The total electric field at a point on the axis of a uniformly charged disk, whi

Question

The total electric field at a point on the axis of a uniformly charged disk, which has a radius R and a uniform charge density of sigma, is given by the following expression, where x is the distance of the point from the disk. E_x = 2 pi k_e sigma [1 - x/(R^2 + x^2)^1/2] Consider a disk of radius R = 2.70 cm having a uniformly distributed charge of +5.13 mu C. Using the expression above, compute the electric field at a point on the axis and 2.85 mm from the center. N/C Explain how the answer to part (a) compares with the field computed from the near-field approximation E = sigma/2t_0. Using the first expression above, compute the electric field at a point on the axis and 28.5 cm from the center of the disk. Explain how the answer to part (c) compares with the electric field obtained by treating the disk as a +5.13-mu C charged particle at a distance of 28-5 cm.

Explanation / Answer

part a:

values given are:

R=2.7 cm=0.027 m

charge=q=5.13 uC=5.13*10^(-6) C

then charge density=sigma=q/(pi*R^2)=5.13*10^(-6)/(pi*0.027^2)=0.00224 C/m^2

x=2.85 mm=2.85*10^(-3) m

then electric field

=2*pi*9*10^9*0.00224*(1-(2.85*10^(-3)/sqrt((0.027)^2+(2.85*10^(-3))^2)))

=1.1337*10^8 N/C

part b:

using near field approximation,

electric field value=0.00224/(2*8.85*10^(-12))=1.2655*10^8 N/C

so approximated value is greater than the actual value by a percentage of

(1.2655-1.1337)*100/1.1337=11.626%

part c:

using x=28.5 cm=0.285 m

electric field value=2*pi*9*10^9*0.00224*(1-(0.285/sqrt((0.027)^2+(0.285)^2)))=5.6463*10^5 N/C

part d:

for a point charge of 5.13 uC,

electric field at a distance of 28.5 cm is given by

9*10^9*5.13*10^(-6)/0.285^2
=5.6842*10^5 N/C

so approximated value is higher than the actual value by a percentage of

(5.6482-5.6463)*100/5.6463=0.03365%

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