What is the Electric Field at 2.6 mm from the Axis of Coaxial Cylinders?

[tanzerino]

1. Charge of uniform density 76nC/m^3 is distributed throughout a hollow cylindrical region formed by two coaxial cylindrical surfaces of radii 1.5 mm and 3.7 mm. Determine the magnitude of the electric field at a point which is 2.6 mm from the symmetry axis.



I tried denistyxr/3epsilon but id don't think it is the correct equation to use

 

[Mindscrape]

The correct equation to use is gauss's law.

$$\int\int \mathbf{E} \cdot d\mathbf{a}=\int\int\int \rho/\epsilon_0 dV$$

 

[tanzerino]

i don't understand how to do the triple integration however i remember that E=k int(dq/r^2)
E.dA is phi i want E

 

[Mindscrape]

The density is uniform so you don't actually have to worry about triple integration, you just have to get the right volume. Same for the double integration, you just have to get the right Gaussian surface area. In other words

$$EA=\frac{\rho}{\epsilon_0}V$$

Try to work it out and we can help you where you go wrong.

 

[tanzerino]

i tried this rule and EA=densityXv/epsilon

so E=Q/Aepsilon =Q/4pir^2Xepsilon ? doesn't give an answer

 

[tanzerino]

i got it .it will be EA=densityXv/epsilon then EX2*pi*r*h=densityXpi*r^2*h/epsilon
then 2E=density*r/epsilon then E=density*r/2epsilon

 

[SheepSlayer]

how to set it up:
a=gaussian radius
h=height of cylinder
r=inner radius

E(2∏ah)=[ρ∏h(a²-r²)]/ε₀

E=[ρ∏h(a²-r²)]/[(2∏ah)ε₀]

E=[ρ(a²-r²)]/[(2a)ε₀]