{s}}=\sf{r\,\hat i-a\sin\theta\,\hat j+a\cos\theta\,\hat k}$

Since OPQ is a right triangle, right angled at O,

• $\sf{s=\left(r^2+a^2\right)^{\frac{1}{2}}\quad\quad\dots(2)}$

Then unit vector along $\vec{\sf{s}}$ will be,

$\longrightarrow\mathsf{\hat s}=\dfrac{\vec{\mathsf{s}}}{\mathsf{s}}$

$\longrightarrow\sf{\hat s=\dfrac{r\,\hat i-a\sin\theta\,\hat j+a\cos\theta\,\hat k}{s}}$

Here $\vec{\sf{dE}}$ is the electric field at the point P due to the element and is directed along $\vec{\sf{s}}.$ So, $\sf{\left(k=\dfrac{1}{4\pi\epsilon_0}\right)}$

$\longrightarrow\vec{\sf{dE}}=\sf{\dfrac{k\ dq}{s^2}\,\hat s}$

$\longrightarrow\vec{\sf{dE}}=\sf{\dfrac{k\ dq}{s^2}\cdot\dfrac{r\,\hat i-a\sin\theta\,\hat j+a\cos\theta\,\hat k}{s}}$

$\longrightarrow\vec{\sf{dE}}=\sf{\dfrac{k\ dq}{s^3}\left(r\,\hat i-a\sin\theta\,\hat j+a\cos\theta\,\hat k\right)}$

From (1),

$\longrightarrow\vec{\sf{dE}}=\sf{\dfrac{kq\ d\theta}{2\pi s^3}\left(r\,\hat i-a\sin\theta\,\hat j+a\cos\theta\,\hat k\right)}$

$\longrightarrow\vec{\sf{dE}}=\sf{\dfrac{kq}{2\pi s^3}\left(r\,d\theta\,\hat i-a\sin\theta\,d\theta\,\hat j+a\cos\theta\,d\theta\,\hat k\right)}$

Hence the net electric field will be,

$\displaystyle\longrightarrow\vec{\sf{E}}=\sf{\int\limits_0^{2\pi}\dfrac{kq}{2\pi s^3}\left(r\,d\theta\,\hat i-a\sin\theta\,d\theta\,\hat j+a\cos\theta\,d\theta\,\hat k\right)}$

$\displaystyle\longrightarrow\vec{\sf{E}}=\sf{\dfrac{kq}{2\pi s^3}\left(r\,\hat i\int\limits_0^{2\pi} d\theta-a\,\hat j\int\limits_0^{2\pi}\sin\theta\,d\theta+a\,\hat k\int\limits_0^{2\pi}\cos\theta\,d\theta\right)}$

$\displaystyle\longrightarrow\vec{\sf{E}}=\sf{\dfrac{kq}{2\pi s^3}\left(r\,\hat i\Big[\theta\Big]_0^{2\pi}+a\Big[\cos\theta\Big]_0^{2\pi}\,\hat j+a\Big[\sin\theta\Big]_0^{2\pi}\,\hat k\right)}$

$\displaystyle\longrightarrow\vec{\sf{E}}=\sf{\dfrac{kq}{2\pi s^3}\left(2\pi r\,\hat i\right)}$

$\displaystyle\longrightarrow\vec{\sf{E}}=\sf{\dfrac{kqr}{s^3}\,\hat i}$

From (2),

$\displaystyle\longrightarrow\vec{\sf{E}}=\sf{\dfrac{kqr}{\left(r^2+a^2\right)^{\frac{3}{2}}}\,\hat i}$

and so it's magnitude is,

$\displaystyle\longrightarrow\sf{E=\dfrac{kqr}{\left(r^2+a^2\right)^{\frac{3}{2}}}\quad\quad\dots(3)}$

or,

$\displaystyle\longrightarrow\sf{E=\dfrac{kqr}{r^3\left(1+\dfrac{a^2}{r^2}\right)^{\frac{3}{2}}}}$

$\displaystyle\longrightarrow\sf{E=\dfrac{kq}{r^2\left(1+\dfrac{a^2}{r^2}\right)^{\frac{3}{2}}}\quad\quad\dots(4)}$

At the center,

• $\sf{r=0 }$

Then (3) becomes,

$\sf{\longrightarrow E=0}$

I.e., no electric field is experienced at the center of the ring.

At a very large distance away from the center,

• $\sf{r\to\infty}$

so that $\sf{a<<r}$ and thus $\sf{\dfrac{a^2}{r^2}}$ can be neglected.

Then (4) becomes,

$\displaystyle\longrightarrow\sf{E=\dfrac{kq}{r^2}}$

This is same as the expression for electric field due to a point charge, i.e., for large distances the ring behaves like a point charge.