Question

3. A uniformly charged ring of radius 10.0 cm has a total T charge of 75.0 mC. Find the electric field on the axis of the ring at (a) 1.00 cm, (b) 5.00 cm, (c) 30.0 cm, and (d) 100 cm from the center of the ring.

Answer 1

To solve this problem we will apply the concepts related to the electric field in a ring. This concept is already standardized in the following mathematical expression, which relates the coulomb constant, the distance to the axis, the distance of the two points. Mathematically it is described as,

$E = \frac{k_exq}{(x^2+r^2)^{(3/2)}} \hat{i}$

Here,

$k_e$ = Coulomb constant

q = Charge

x = Distance to the axis

r = Distance between the charges

Our values are given as,

$r = 10.0cm = 10*10^{-2} m$

$q = 75\mu C = 75*10^{-6} C$

$x_a = 1.00cm$

$x_b = 5.00cm$

$x_c = 30cm$

$x_d = 100cm$

So applying this to our 4 distances, we have

PART A)

$E_a = \frac{(9*10^9)(1.00*10^{-2})(75*10^{-6})}{((1.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}$

$E_a = 6.64*10^6N/C \hat{i}$

PART B)

$E_b = \frac{(9*10^9)(5.00*10^{-2})(75*10^{-6})}{((5.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}$

$E_b = 24.1*10^6N/C \hat{i}$

PART C)

$E_c = \frac{(9*10^9)(30.00*10^{-2})(75*10^{-6})}{((30.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}$

$E_c = 6.39*10^6N/C \hat{i}$

PART D)

$E_d = \frac{(9*10^9)(100.00*10^{-2})(75*10^{-6})}{((100.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}$

$E_d = 0.664*10^6N/C \hat{i}$