Practice Problems: Gauss's Law Solutions

1. (easy) A student measures the electric flux through a closed spherical surface of volume V to be X. She then removes the charge from inside the spherical surface and places it in a closed cylindrical surface of volume V/2. She then claims that the flux through the cylindrical surface is 2X. Explain why the student is wrong.
According to Gauss's Law, the magnitude of the electric flux is independent of the shape or size of the Gaussian surface. Thus, the flux through both the spherical surface and the cylindrical surface must be the same.

2. (easy) A pyramid with a 6.0 m square base has a height of 4.0 m. If it was placed in a vertical E-field (uniform magnitude of 52.0 N/C), determine the flux through one of the four sides.
Ftotal = 0 = Fbottom + Fsides
0 = -EAbottom + 4(Fone side)
EAbottom = 4(Fone side)
52.0(6)2 = 4(Fone side)
(Fone side) = 468 Nm2/C

Alternative Solution:

3. (easy) A circular plane, with a radius of 2.2 m, is immersed in an E-Field with a magnitude of 800 N/C. The field makes an angle of 20° with the plane. What is the magnitude of the flux through the plane?
|Φ| = EAcosθ
|Φ| = 800(3.14)(2.2)²cos70 = 4.2x10³ Nm²/C

4. (easy) The net electric flux through an 3D closed surface is positive 2.2x10³ Nm²/C. How much charge must be inside the surface? What is the sign on the charge?
Φ = q_in/ε_o
2.2x10³ = q_in/8.85x10^-12
q_in = 1.9x10^-8 C (it is positive)

5. (moderate) An imaginary rectangular shaped box is placed on the x-axis of a coordinate system. The left edge of the box is a distance of 0.4 m to the right of the origin. The box has a depth of 0.4 m, a width of 0.6 m, and a height of 0.4 m. A non uniform E-field (measured in N/C) passes through the box and varies along the x-axis according to the following:

E = (2.0x2 + 3.0) directed perfectly along the x-axis.
Determine the net flux through the surface.

Ftotal = Fleft + Fright

Ftotal = -EleftA + ErightA
Ftotal = -(2(0.4)2+ 3.0)(0.4)(0.4) + (2(1.0)2+3.0)(0.4)(0.4)
Ftotal = 0.27 Nm2/C 

6.(moderate) A Gaussian cube, 0.5 m along each edge, sits on the axis of a coordinate system such that three of its edges are along the postive x, y, and z axes. Determine the net electric flux through the top face of the cube if there is a uniform E-field of -0.5i + 0.3j  acting in that region of space.
Because the flux is calculated using a dot product, the only component of the E-Field that gives any flux through the top face is the y-component.
Φ = EAcosθ = 0.3(0.5)²cos0= 0.075 Nm²/C

7. (moderate) The concept of flux can also be applied to gravitational fields. Gauss's Law for gravity is:

(1/4πG)Φ_g = -m
where m is the mass contained within a Gaussian surface and the gravitational flux is
Φ_g = ∫g·dA

g is the gravitational field through the surface. Show that this law is equivalent to Newton's Law of Universal Gravitation.
Assume a point mass surrounded by a spherical Gaussian surface of radius R. The gravitational field lines point inward toward the point mass, creating a negative flux.
Φ_g = ∫g·dA = -4πGm
g is constant at all points on the Gaussian surface.
gAcos180 = -4πGm
-g(4πR²) = -4πGm
g = Gm/R²
This is the gravitational field equation used in Newton's Law.