Question

show thats the maximum magnitude e max of the elctric field along the axis of a uniformly charged ring occurs at x=a/2 and the vakues of q divide by6

Answer 1

Answer

Electric field on the axis of ring is given by

$E=\frac{kqx}{(x^2+R^2)^{1.5}}$

For maximum of electric field differentiate above electric field with respect to x

$\frac{dE}{dx}=0$

then we get

$x=\pm\frac{R}{\sqrt{2}}$ Here electric field is maximum.

Putting value of x then we get

$E_{max}=\frac{kqx}{((\frac{R}{\sqrt{2}})^2+R^2)^{1.5}}=\frac{2kq}{3\sqrt{3}R^2}$