Electrodynamics and electrical fields

[Niles]

[SOLVED] Electrodynamics and electrical fields

The question is: A long cylinder carries a charge density that is proportional to the distance from the axis ρ =kr, where k is a constant, r is the distance from the axis. Find electric field inside the cylinder.


My attempt: Ok, first of all we know from Gauss' law that \oint_S {{\bf{E}} \cdot d{\bf{a}}} = \frac{1}{{\varepsilon _0 }}Q.

We also know that Q = \int {\rho dV = } \int {k \cdot s \cdot s \cdot dsd\phi dz = \frac{2}{3}\pi kls^3 }. Here I have used a Gaussian cylinder of length l and radius s.

Now I must find \oint_S {{\bf{E}} \cdot d{\bf{a}}}.

First question: The electric field E is given by {\bf{E}} = \frac{1}{{4\pi \varepsilon _0 }}\frac{q}{{r^2 }}{\bf{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} <br /> \over r} }}. Is the unit vector r in this case the spherical radius-vector?

For this cylinder, I write the parametric and differentiate it with respect to phi and z. Then I find the normal-vector as the cross-products of these two.

Question two: When I find the element da as above, what do I do with E, which is in spherical coordinates?

I hope you can help me.

 

[foxjwill]

Ok, you're making this way to hard for yourself. Assume our cylinder has radius a and height h, and choose our Gaussian surface to be a cylinder of radius r co-axial to the original cylinder. The electric field at any point on the Gaussian surface for a radius r \leq a has constant magnitude and is directed parallel to the surface normal there, so we can simplify the integral to

{\int_S E d{A}}​

and then to

E{\int_S d{A}}.​

Now, since {\int_S d{A}} is just the surface area of the Gaussian surface, which we know to be 2\pi r h, we can substitute that in, and we are left with the very simple equation

E 2\pi r h = \frac{q_{\mbox{enc}}{{\varepsilon _0 }}.​

I will leave it to you to include the density function in that.

 

[Niles]

Ok, you're making this way to hard for yourself. Assume our cylinder has radius a and height h, and choose our Gaussian surface to be a cylinder of radius r co-axial to the original cylinder. The electric field at any point on the Gaussian surface for a radius r \leq a has constant magnitude and is directed parallel to the surface normal there, so we can simplify the integral to

{\int_S E d{A}}​

and then to

E{\int_S d{A}}.​

Now, since {\int_S d{A}} is just the surface area of the Gaussian surface, which we know to be 2\pi r h, we can substitute that in, and we are left with the very simple equation

E 2\pi r h = \frac{Q}{{\varepsilon _0 }}.​

Hi foxjwill.

First, thanks for taking the time to reply. I can see that using the definition of the dot-product simplifies things greatly - but still I would like to know how it can be done "my way". I have just studied vector-calculus most of this week, and I am kinda disappointed in myself that I cannot figure it out.

The thing that annoys me is particularly the r unit-vector in E. Is this in Cartesian or Spherical coordinates?

 

[foxjwill]

The thing that annoys me is particularly the r unit-vector in E. Is this in Cartesian or Spherical coordinates?

Whichever is more convenient for the particular problem, as long as the same point is described. I'd suggest actually using cylindrical coordinates.

 

[Niles]

Ok, so <br /> {\bf{E}} = \frac{1}{{4\pi \varepsilon _0 }}\frac{q}{{r^2 }}\rho}<br />

I get that the normal-vector has the rho-component 3*cos(theta). I integrate this over theta and z, which are respectively 0..2Pi and 0..l - and this gives 0, which is not correct.

Where is my error?

 

[Niles]

I worked it out. d\vec{A} = \vec{n}*dA, and the vectors go out since they are parallel.

Thanks.