Net charge of cube

irnubcake

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. > < "

Related Introductory Physics Homework Help News on Phys.org

variation

Hello,

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

Goodluck

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