Question

For a uniformly charged ring of radius r, the electric field on its axis has the largest magnitude at a distance h from its centre. The value of h is: a) r / √2 b) r / 2 c) r / √3 d) r / 3

Answer 1

For a uniformly charged ring of radius r, the electric field on its axis has the largest magnitude at a distance h from its centre. Then the value of h is r / √3.

The correct answer is c) r / √3.

Step-by-step explanation:

he electric field on the axis of a uniformly charged ring is given by the equation
`E = kQz / (z^2 + r^2)^(3/2)`, where E is the magnitude of the electric field, k is the electrostatic constant, Q is the total charge on the ring, z is the distance along the axis from the center of the ring, and r is the radius of the ring.

To find the point where the electric field has the largest magnitude, we need to find the maximum value of E. To do this, we can take the derivative of E with respect to z and set it equal to zero.

Differentiating E with respect to z, we get:
`dE/dz = (kQ / (z^2 + r^2)^(3/2)) - (3kQz^2 / (z^2 + r^2)^(5/2))`

Setting dE/dz equal to zero, we have:

`(kQ / (z^2 + r^2)^(3/2)) - (3kQz^2 / (z^2 + r^2)^(^5^/^2^)) = 0`

Multiplying through by
`(z^2 + r^2)^(^5^/^2^)`), we get:

`kQ - 3kQz^2 = 0`

Simplifying, we find:

`z^2 = r^2 / 3`

Taking the square root of both sides, we get:

z = r / √3

Therefore, the value of h is r / √3.

So, the correct answer is option c) r / √3.