Phy 132 - Assignment 3
A. 1. a. Stays the same. It follows from Gauss’s law that E = (constant)q/r2, where r
is the distance from the center. Since r is unchanged, E is unchanged.
b. . Increases. Again, E = (constant)q/r2, where r is the distance from the center.
Since r decreases, E is increases.
2. a. “Uniformly distributed” means (Q/L for the whole wire) = (q/l for any part).
b.
B. 1. a. 1. One of the 8.854 x 10-12 charges is inside surface A. By Gauss’s law, the
flux through the surface is q/ε0 which equals 1.
b. 1. There is one charge inside this surface too.
c. 2. With two of those charges inside surface C, q/ε0 equals 2.
d. 3. Three of the charges are inside this one.

2. Flux: Φ = E cos A, where A is perpendicular to the surface and points outward.
On each side, A = (.2 m)(.2 m) = .04 m2
a.
v & vi. For the front or back, ⃑ points out of the page or into the page. Both of these
directions are perpendicular to ⃑ , so cos 90° = 0, making the flux zero.
b. From Gauss’s law, charge inside = εoΦ = εo(-38.1 + 38.1 + 22.0 + (-22.0) + 0 + 0)
= 0.
C. a.
b.. Φ = 0 on the left side of the box because E = 0 at the slab’s center.
Φ = 0 for top, bottom, front & back. because ⃑ is perpendicular to ⃑ . (cos = 0)
Through right side, Φ = EA, where A is the area of its right side.
So, total flux through the box is EA
c.The charge enclosed in this box is q = V. (From the definition of charge density,
 = q/V.)
The volume of the box is V = Ax.
Putting these things together in Gauss’s law:
D.
a. Since the field points radially outward, the cylinder shown is a good choice for the
Gaussian surface: ⃑ is constant in magnitude and perpendicular to the surface
everywhere.
b. Φ = 0 on the left and right ends because ⃑ is perpendicular to ⃑ . (⃑ · ⃑ = 0
because cos = 0)
Flux through curved part is EA, where A is the area of the curved part. (2πrl)
So, total flux through the surface is E(2πrl ).
E.
a. The Gaussian surface is a cylinder.

b. Φ = 0 on the left and right ends because ⃑ is perpendicular to A . (⃑ · ⃑ = 0
because cos = 0)
Flux through curved part is EA, where A is the area of the curved part. (2πrl)
So, total flux through the surface is E(2πrl ).
c. The charge enclosed is q = V. (From the definition of charge density,  = q/V.)
The volume which contains charge is V = πR2l .
F. 1. a. The flux is tripled since Φ = q/ε0
b. No effect since Φ depends on the enclosed charge only. (E goes down, but there
is a compensating increase in A.)
c. No effect. (Depends on q only.)
d. No effect. Gauss’s law just says Φ depends on how much charge in inside the
surface. It doesn’t matter where inside.
2.