- #1
Fernandopozasaura
- 13
- 3
Homework Statement
It is known that the potencial is given as V = 80 ρ0.6 volts. Assuming free space conditions, find a) E, b) the volume charge density at ρ=0.5 m and c) the total charge lying withing the closed surface ρ=0.6, 0<z<1
Homework Equations
E[/B]=-∇V
The Attempt at a Solution
(this is author's solution)
a) E field: E=-∇V=-∂V/∂ρ aρ = -48ρ-0.4
b)ρv=-28.8 ε0 ρ-1.4 = -673 pC/m3
So, we get an expression for ρv that depends on ρ.
c) the total charge lying within the closed surface ρ = 0.6, 0 < z < 1. The author says "the easiest way to do this calculation is to evaluate Dρ at ρ = .6 (noting that it is constant), and then multiply by the cylinder area."
Using part a, we have Dρ for ρ=0.6
= −480(.6)−.4 = −521 pC/m2. Thus
Q = −2π(.6)(1)521 × 10−12C = −1.96 nC.
And here is where I need some help becouse I understand that the author is saying that the charge is located only in cylinder's surface and so, he uses the *surface* of cylinder to get the total charge.
I worked out the solution thinking that the charge is distributed in the whole volume, not only in the surface. I thought this because the expression that gives us charge density is function or ρ and because it is expressed in C/m3 (charge per unit of volume).
I've done a volume integration from 0 < Φ < 2π, 0 < z < 1 and 0 < ρ < 0.6 meters.
There's nothing in exercise that make you think that the cylinder is conductive in which case the charge would be on surface.
I guess I'm wrong, but why?
Thanks for your answers.
Fernando
Author's solution says that