The box-like Gaussian surface of Fig. 1 encloses a net charge of +24.0ε0 C and lies in an electric field given by N/C, with x and z in meters and b a constant. The bottom face is in the xz plane; the top face is in the horizontal plane passing through y2 = 1.00 m. For x1 = 1.00 m, x2 = 4.00 m, z1 = 1.00 m, and z2 = 3.00 m, what is b?
Fundamentals of Electricity - jojo, Monday, November 30, 2009 at 7:32am
The electric field 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 is also uniform over the entire face and is directed into that face (see Fig. 3). The two faces in question are inclined at 300 from the horizontal, while and are both horizontal, and have respective magnitudes of 2.50∙104 N/C and 7.00 ∙104 N/C. Assuming that no other electric field lines cross the surface of the parallelepiped, determine the net charge contained within.
Fundamentals of Electricity - drwls, Monday, November 30, 2009 at 8:41am
We can't see your figures so I haven't a clue what 'b' is.
I also don't know that the å in
+24.0å0 is supposed to represent.
Obviously this is a Gauss' Law exercise. The flux of E out of the parallelipiped is proportional to the charge inside. There will be a å (permittivity) term in the equation, which is probably not the same meaning as yours.
Fundamentals of Electricity - Anonymous, Saturday, March 12, 2016 at 12:00am
Fundamentals of Electricity - Anonymous, Saturday, March 12, 2016 at 12:02am