Electric flux through a square surface at 35° to the field

[Rave Grrl]

I used the formula (E * Area * cos(theta)) and got the wrong answer (.02529) and I really don't understand why. I need help

The square surface shown measures 4.0 mm on each side. It is immersed in a uniform electric field with magnitude E = 1930 N/C. The field lines make an angle of 35° with a normal to the surface, as shown. Take the normal to be "outward," as though the surface were one face of a box. Calculate the electric flux through the surface.

 

[dextercioby]

It looks correct (probably not the final #).U may have to check the units ($\left(1\mbox{mm})\right)^{2}=10^{-6} \mbox{m}^{2}$...

Daniel.

 

[HallsofIvy]

35 degrees to the normal is 55 degrees to the surface. Did you try that?

 

[dextercioby]

Halls,she needs the scalar product between the electric field vector and the normal to the surface,which make an angle of 35° between them...

Daniel.

 

[Rave Grrl]

I tried 55, and it didn't work.

Is electrical flux supposed to be negative? Thats all I can think of. I don't know if that's right.

 

[dextercioby]

Depends on the sense of the electric field.If it enters the page or not.Your problem is pretty vague in this issue,because you haven't provided the drawing/picture.

Daniel.