# Q 1.15(b) Consider a uniform electric field $E=3 \times 10^{3}\ \widehat{i}\ N/C$ .What is the flux through the same square if the normal to its plane makes a $60^{\circ}$ angle with the x-axis?

Now, Since the normal of the square plane makes a $60^{\circ}$ angle with the x-axis

$cos\Theta = cos(60^{0}) = 0.5$

Therefore, flux through this surface:

$\phi = E.A = EAcos\Theta$

$\implies \phi = (3\times10^3)(0.01)(0.5) Nm^2/C = \boldsymbol{15\ Nm^2/C}$

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