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Learning Introductory Physics with Activities

Paul J. Emigh, Rebecka Tumblin, Kathryn Hadley, Danielle Skinner

Section 22.6 Practice - Symmetry

Calculation 22.6.1. Flux Through a Rectangle.

A rectangular surface measures 2.0 m x 4.0 m and lies in the xy-plane. Find the electric flux through the surface if the E-field is given by: \(\vec{E} = (2.0 \hat{x} - 3.0 \hat{z}) \frac{\mathrm{N}}{\mathrm{C}}\)
Answer.
\(\Phi_E = -24 \frac{\mathrm{Nm}^2}{\mathrm{C}}\)

Calculation 22.6.2. Greatest Total Flux.

Figure 22.6.1. Charges in Gaussian surfaces.
Which rectangular Gaussian Surface in the figure above has the greatest total flux?
Answer.
A.

Calculation 22.6.3. Gauss’s Law for a Dipole.

Would Gauss’s law be helpful for determining the electric field of a dipole? Why?
Answer.
No. Gauss’s Law tells us that the net electric flux through a surface is equal to the charge enclosed (divided by \(\epsilon_o\)). To use Gauss’s Law, you need to take advanage of the symmetry of the charge distribution. If a Gaussian surface is drawn around both charges, the charge enclosed is zero, so the net flux is zero. There is not a useable symmetry for the dipole.

Calculation 22.6.4. Different Gaussian Surfaces.

Two concentric spherical surfaces enclose a point charge q. The radius of the outer sphere is twice that of the inner one. Compare the electric fluxes crossing the two surfaces.
Answer.
Since the electric field has a \(1/r^2\) dependence, the fluxes are the same since \(A = 4\pi r^2\text{.}\)

Calculation 22.6.5. Net Electric Flux.

A point charge of 10 \(\mu\)C is at an unspecified location inside a cube of side 2 cm. Find the net electric flux though the surfaces of the cube.
Answer.
\(\Phi_E = 1.13 \times 10^{6} \ \frac{\mathrm{Nm}^2}{\mathrm{C}}\)

References References

[1]
Practice activities provided by Kathy Hadley: http://khadley.com/Courses/Physics/ph_213/topics/gauss/.
[2]
Practice activities provided by Ling, S. J., Moebs, W., & Sanny, J. (2016). Electric Potential, Gauss’s Law. In University Physics Volume 2. OpenStax.
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