Question
A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge.
✓
Answer
Net Electric Field at point $\text{P} =\int^{2\pi\text{a}}_{\circ}\text{dE}\cos\theta$
dE = Electric field due to a small element having charge dq
$ = \frac{1}{4\pi\varepsilon_\circ}\frac{\text{dq}}{\text{r}^{2}}$
Let $\lambda$ = Linear charge density
$ = \frac{\text{dq}}{\text{dl}}$
$\text{dq} = \lambda\text{dl}$
Hence $\text{E} = \int^{2\pi\text{a}}_{\circ}\frac{1}{4\pi\varepsilon_\circ}.\frac{\lambda\text{dl}}{\text{r}^{2}}\times\frac{x}{r} , $
where $\cos\theta =\frac{\text{x}}{\text{r}}$
$ = \frac{\lambda\text{x}}{4\pi\varepsilon_\circ\text{r}^{3}}(2\pi\text{a})$
$ =\frac{1}{4\pi\varepsilon_\circ}\frac{\text{Qx}}{(\text{x}^{2} + \text{a}^{2})^{\frac{3}{2}}}$
where total charge $\text{Q} = \lambda\times2\pi\text{a}$
At large distance i.e. x >> a
$\text{E}\simeq\frac{1}{4\pi\varepsilon_\circ}.\frac{\text{Q}}{\text{x}^{2}}$
This is the Electric field due to a point charge at distance x.
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