At each point on the surface of the cube shown in the figure the electric field is parallel to the z axis. The length of each edge of the cube is 2.4 m. On the top face of the cube the electric field E-27k N/C and on the bottom face it is E = +23k N/C Determine the net charge contained within the cube.

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At each point on the surface of the cube shown in the figure the electric field is parallel to the z axis. The length of each edge of the
É
cube is 2.4 m. On the top face of the cube the electric field - - 27k N/C and on the bottom face it is E
+23k N/C
Determine the net charge contained within the cube.
Number
i
T
Units
Transcribed Image Text:At each point on the surface of the cube shown in the figure the electric field is parallel to the z axis. The length of each edge of the É cube is 2.4 m. On the top face of the cube the electric field - - 27k N/C and on the bottom face it is E +23k N/C Determine the net charge contained within the cube. Number i T Units
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Step 1

Given:

Edge of the cube = 2.4 m

The electric field at the top face, E  = -27 k^ N/C

The electric field at the bottom face, E = +23 k^ N/C

Solution:

The total charge enclosed by the cube is determined by the total flux acting through the cube. There is no electric field that acts on the sides of the cube other than the top and bottom surfaces of the cube.

ϕ1 =EA = -27 × 2.42

ϕ1=-155.52 Nm2/C

Similarly for the bottom surface,

ϕ2= EA = 23(2.4)2 = 132.48 Nm2/C

The net flux will be 

ϕnet=ϕ1-ϕ2=-155.52 -132.48=-288 Nm2/C

 

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