Gauss's law problem

[purplex76]

An infinitely long solid cylinder radius R1 lies with it's central cylindrical axis lying along the x axis. it is made of a non-conducting material. It has a volume charge density that varies with readius as follows... p(r)=A.r (C/m^3)
where A is a constant. Consider a cylindrical Gaussian surface of length L, radius r, concentric with the x axis.

1) Derive a formula for the amount of charge enclosed by this Gaussian surface for r is greater than or equal to R1, and for r is less than or equal to R1

2) Use gauss's Law to find an expression for the electric field as a function of r in these two regions

3) graph the magnitude of the electric filed for these two regions.


i would appreciate any help with this question because it is really stumping me.....Thanks!

[Tide]

What exactly have you tried so far and what is your thinking?

[Theelectricchild]

Oh wait one more thing, please post this type of question under "college level help" thank you.

[ukamle]

Solution to the Gauss law problem:

Volume charge density= Ar
a -> radius of cylinder
For r > a

Let the radius of cylindrical Gaussian surface be r

E . 2. pi. r. l = integral {( 2*pi.A.l.r. dr / e0 ), 0 , a}

[integral { (), ,} denotes-- () - integral funciton then the limits]

e0 -> permittivity of free space

[ E = A. (a^2)/ (2*r) ] ............... Solution

For r<=a

E . 2*pi.r.l = integral { (2*pi.A.l.r dr/ e0), 0, r}


[ E=A.r/2 ] .............. Solution

[purplex76]

that doesn't make sense to me. Isn't the problem more complex than 2 integrals, because i got no credit for the integrals i put down, being somewhat similar to the ones you replied with.

[ukamle]

:frown: \mbox{i m sorry i forgot to divide by } \epsilon_0\mbox{. Divide the solutions by} \epsilon_0 \mbox{. I feel, that is the correct solution.}

[ukamle]

You may also use the differential form of Gauss law for cylindrically radial field. It goes something like this:
\frac{d(E.r)}{dr} = \frac{\rho r}{\epsilon_0}

Make \rho as a function or r and integrate over proper limits.