Given a uniform electric field $\to E = 5 \times 10^{3}^i N / C$ , find the flux of this field through a square of $10 c m$ on a side whose plane is parallel to the $Y - Z$ plane. What would be the flux through the same square if the plane makes a $30^{\circ}$ angle with the $X-axis?

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Solution

$\to E = 4 \times 10^{3}^i (\frac{N}{C}) F l u x = \to E . \to S = E S c o s θ$ S is the area through which electric field lines crosses
When plane is parallel to Y-Z plane then $θ = 0$ here S = a X a = a^2 a= 5cm
Flux = $E S = 4 \times 10^{3} \times {(5 \times 10^{- 2})}^{2} = 4 \times 10^{3} \times 25 \times 10^{- 4} = 10$
when plane makes angle 30 degree with X axis then perpendicular to plane makes angle 60 degree to X axis
hence
$θ = 60^{0} F l u x = E S = 4 \times 10^{3} \times {(5 \times 10^{- 2})}^{2} \times c o s 60^{0}$
= $4 \times 10^{3} \times 25 \times 10^{- 4} \times \frac{1}{2} = 5$

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