Question

A charge of -6.50 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.25 cm.(a) Find the magnitude of the electric field this disk produces at a point P on the axis of the disk a distance of 2.00 cm from its center.(b) Suppose that the charge was pushed away from the center and distributed uniformly on the outer rim of the disk. Find the magnitude of the electric field at point P.(c) If the charge is brought to the center of the disk, find the magnitude of the electric field at point P.

Answer 1

Answer:

(a) E = 17207.86 N/C

(b) E = 88704.39 N/C

(c) E = 744783.65 N/C

Explanation:

(a) To find the electric field at point B you use the following formula, which is obtained for the integration of a differential dE, produced by a differential element of the disk:

$E=k\sigma 2\pi[1-\frac{z}{\sqrt{z^2+R^2}}}]$   (1)

k: Coulomb constant = 9.98*10^9 Nm^2/C^2

R: radius of the disk = 1.25cm = 0.0125m

z: perpendicular distance to the disk = 2.00cm = 0.02m

σ: charge density = Q/A = (-6.50*10^-9C)/(π(0.0125m)^2)=1.32$E=k\sigma 2\pi[1-0]=744783.65N/C$*10^-5 C/m^2

$E=(8.98*10^9Nm^2/C^2)(1.32*10^{-5}C/m^2)(2\pi)[1-\frac{0.02m}{\sqrt{(0.02m)^2(0.0125m)^2}}]\\\\E=17207.86N/C$

(b) In  this case you can take the distribution of charge as the linear charge density of a ring. The electric field for this case is given by the following formula (again, it is obtained by integration of a differential dE):

$E=k\frac{\lambda 2\pi Rz}{(z^2+R^2)^{3/2}}$   (2)

λ: linear charge density = (-6.50*10^-9C)/(2π(0.0125m))=8.27*10^-8 C/m

by replacing the values of the parameters you obtain:

$E=(8.98*10^9Nm^2/C^2)\frac{(8.27*10^{-8}C/m)(2\pi)(0.0125m)(0.02)}{((0.02m)^2+(0.0125m)^2)^{3/2}}\\\\E=88704.39N/C$

(c) If the charge is in the center of the disk z=0 and you have in equation (1):

$E=k\sigma 2\pi[1-0]=744783.65N/C$