MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics: 8.02

Problem Solving 2: Gauss’s Law Section ________________ Table Number ________________

Group Names ____________________________________

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REFERENCE: Reading Course Notes: Sections 3.6-3.7, 3.10. Introduction: When attempting Gauss’s Law problems, we described a problem solving strategy summarized below (see also, Section 3.6, 8.02 Course Notes):

Summary: Methodology for Applying Gauss’s Law

E ⋅ dA

closedsurface S

∫∫ =qin

ε0

Step 1: Identify the ‘symmetry’ properties of the charge distribution. Step 2: Determine the direction of the electric field Step 3: Decide how many different regions of space the charge distribution determines For each region of space… Step 4: Choose a Gaussian surface through each part of which the electric flux is either

constant or zero Step 5: Calculate the flux through the Gaussian surface (in terms of the unknown E) Step 6: Calculate the charge enclosed in the choice of the Gaussian surface Step 7: Equate the two sides of Gauss’s Law in order to find an expression for the

magnitude of the electric field Then… Step 8: Graph the magnitude of the electric field as a function of the parameter specifying

the Gaussian surface for all regions of space. You should now apply this strategy to the following problem.

Problem 1: Concentric Cylinders

A long very thin non-conducting cylindrical shell of radius b and length L surrounds a long solid non-conducting cylinder of radius a and length L with b > a . The inner cylinder has a uniform charge +Q distributed throughout its volume. On the outer cylinder we place an equal and opposite to charge, −Q . The region a < r < b is empty. Question 1 (this is Step 1 of your methodology above): What is the ‘symmetry’ property of the charge distribution here (which of the three below)? Spherical Cylindrical Planar Question 2 (Step 2 of your methodology): What is the direction of the electric field (again, which of the three choices below)? Radial (in/out) Angular (CW/CCW) Perpendicular to page Question 3 (Step 3 of the methodology): Define the different regions of space that you will need to find expressions for the electric field. What determines those different regions? (How many different formulae for

E are you going to have to calculate?)

Answer:

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics: 8.02

Question 4 (Step 4 of your methodology): For each region of space, make a figure showing clearly your choice of a Gaussian surface. What variable did you choose to parameterize your Gaussian surface (for example, for a sphere you’d use the radius r )? What is the range of that variable? Answer:

Question 5 (Step 5 of your methodology): In the region r < a , calculate the flux

E ⋅ dA

closedsurface S

∫∫

through your choice of the Gaussian surface (that is, just write down the left hand side of Gauss’s Law). Your expression should include the unknown electric field for that region. Answer: Question 6 In the region r < a , what is the charge enclosed in your choice of Gaussian surface? (this should be in terms of Q , r and a , not

E ).

Answer: Question 7 (Step 7 of your methodology): In the region r < a , equate the two sides of Gauss’s Law that you calculated in questions 6 and 7, and solve to find an expression for the magnitude of the electric field as a function of r . Answer:

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics: 8.02

Question 8 (Step 6 and 7 or your methodology): Repeat the same procedure in order to calculate the electric field as a function of r in the region a < r < b . Answer: Question 9 (Step 6 and 7 or your methodology): What is the electric field in the region r > b . Answer: Question 10 (Step 8 of your methodology): Make a graph of the magnitude of the electric field as a function of r for all regions of space.

Problem 2 A semi-infinite slab of charge with charge density ρ extends from x = −d to x = d . Find a vector expression for the electric field everywhere i.e. in the regions (i) x < −d , (ii) −d < x < d , (iii) x > d .

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics: 8.02

Problem 3: A non-conducting sphere is constructed of two materials. The inner portion, radius a , has a non-uniform volume charge density given by:

ρ1(r) = α

r; r < a

where α > 0 is a constant. The outer portion, with inner radius a and outer radius b has a uniform charge density.

(a) If the electric field outside the sphere r > b is everywhere zero, what is the uniform charge density ρ2 of the outer portion of the sphere, a < r < b ?

(b) Find a vector expression for the electric field everywhere in space, i.e. in the regions (i)

r < a , (ii) a < r < b , and (iii) r > b .

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics: 8.02