Find the net electric flux through the cube shown in the figure above.

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Consider the following image. (Take q1 = +3.15 nC and q2 = 8.30 nC.)

image
(a) Find the net electric flux through the cube shown in the figure above.
 N · m2/C

(b) Can you use Gauss's law to find the electric field on the surface of this cube?
    

Explain.

 

asked Feb 10, 2012 in Physics by apush ~Expert~ (552 points)
    

1 Answer

+4 votes
 
Best answer

The formula for electric flux is:

electric flux = (charge in) / (permittivity of free space)

where,  permittivity of free space =  8.854*10^-12  C^2/N*m^2

and the charge in the cube is = +3.15 nC = 3.15*10^-9 C

plugging these values into the equation we get:

 

electric flux = (3.15*10^-9 C) / (8.854*10^-12  C^2/N*m^2) = 355.77 N*m^2/C

so:

electric flux = 355.77 N*m^2/C

 

Part B,

No, Gauss's law cannot be used to find the electric field on the surface, it is only used to calculate the electric flux through a closed surface.

answered Feb 18, 2012 by Roar3215 ~Expert~ (1,318 points)
selected Feb 18, 2012 by apush

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