Question

A circular ring carries a charge q the variation if electric field is x measured from centre along axis the maximum electric field on the axis is

Answer 1

electric field along axis of circular ring is given by, $E(x)=\frac{kqx}{(x^2+r^2)^{3/2}}$

where, x is observation point along axis of circular ring and r is the radius of ring.

differentiating electric field with respect to x,

$\frac{dE(x)}{dx}=kq\left[\frac{(x^2+r^2)^{3/2}-x\frac{3}{2}(x^2+r^2)^{1/2}2x}{(x^2+r^2)^3}\right]$

$=kq\sqrt{x^2+r^2}\left[\frac{x^2+r^2-3x^2}{(x^2+r^2)^3}\right]$

$=kq\sqrt{x^2+r^2}\left[\frac{r^2-2x^2}{(x^2+r^2)^3}\right]$

for maximum electric field, $\frac{dE(x)}{dx}=0$

r² = 2x²

or, x = ± r/√2

hence, there are two optimum points,

for getting which point value of x for maximum again differentiating with respect to x. and putting both the value.

you will get, $\frac{d^2E}{dx^2}<0$ at x = r/√2

so, maximum value of electric field at X = r/√2

so, $E_{max}$ = kq(r/√2)/(r²/2 + r²)^{3/2}

= kqr/{√2(r² + 2r²)^{3/2}/2√2}

= 2kqr/(3r²)^{3/2}

= 2kqr/(3√3r³)

= 2kq/3√3r²