 # Consider a uniform electric field `
Question:

Consider a uniform electric field $\mathbf{E}=3 \times 10^{3} \hat{\mathrm{N}} / \mathrm{C}$.

(a) What is the flux of this field through a square of $10 \mathrm{~cm}$ on a side whose plane is parallel to the $y z$ plane?

(b) What is the flux through the same square if the normal to its plane makes a $60^{\circ}$ angle with the $x$-axis?

Solution:

(a) Electric field intensity, $\vec{E}=3 \times 10^{3} \hat{\mathrm{N}} \mathrm{N} / \mathrm{C}$

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

Side of the square, $s=10 \mathrm{~cm}=0.1 \mathrm{~m}$

Area of the square, $A=\mathrm{s}^{2}=0.01 \mathrm{~m}^{2}$

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, $\theta=0^{\circ}$ Flux $(\Phi)$ through the plane is given by the relation,

$\Phi=|\vec{E}| A \cos \theta$

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

$=30 \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}$

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

Flux, $\Phi=|\vec{E}| A \cos \theta$

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

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