1. Physics 2112 Unit 4: Gauss’ Law
Today’s Concepts:
A) Conductors
B) Using Gauss’ Law

2. Use #1: Determining E without calculus
E from infinite line of charge.
We did this before with calculus. Remember?
Let’s do it again using Guass Law

3. Use #2: Determine Charge on Surfaces
E = 0 inside any conductor at equilibrium
(If E ≠0, then charge feels force and moves!)
Excess charge on conductor only on surface at equilibrium
E = 0
Why?
Apply Gauss’ Law
Careful here… inside means part of conductor, does NOT include cavity of a conductor!
Take Gaussian surface to be just inside conductor surface

4. Gauss’ Law + Conductors + Induced Charges
How Does This Work?
Charges in conductor move to surfaces to make Qenclosed = 0.
We say charge is induced on the surfaces of conductors
ALWAYS TRUE!

5. Example 4.1
An infinite line of charge with linear density λ1 = -5.1μC/m is positioned along the axis of a neutral conducting shell of inner radius a = 2.7 cm and outer radius b = 5.0 cm and infinite length.
A) What is λb, the linear charge density on the outer surface of the conducting shell ?
B) What is Ex(R), the electric field at point R, located a distance dR = 0.9 cm from the origin and making an angle of 30o with respect to the y-axis as shown?

6. Example 4.2
An infinite line of charge with linear density λ1 = -5.1μC/m is positioned along the axis of a thick conducting shell of inner radius a = 2.7 cm and outer radius b = 5.0 cm and infinite length. The conducting shell is uniformly charged with a linear charge density λ 2 = +2.0 μC/m.
x R
C) What is Ex(R), the electric field at point R, located a distance dR = 7.0 cm from the origin and making an angle of 30o with respect to the y-axis as shown?

7. Gauss’ Law
ALWAYS TRUE!
In cases with symmetry can pull E outside and get
In General, integral to calculate flux is difficult…. and not useful!
To use Gauss’ Law to calculate E, need to choose surface carefully!
1) Want E to be constant and equal to value at location of interest OR
2) Want E dot A = 0 so doesn’t add to integral

8. CheckPoint: Charged Conducting Sphere & Shell 1
A positively charged solid conducting sphere is contained within a negatively charged conducting spherical shell as shown. The magnitude of the total charge on each sphere is the same. Which of the following statements best describes the electric field in the region between the spheres?
The field points radially outward
The field points radially inward
The field is zero

9. Gauss’ Law Symmetries ALWAYS TRUE!
In cases with symmetry can pull E outside and get
Spherical
Cylindrical
Planar

10. CheckPoint: Gaussian Surface Choice
You are told to use Gauss' Law to calculate the electric field at a distance R away from a charged cube of dimension a. Which of the following Gaussian surfaces is best suited for this purpose?
a sphere of radius R+1/2a
a cube of dimension R+1/2a
a cylinder with cross sectional radius of R+1/2a and arbitrary length
This field cannot be calculated using Gauss' law
None of the above

11. CheckPoint: Charged Sphericlal Shell
A charged spherical insulating shell has inner radius a and outer radius b. The charge density on the shell is ρ. What is the magnitude of the E-field at a distance r away from the center of the shell where r < a?
ρ/εo
zero
ρ(b3-a3)/(3εor2)
none of the above

12. Example 4.3
A charged spherical insulating shell has inner radius a and outer radius b. The charge density on the shell is ρ.
What is the magnitude of the E-field at a distance r away from the center of the shell where a< r < b?

13. Quick Review: Infinite Sheet of Charge

14. CheckPoint: Infinite Sheets of Charge
In both cases shown below, the colored lines represent positive (blue) and negative (red) charged planes. The magnitudes of the charge per unit area on each plane is the same. In which case is the magnitude of the electric field at point P bigger?
Case A
Case B
They are the same

15. Example 4.4 y
Point charge +3Q at center of neutral conducting shell of inner radius r1 and outer radius r2.
What is E for the three regions of:
r < r1
r1< r < r2
r2 < r ? r2
neutral
conductor
+3Q r1 x

16. Calculation y
Point charge +3Q at center of neutral conducting shell of inner radius r1 and outer radius r2.
What is E everywhere? r2
neutral
conductor
+3Q r1 x
r < r1
r > r2
r1 < r < r2