#4 The electric field El at one face of a parallelepiped ls uniform over the ent
ID: 1394325 • Letter: #
Question
#4 The electric field El at one face of a parallelepiped ls uniform over the entire face and is directed out of the face. At the opposite face, the electric field E s also uniform over the entire face. The two faces in question are inclined at 30.0° from the horizontal, while and are both horizonta has a magnitude of 9.75 × 104 N/C. 1 has a magnitude of 2.50 × 10' N/07, and (a.) Assuming that no other electric field lines cross the surface of the parallelepiped, determine the net charge contained within (b.) 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? 6.00 cm 5.00 cm 30°Explanation / Answer
A = 0.05^2 = 0.0025 m^2
flux through the 1 st surface = flux 1 = E1*A*cos60 = 2.5*10^4*0.0025*cos60 = 31.25 wb
flux through the 2nd surface = flux 2 = E2*A*cos120 =
9.75*10^4*0.0025*cos120 = -121.875 wb
total flux = -90.625 Wb
from gauss law total flux = Qinside/eo
Qin / eo = 90.625
Qin = 90.625*8.84*10^-12 = 8.01*10^-10 C
(b)
the electric field is produced only by the charges within the parallelopiped
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