Question

A disk with a uniform positive surface charge density lies in the x-y plane, centered on the origin. The disk contains 2.5 x 10-6 C/m2 of charge, and is 7.5 cm in radius. What is the electric field at z?

Answer 1

Answer:

E=3 x 10^4 N/c

Explanation:

The electric field strength can be found out disk with a uniform positive surface charge density by

$E= (\sigma/\2epsilon_o)(1-z/ \sqrt(z^2+r^2))$

σ= charge density

r= radius of the disk

z= position in which we have to find electric field = 15 cm

ε_0= constant ( vacuum permitivity)

putting values we get

$E= \frac{2.5\times10^{-6}}{2\times2.5\times10^{-6}}(1-\frac{0.15}{\sqrt{0.15^2+0.075^2} })$

solving we get

E=30000 N/c

E=3 x 10^4 N/c