Question

A uniformly charged disk of radius $35.0cm$ carries charge with a density of $7.90\times {10}^{-3}C/m^2$. Calculate the electric field on the axis of the disk at (a) $5.00cm$, (b) $10.0cm$, (c) $50.0cm$, and (d) $200cm$ from the center of the disk.

Answer 1

The electric field for the disk is given by

$E=2\pi k_e\sigma (1-\frac{x}{\sqrt{x^2+R^2}})$
in the positive $x$ direction (away from the disk). Substituting,
$E=2\pi (8.99\times {10}^{9}N\cdot m^2/C^2)(7.90\times {10}^{-3}C/m^2)\times (1-\frac{x}{\sqrt{x^2+(0.350{)}^2}})$

$=(4.46\times {10}^{8}N/C)(1-\frac{x}{\sqrt{x^2+0.123}})$
(a) At $x=0.0500m,E=3.83\times {10}^{8}N/C=383MN/C$
(b) At $x=0.100m,E=3.24\times {10}^{8}N/C=324MN/C$
(C) At $x=0.500m,E=8.07\times {10}^{7}N/C=80.7MN/C$
(d) At $x=2.000m,E=6.68\times {10}^{8}N/C=6.68MN/C$