Question

Charge is uniformly distributed around a ring of radius R = 7.40 cm, and the resulting electric field magnitude E is measured along the ring's central axis (perpendicular to the plane of the ring). At what distance from the ring's center is E maximum?

Answer 1

`x = 5.23 cm`

Step-by-step explanation:

Here if the radius of the ring is R then electric field on its axis is given by the formula

`E = (kqx)/((x^2 + R^2)^(3/2))`

now we know that if the electric field is maximum at a point on its axis then its differentiation with respect to the distance will be zero

so we have

`(dE)/(dx) = 0`

`0 = kq((1(x^2 + R^2)^(3/2) - x.3x(x^2 + R^2)^(1/2))/((x^2 + R^2)^3))`

so we have

`(x^2 + R^2) - 3x^2 = 0`

`x = (R)/(\sqrt2)`

so here we have

R = 7.40 cm

`x = 5.23 cm`