At each point on the surface of the cube


At each point on the surface of the cube shown, the electric field is parallel to the z axis. The length of each edge of the cube is 3.0 m. On the top face of the cube the field is \vec E=-34 \vec k N/C and on the bottom face it is \vec E=+20 \vec k N/C. Determine the net charge contained within the cube.

At each point on the surface of the cube

Image from: J. Walker, D. Halliday, and R. Resnick, Fundamentals of physics: [extended], 10th ed. United States: Wiley, John & Sons, 2013.

Solution:

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We can figure out the flux through the top surface and the bottom surface of the cube. The flux through the top surface is:

\Phi=(E)(A)

(Where \Phi is flux, E is electric field, and A is the area of the face)

\Phi=(-34)(3^2)=-306 \dfrac{N\cdot m^2}{C}

The flux through the bottom surface is:

\Phi=(E)(A)

\Phi=(20)(3^2)=-180 \dfrac{N\cdot m^2}{C}

(Our answer is negative because we are looking at the bottom face, which has a negative -k component since our area vector is facing down)

 
The net flux can be calculated by adding up the bottom and top face. Note that the other faces do not have a flux going through their surfaces since the electric field is parallel to the z-axis.

Total flux = -306 + (-180) = -486 \dfrac{N\cdot m^2}{C}

 

We can now use the following equation to figure out the charge enclosed.

\Phi=\dfrac{q}{\epsilon_0}

(Where \Phi is the net flux, \epsilon_0 is permittivity of free space and q is the charge enclosed)

q=(\Phi)(\epsilon_0)

q=(-486)(8.85\times10^{-12})

q=-4.3\times10^{-9} C

 

Final Answer:

q=-4.3\times10^{-9} C

 

This question can be found in Fundamentals of Physics, 10th edition, chapter 23, question 6.

 

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