The electric field E_1 at one face of a parallelepiped is uniform over the entire face and is directed out of the face. At the opposite face, the electric field E_2 is also uniform over the entire face and is directed into that face (the figure ). The two faces in question are inclined at 30.0 degrees from the horizontal, while E_1 and E_2 are both horizontal; E_1 has a magnitude of 2.90×10^4 N/C, and E_2 has a magnitude of 8.60×10^4 N/C.

Question

The electric field E_1 at one face of a parallelepiped is uniform over the entire face and is directed out of the face. At the opposite face, the electric field E_2 is also uniform over the entire face and is directed into that face (the figure ). The two faces in question are inclined at 30.0 degrees from the horizontal, while  E_1 and  E_2 are both horizontal;  E_1 has a magnitude of 2.90×10^4 N/C, and  E_2 has a magnitude of 8.60×10^4 N/C.

Assuming that no other electric field lines cross the surfaces of the parallelepiped, determine the net charge contained within.

drwls · Answer

Use Gauss' Law. The sum of the fluxes (E-fields NORMAL TO the surface multiplied by the areas) for all six surfaces is proportional to the charge inside, Q. (Total flux = Q/epsilono)

David · Answer

-6.93*10^-5 N/C