Question

A uniformly charged disk of radius 35.0 cm carries a charge density of 7.50 ✕ 10^-3 C/m². Calculate the electric field on the axis of the disk at the following distances from the center of the disk.

(a) 5.00 cm
MN/C

(b) 10.0 cm
MN/C

(c) 50.0 cm
MN/C

(d) 200 cm
MN/C

Answer 1

here,

the equation for the electric field of a charged disk, which is:

E = (sigma / 2e0) * {1 - [z / sqrt(z^2 + r^2)]}

where,

E = Electric Field
sigma = Charge Density of the Disk (7.50*10^-3 C/m^2)

e0 = Permittivity of Free Space (8.85*10^-12 C^2/Nm^2)

z = Radial distance from the center of the Disk

r = Radius of the Disk (0.350 m)

E = (0.0075 / (2 * 8.85 * 10^-12) * {1 - [z / sqrt(z^2 + 0.35^2)]}

a)

for z = 5 cm = 0.05 m

E = (0.0075 / (2 * 8.85 * 10^-12) * {1 - [0.05 / sqrt(0.05^2 + 0.35^2)]}

E = 3.64 * 10^8 N/C

E = 364.8 MN/C

b)

for z = 10 cm = 0.1 m

E = (0.0075 / (2 * 8.85 * 10^-12) * {1 - [0.1 / sqrt(0.1^2 + 0.35^2)]}

E = 3.07 * 10^8 N/C

E = 307.3 MN/C

c)

for z = 50 cm = 0.5 m

E = (0.0075 / (2 * 8.85 * 10^-12) * {1 - [0.5 / sqrt(0.5^2 + 0.35^2)]}

E = 7.66 * 10^7 N/C

E = 76.6 MN/C

d)

for z = 200 cm = 2 m

E = (0.0075 / (2 * 8.85 * 10^-12) * {1 - [2 / sqrt(2^2 + 0.35^2)]}

E = 6.34 * 10^6 N/C

E = 6.34 MN/C