A thin nonconducting rod with a uniform distribution of positive charge Q is ben

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 = ?? N/C

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

(d) If R = 9.00 cm and Q = 4.50 ?C, what is the maximum magnitude? N/C

Explanation / Answer

By symmetry all electric field vectors cancel
with the exception of the z componen

E = kQz/(r^2+z^2)^(3/2)
Set the first derivative of this function to zero.

dE/dz = kQ[ (r^2+z^2)^-(3/2) - 3z^2(r^2+z^2)^-(5/2)] = 0
multiply both sides of this equation by (r^2+z^2)^(5/2) and divide both by kQ:
dE/dz = (r^2+z^2) - 3z^2 = 0
r^2 +z^2 -3z^2 = 0
r^2 - 2z^2 = 0
r^2 = 2z^2
Z=r/sqrt(2)

a)

when z=0 electricfield will be zero

b)

z=infinity

in this case alsoelectricfield will be equal to zero

c)

Z=R

E=KQ/sqrt(2)*R2

d)

substitute Z=r/sqrt(2) you can get maximum electricfield substitute given values in the problem

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.