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Net charge of cube

  • Thread starter irnubcake
  • Start date
5
0
1. Homework Statement

At each point on the surface of a cube the electric field is parallel to the z axis. The length of each edge of the cube is 3.5 m. On the top face of the cube the electric field is in the negative z direction and has a magnitude of 37 N/C magnitude. On the bottom face of the cube the field is in the positive z direction and has a magnitude of 15 N/C. Determine the net charge contained within the cube.

3. The Attempt at a Solution

The electric flux at the top face = EA = (-37 N/C)(3.5m)^2 = -453.25 N.m^2/C
At the bottom face = (15 N/C)(3.5m)^2 = 183.75 N.m^2/C

Gauss's law: flux * permittivity constant = charge
Applied on cube: (net flux)(permittivity constant) = net charge
(-269.5 Nm^2/C)(8.85*10^-12) = -2.385075 * 10^-9 C
= -2385075 uC (micro)
Both value and units are incorrect. > < "
 
39
0
Hello,

Basically, you used the correct physical law.
But some mistake is done when the eletric flux was calculated.
The top surface:[tex]\vec{A}_t=3.5^2\hat{z}(m^2)[/tex] and [tex]\vec{E}_t=-37\hat{z}(N^2/C)[/tex].
The bottom surface:[tex]\vec{A}_b=-3.5^2\hat{z}(m^2)[/tex] and [tex]\vec{E}_b=+15\hat{z}(N^2/C)[/tex]
You may use these quantities and try again.
Also notice the constant in Gauss's law:[tex]\oint\vec{E}\cdot d\vec{A}=\frac{Q_\text{total}}{\epsilon_0}[/tex]


Goodluck
 
Last edited:

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