A uniformly charged rod (length = 2.0 m, charge per unit length = 5.0 nC/m) is bent to form one quadrant of a circle. What is the magnitude of the electric field at the center of the circle?
A) 62 N/C
B) 56 N/C
C) 50 N/C
D) 44 N/C
E) 25 N/C

Question

Quizplus · Accepted Answer

The electric field at the center of the circle can be found by integrating the electric field due to small segments of the charged rod along the curved portion of the circle. The electric field due to a small segment can be given as:
dE = (k * dq) / r^2
Here, k is the Coulomb's constant, dq is the charge on the small segment of length dl and r is the distance between the small segment and the center of the circle.

Let's assume that the circle has a radius of r. The length of the curved portion of the rod is half of the circumference of the circle:
L = Πr/2

Therefore, the contribution to the electric field at the center due to a small element of length dl is:
dE = (k * dq) / r^2 = (k * λ * dl) / r^2
where λ is the charge per unit length of the rod.

Now, we can integrate this expression over the length of the arc to get the total electric field at the center:
E = ∫ dE = ∫ (k * λ * dl) / r^2
       = (k * λ / r^2) * ∫ dl
       = (k * λ / r^2) * L

Substituting the given values, we get:
E = (9 x 10^9 N.m^2/C^2) * (5 x 10^-9 C/m) * (Π / 4) / (r^2)
  = (1.125 Π / r^2) N/C

For a quadrant of the circle, r = 2/Π, and therefore:
E = (1.125 Π / (2/Π)^2) N/C = 50 N/C

Therefore, the answer is (C) 50 N/C.

A Uniformly Charged Rod (Length = 2