Question

Electric field due to charged ring on axis at a point A very close to centre is same as at point B far away from centre.The radius of ring is:-

Answer 1

Answer:

Here the distance on the axis of ring must be equal to the radius of the ring

Explanation:

As we know that the electric field on the axis of the ring is given as

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

now if electric field is required at a point close to its center

then we have

$x < < < R$

so we have

$E_1 = \frac{kq x}{R^3}$

Similarly if electric field is required at large distance from the center so we have

$x > > > R$

$E_2 = \frac{kq}{x^2}$

now we have

$E_1 = E_2$

$\frac{kq x}{R^3} = \frac{kq}{x^2}$

$x^3 = R^3$

$x = R$

#Learn

Topic : Electric Field

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