gauss law problem

Question

An infinite line of charge with linear density λ1 = 7.4 μC/m is positioned along the axis of a thick insulating shell of inner radius a = 2.8 cm and outer radius b = 4.2 cm. The insulating shell is uniformly charged with a volume density of ρ = -792 μC/m3.

 

 

Im confused in this one because it has two different charge densities. I need to find the electric field at a<r<b and just set it up without plugging in the values. i need to express my answer in rho a lambda r b and Eo

 

 

i know its EAs = Q_en/E₀ Im thinking it is E (4pir²)= [-rho*l(pir^2-pia^2)]+lambda*l]/E_o 

im not sure if its right and i dont think so cause i dont know l 

Answer 1

Start with Gauss' law.  For the Gaussian surface, use a hollow cylinder of radius r (a < r < b) centered at the same axis.  (It looks like you tried to use a sphere for the Gaussian surface, which will not work here.  You need to use a cylindrical surface.)

 

EA = Q/ε₀

 

A = surface area of the Gaussian surface cylinder without the end surfaces (radial electric field)

 

Let L = length of the cylinder.  It is actually infinity, but you will find that L drops out of the math later.

 

A = 2πrL

 

Q = (the charge on the line) + (the charge enclosed inside the shell)

 

Q = λL + ρV = λL + ρ[π(r² - a²)L],

 

where V = π(r² - a²)L is the volume of the shell with charge in its interior over a radius range from a to r.

 

Put this all together in Gauss' law and get

 

E(r)(2πrL) = (1/ε₀){λL + ρ[π(r² - a²)L]}

 

Note that L cancels out, so we do not have to worry about an infinity.  Then

 

E(r) = [1/(2πε₀r)] [λ + ρπ(r² - a²)]

 

Be sure to carefully check my math.