The electric field E 1 at one face of a parallelepiped is uniform over the entir

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 (Figure 1) ). The two faces in question are inclined at 30.0 from the horizontal, while E 1 and E 2are both horizontal; E 1 has a magnitude of 2.30×104 N/C , and E 2 has a magnitude of 8.60×104 N/C

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

q=?? C

2)Is the electric field produced only by the charges within the parallelepiped, or is the field also due to charges outside the parallelepiped? How can you tell?

thanks

Explanation / Answer

According to Gauss Law of Electric Flux -
The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity.
E dA =Q/0
E x A=Q/0
E*A*cos(60) = Q/0
Q=E*A cos(60)* 0
Q = (2.3 - 8.6)10^4 * 6 * 5 *(10^-2)^2 * cos(60)*(8.8510^-12)  =4.7104 Q=(4.7104)(6102)(5102)(sin(30°))( Charge Q= - 8.36*10^-10 C


(2) Electric field produced will be only by the charges within the parallelepiped , this is what Gauss law state's.

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