The electric field is measured all over a cubical surface, and the pattern of fi

Question

The electric field is measured all over a cubical surface, and the pattern of field detected is shown in the figure above. On the right side of the cube, the electric field has magnitude E_1 = 250 V/m, and the angle between the electric field and the surface of the cube is theta = 12 degrees. On the bottom of the cube, the electric field has the same magnitude E_1, and the angle between the electric field and the surface of the cube is also theta = 12 degrees. On the top of the cube and the left side of the cube, the electric field is zero. On half of the front and back faces, the electric field has magnitude E_1 and is parallel to the face; on the other half of the front and back faces, the electric field is zero. One edge of the cube is 55 cm long. What is the net electric flux on this cubical surface? Net electric flux = V Amiddot m What is the net charge inside this cubical surface? Q = C

Explanation / Answer

flux throuugh right face phi 1= E1*A*sin12

A = area = L^2 = 0.55^2

phi1 = 250*0.55^2*sin12 = 15.72

flux through bottom face

phi2 = E1*A*sin12 = 250*0.55^2*sin12 = 15.72

net flux = 15.72 + 15.72= 31.44

+++++++++++++++++

from coloumbs law

net flux = Qin/eo

31.44 = Qin/(8.84*10^-12)

Qin = 2.78*10^-10 C

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