Sorry about that, I'm new to posting so I'm unsure how much I should post of a solution.
Ok so separation of variables (in cartesian coordinates):
\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = 0
where V(x,y) = X(x)Y(y).
Substituting...
Well I took a stab at it, and I think I have an idea where the solution should go. If you the standard Laplace's Equation: \nabla^2V=0. Where in this case we are dealing with cartesian coordinates.
I'm hoping you know the general solutions are going to be sines and cosines for the y-direction...
So attached is my solution. I am off by a minus sign, but overall everything else works out nicely. The integral again isn't all that pretty but does simplify nicely. And if anyone does find the negative sign mess up (should be with applying the arcsin or somewhere in there) please let me know.
Also, a nice site I found for a hemisphere line charge if you want more of a visualization of what is going on:
http://www.phys.uri.edu/~gerhard/PHY204/tsl329.pdf
First off you want to treat each case separately. By symmetry of the problem we know that the E-field will only be in the y-direction. Since the charge is distributed uniformly along the arc, you can deal with it as a line charge density which is what you did:
dQ_1=\rho_l(\rho d\phi) now...
Well since this is a charge Q distributed over the entire ring, it cannot be thought of as a point charge, unless you're looking at the far field. If z>>a,b the potential will due to the disk charge Q will behave like a point charge. So in this case, it is easier to find the potential, V (and we...
So you're close. In this case we're integrating over r so instead of R = (r^2+h^2)^{1/2} you'll use R = (r^2+z^2)^{1/2}. This is because you're not solving for z = h yet. So when it's written like this, it is in cylindrical coordinates (a function of r and z). So yes, use a triangle with r being...
That is true you don't know what \rho_s or Q_{enc} is but you do know that there is going to be a charge distribution on the ring. You know that it will have a total charge Q enclosed. If you recall from your solution, it is indeed in terms of Q which is the charge on the ring.
So you want to...
Are you trying to solve for the capacitance or the E-field? I'm assuming the capacitance since that was one of your relevant equations and typically for a parallel plate capacitor, that's what you're looking for.
A few equations to get you started:
E_n*\epsilon(x) = \rho_s where E_n is the...
The figure is that pie shaped figure in the beginning that I drew on the attachment. That zoomed in circle is also similar to what they show as well with the dr strip and radius, r.
Using Ohm's Law is how you relate the potential difference between the flat ends and the current flowing through...
First you need to calculate the charge using the charge density. We know that:
Q = \int_{a}^{b}\int_{0}^{2\pi}\rho_lrd\phi
From there you should find the potential of a disk along the z-axis. The integral should be set up as:
d\phi = \frac{\rho_l*2*\pi*rdr}{4*pi*\epsilon_o\sqrt{z^2+r^2}}.
If...
You are correct that the first E-field (over surface s1) is E = \frac{\rho*R_1}{2*\epsilon_o} and E would be in the radial direction.
The reason the E-field is not zero is due to the charge density being distributed over the entire volume of the inner wire.
The second surface would enclose all...
Indeed it was a difficult problem... I've posted another one that's similar that involves resistance rather than inductance. Thanks again for all your help :)
Homework Statement
Consider a segment of a toroidal (doughnut-shaped) resistor with a horizontal cross-section (see attachment for the figure). Show that the resistance between the flat ends having a circular cross-section is given by
R = \frac{\phi_o}{σπ(√b-√a)^2}
Homework Equations...
So since the integral is essentially the ∫(√a^2-x^2)/x dx you have to use the integral of the form X = a + bx + cx^2 which in this case a = a^2 and c = -1... so the integral table for this is ∫√X/x dx = √X + b/2∫dx/√X + a∫dx/x√X...it's not a very nice integral but if you follow through it will...