Question

A flat disk of radius 0.50 m is oriented so that the plane of the disk makes an angle of 30 degrees with a uniform electric field. If the field strength is 713.0 N/C find the electric Tiux through the surface A) 560 Nm2/C B) 620 N·m2/C C) 160 n N.m2/C D) 280 N.m2/C

Answer 1

Answer:

The electric flux is $280\ \rm N.m^2/C$

Explanation:

Given:

• Radius of the disc R=0.50 m
• Angle made by disk with the horizontal $\theta=30^\circ$
• Magnitude of the electric Field $E=713.0\ \rm N/C$

The flux of the Electric Field E due to the are dA in space can be found out by using Gauss Law which is as follows

$\phi=\int E.dA$

where

• $\phi$ is the total Electric Flux
• E is the Electric Field
• dA is the Area through which the electric flux is to be calculated.

Now according to question we have

$=EA\cos\theta \\=713\times 3.14\times 0.5^2 \times \cos60^\circ\\=280\ \rm N.m^2/C$

Hence the electric flux is calculated.