Question

A square that has 10-cm-long edges is centered on the x aus in a region where there exists a uniform electric field given by E-200 KN/C)i. (a) What is the electric flux of this electric field = through the surface of a square if the normal to the surface is in the +r direction? (b) What is the electric flux through the same square surface if the normal to the surface makes a 60° angle with the y axis and an angle of 90° with the z axis?

Answer 1

(a) When the electric field vector is normal to the surface of the square, the electric flux is

$$\Phi=EA=1\cdot10^{-200}\cdot(0.1\cdot0.1)=\\=1\cdot10^{-204}\text{ kN}/\text{Cm}^2$$

(b) If the electric field is directed along x-axis, the angle of the plane of the square to the x-axis is 60°, the flux is

$$\Phi=EA\text{ sin}60°=\\=8.66\cdot10^{-205}\text{ kN}/\text{C/m}^2$$