Question

Consider a uniform electric field $E=3\times {10}^3\hat{i}N{C}^{-1}$. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a ${60}^o$ angle with the x-axis?

Answer 1

(a)

Electric field intensity, $\overline{E}=3\times {10}^3$N/C

Magnitude of electric field intensity, $|E|=3\times {10}^3$N/C

Side of the square, $s=10cm=0.1m$

Area of the square, $A=s^2=0.01m^2$

The plane of the square is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane and electric field, $\theta =0^o$

Flux $\left(\varphi \right)$ through the plane is given by the relation,

$\varphi =|\overline{E}|Acos\theta$

$=3\times {10}^3\times 0.01\times cos{0}^o$

$=30N{m}^2/C$

(b)

Plane makes an angle of ${60}^o$ with the x-axis. Hence, $\theta ={60}^o$

Flux, $\varphi =|\overline{E}|Acos\theta$

$=3\times {10}^3\times 0.01\times cos{60}^o$

$=30\times \frac{1}{2}=15N{m}^2/C$