A rectangular surface of sides 10 cm and 15cm is placed inside a uniform electric field of $25 V m^{- 1}$ , such that normal to the surface makes an angle of 60° with the distance of electric field. Find the flux of electric field through the rectangular surface $(i n N m^{2} C^{- 1})$

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answer is 0.19.

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Detailed Solution

At the surface, $\vec{E} = 25 V / m$

$\vec{A} = 10 \times 10^{- 2} \times 15 \times 10^{- 2} = 1.5 \times 10^{- 2} m^{2}$

Now we know that $\int ϕ = \int Eda = E \cdot A = EAcos θ$

$ϕ = E A \cos θ = 25 \times 1.5 \times 10^{- 2} \times \cos 60^{\circ} = 0.1875 {Nm}^{2} / C ≅ 0.19 {Nm}^{2} / C$

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