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Consider a uniform electric field $E =3 \times 10^{3} i\; N / C .$
$(a)$ What is the flux of this field through a square of $10 \;cm$ 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, $E =3 \times 10^{3} \hat{ i } \,N / C$
Magnitude of electric field intensity, $| E |=3 \times 10^{3} \,N / C$
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, $\theta=0^{\circ}$
Flux ( $\phi$ ) through the plane is given by the relation, $\phi=| E | A \cos \theta=3 \times 10^{3} \times 0.01 \times \cos 0^{\circ}=30 \,N\, m ^{2} / C$
$(b)$ Plane makes an angle of $60^{\circ}$ with the $x$ – axis. Hence, $\theta=60^{\circ}$ Flux, $\phi=| E | A \cos \theta$$=3 \times 10^{3} \times 0.01 \times \cos 60^{\circ}$
$=30 \times \frac{1}{2}=15\; N \,m ^{2} / C$
