Solving Gauss's Law Problem for a Solid Cylinder

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The discussion centers on applying Gauss's Law to determine the electric field around an infinitely long solid cylinder with a non-uniform volume charge density, p(r) = A.r. For r greater than or equal to the cylinder's radius R1, the derived electric field is E = A(R1^2)/(2*r*ε0), while for r less than or equal to R1, it is E = A*r/(2*ε0). Participants express confusion over the integration process and the complexity of the problem, suggesting that the solutions should be divided by ε0 for accuracy. Additionally, an alternative method using the differential form of Gauss's Law is proposed for further clarity. The conversation highlights the challenges in solving this electrostatics problem and the importance of proper integration limits.
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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!
 
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What exactly have you tried so far and what is your thinking?
 
Oh wait one more thing, please post this type of question under "college level help" thank you.
 
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
 
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.
 
gauss law problem

: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.}
 
another method

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.
 
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