1. Chapter 22--Examples

2. Problem
A charge, +q, is surrounded by a thin, spherical shell of radius, a, which has a charge density of –s on its surface. This shell is, in turn, surrounded by another thin shell of radius, b, which has a surface charge of +s. Find the electric fields in
Region 1: r<a
Region 2: a<= r <=b
Region 3: r> b +q a b -s +s

3. Step 1: Pick your shape
I choose spherical! So

4. Region 1: r<a +q a b -s +s
qenclosed=q

5. Region 2: a<=r<=b+q a b -s +s
qenclosed=q+(4pa2)*(-s)

6. Region 3: r>b +q a b -s +s
qenclosed=q+(4pa2)*(-s)+ (4pb2)*(s)

7. Problem
An electric filed given by E=4i-3(y2+2)j pierces the Gaussian cube shown below. (E is in newtons/coulomb and y is meters). What net charge is enclosed by the Gaussian cube?
X=1.0 m
X=3.0 m x y z

8. First, let’s get a sense of direction4 y
Planes
y-z: Normal to +x
x-z: Normal to –y
y-z: Normal to –x
x-z: Normal to +y
x-y: Normal to +z
x-y: Normal to -z 6 1 3 x 2 z 5
X=3.0 m
X=1.0 m

9. Need to find qenclosed

10. Integrating each side (start with surface 1)
Region 3, in which the normal vector points in the opposite direction, will have a value of -16

11. The rest of the sides
Since E is perpendicular to sides 5 & 6, the result is zero.

12. Problem
The figure below shows a cross-section of two thin concentric cylinders with radii of a and b where b>a. The cylinders equal and opposite charges per unit length of l.
Prove that E = 0 for r>a
Prove that E=0 for r>b
Prove that, for a<r<b, a b -l l

13. First, I choose a shape
I choose cylindrical! So

14. For r<a
qenclosed =0 a b -l l

15. For a<r<b
qenclosed =lL a b -l l

16. For r>b qenclosed =lL-lL=0 This is the principle of a coaxial cable

17. Problem
A very long, solid insulating cylinder with radius R has a cylindrical hole with radius, a, bored along its entire length. The axis of the hole is a distance b from the axis of the cylinder, where a<b<R. The solid material of the cylinder has a uniform charge density, p.
Find the magnitude and direction of the electric field inside the hole and show that E is uniform over the entire hole. R b a

18. First, let’s do a solid cylinder of radius, R

19. Now what if we have an off-axis cylinder
We learned in Phys 250, that we can “translate” coordinates by r’=r-b
Where
b is the direction and distance of the center of the off-axis cylinder
r is a vector from the origin b r’ r

20. Ehole= Esolid cylinder-Eoff-axis hole
All of these are constants and do not depend on r.