Question

The square surface shown in the figure measures 3.6 mm on each side. It is immersed in a uniform electric field with magnitude E = 1400 N/C and with field lines at an angle of 35° with a normal to the surface, as shown. Take that normal to be "outward," as though the surface were one face of a box. Calculate the electric flux through the surface.

Answer 1

I got 0.0149, is that right?

Answer 2

@ganeshie8

Answer 3

@Elsa213

Answer 4

the requested flux, is given by the subsequent computation:
\[\Large \begin{gathered}
\Phi \left( {\mathbf{E}} \right) = ES\cos 35 = \hfill \\
\hfill \\
= 1400 \cdot {\left( {3.6 \cdot {{10}^{ - 3}}} \right)^2}\cos 35 \cong 0.0149\frac{{N{m^2}}}{C} \hfill \\
\end{gathered} \]
so your answer is correct!

Answer 5

please wait

Answer 6

we have to make this angle: \[\theta = 180 - 35 = 145\], since we have this drawing:

Answer 7

so the right result is:
\[\Large \begin{gathered}
\Phi \left( {\mathbf{E}} \right) = ES\cos 145 = \hfill \\
= 1400 \cdot {\left( {3.6 \cdot {{10}^{ - 3}}} \right)^2}\cos 145 \cong - 0.0149\frac{{N{m^2}}}{C} \hfill \\
\end{gathered} \]

Answer 8

I understand, thank you very much!

Answer 9

sorry for my error above!!
:)

Answer 10

Np

Answer 11

thanks! :)