A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a…

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Question “A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a…”

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

Let the non-conducting rod with charge in the form a circle be contained in the plan and the electric field is to exist along the perpendicular line .

An electric field is created by non-conducting with uniform charge distribution in circular shape with radius is.

Here is the electric field. It is the total amount of charge in the circular cylinder. This is the point at the field is measured. The radius of the circular cylinder is k.

The condition determines the value that provides maximum electric field.

is the electric field expression that meets the above conditions.

(a)

To find the electric field created by a circular distribution of ring charges at any point along its axis that passes through its center perpendicularly, use the expression

The

will make the expression zero.

Therefore, no electric field acts at the

.

(b)

The electric field that contains the value can also be found by using the assumption that is the

.

The denominator can also be changed as follows:

The expression for the electric force is,

Therefore, the electric field is ininversely proportional the square of the point z is,

.

(c)

The

direction has the minimum conditions for the maximum electricity value.

The maximum electric field value is therefore

On simplification

Consider the positive value of this alone

(d)

is the expression for the electric fields.

Substitute

for

,

for

,

for

,

for

,

for

is the maximum electric field that can be produced.

Ans: Part A

The electric field at the point is

Part b

The electric field at large

values, i.e.

.

Part c

The maximum electric field is at

.

Part d

is the maximum electric field that can be produced.

Conclusion

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