Question

Consider a uniform electric field $E = 3 × 10^3 \,\hat{i} NC^{-1}$.
(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 $60^0$ angle with the x-axis?

Answer 1

(a) Electric field intensity,$E = 3 × 10^3 \,\hat{i} NC^{-1}$
Magnitude of electric field intensity, $|E| = 3 × 10^3 NC^{-1}$
Side of the square, $s = 10 cm = 0.1 m$
Area of the square, $A = s^2 = 0.01 m^2$
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, $\theta = 0^0$
Flux (Φ ) through the plane is given by the relation,
$Φ = |E| Acos θ$
$= 3 × 10^3 × 0.01 × cos 0^0$
$= 30 Nm^2C^{-1}$

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(b) Plane makes an angle of $60^0$ with the x – axis. Hence, $θ = 60^0$
$Flux, Φ = |E| Acos θ$
$= 3 × 10^3 × 0.01 × cos 60^0$
$= 30 ×\frac{1}{2}$
$= 15 N m^2C^{-1}$