- #1

Fernandopozasaura

- 13

- 3

## Homework Statement

It is known that the potencial is given as V = 80 ρ

^{0.6}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 a

_{ρ}= -48ρ

^{-0.4}

b)ρ

_{v}=-28.8 ε

_{0}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