Question

A disk of radius 2.1 cm has a surface charge density of 5.6 µC/m2 on its upper face. What is the magnitude of the electric field produced by the disk at a point on its central axis at distance z = 9.5 cm from the disk?

Answer 1

Answer:

$=6.3*10^3 N/C$

Explanation:

solution:

from this below equation (1)

$E=$σ/2εo$(1-\frac{z}{\sqrt{z^2-R^2} } )$...........(1)

we obtain:

$=5.6*10^-6 \frac{c}{m^2} /2(8.85*10^-12\frac{c^2}{N.m^2} ).(1-\frac{9.5 cm}{\sqrt{9.5^2-2.1^2} } )$

$=6.3*10^3 N/C$