Question

A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a circle of radius R. The central perpendicular axis through the ring is a z-axis, with the origin at the center of the ring.


(a) What is the magnitude of the electric field due to the rod at z = 0?
______ N/C

(b) What is the magnitude of the electric field due to the rod at z = infinity?
_____ N/C

(c) In terms of R, at what positive value of z is that magnitude maximum?
______ R

(d) If R = 4.00 cm and Q = 9.00 µC, what is the maximum magnitude?
____ N/C

Answer 1

Concepts and reason

The concepts used to solve this problem are electric field, and circular ring of charge.

Find the expression for the electric field produced by a circular distribution of charges along an axis perpendicular to it.

Use the general expression to determine the required electric field at various points along the perpendicular axis.

Fundamentals

Let the non-conducting rod in the form of circle with charge be confined in the plane and the electric field is to be found along the perpendicular axis to it.

The electric field produced by a non-conducting having uniform distribution of charge in a circular shape with radius is

Here, is the electric field, is the total charge in the circular rod, is the point at which the field is measured, is the radius of the circular rod, and k is the constant.

The value of that gives maximum electric field is found using the condition

The electric field expression that satisfies the above condition is .

(a)

The expression to find the electric field produced by a circular ring charge distribution at any point along the axis passing through its center perpendicularly is

At, , the above expression will become zero.

Hence, there is no electric field acting at the point .

(b)

The electric field for the value can be found using the assumption that .

So, the denominator term can be modified as,

Therefore, the expression for the electric field becomes,

Hence, electric field is inversely proportional to the square of the point z is, .

(c)

The condition for the maximum value for electricity along the direction is

Therefore, the maximum value of the electric field is

On simplification

Considering the positive value alone

(d)

The expression for the electric field is

Substitute for , for , for , and for

The maximum electric field produced is .

Ans: Part a

The electric field acting at the point is

Part b

The electric field at large values of i.e. is .

Part c

The electric field is maximum at .

Part d

The maximum electric field produced is .