Recent content by themadhatter1

Hmm interesting. I've got it now. The electric potential inside the shell is the sum of the electric potential on the outer edge and the inner edge. This yields: V=k\frac{3Q}{4r_1} Ok, but I'm still confused on the next question. I would think the electric potential inside the cavity would...

Never mind, what that tells me is that the electric potential inside the conductor is constant, but how do I derive the value?

Yes, I know that vec(E)=-vec(\nabla V) I also know that E=-\frac{dV}{ds} when ds||E If I use the second equation I set E equal to zero and you have v=0 which is wrong. The first one I'm not even sure how to use since I have magnitudes of vectors not vectors.

Homework Statement A hollow spherical conductor, carrying a net charge +Q, has inner radius r1 and outer r2=2r1 radius (see the figure ). At the center of the sphere is a point charge +Q/2. A.Write the electric field strength E in all three regions as a function of r: for r>r2,r1<r<r2...

thanks, I figured it out

Homework Statement I need to solve this DE system for a lab: q_1'=2-\frac{6}{5}q_1+q_2 q_2'=3+\frac{3}{5}q_1-\frac{3}{2}q_2Homework Equations The Attempt at a Solution I know how to use the method of elimination to solve such systems, but this is non homogeneous because of the added constant...

you can rewrite the exponents as (x-y)(x+y). Using the substitution u=x-y and y=x+y you have a parallelogram with bounds v=3u,v=3u-8,v=-2u+1,v=-2u+8. Therefore, the area can be represented by the following integrals...

Homework Statement Evaulate the integral making an appropriate change of variables. \int\int_R(x+y)e^{x^2-y^2}dA where R is the parallelogram enclosed by the lines x-2y=0, x-2y=4, 3x-y=1, 3x-y=8 . Homework Equations The Attempt at a Solution I'm not sure what change of variables I should...

Hmm, that's what I thought you were thinking. I handed in the assignment today and got the answer key. They setup the integral the way you did in the answer key. However, I would argue that part of the region in that integral is not under the cone. Everything under the cone is contained within a...

Ok, I think I was misinterperting the volume described. I took it to mean basically the shadow cast down by the cone on the xy-plane, since that region lies under the cone, above the xy-plane, and in the sphere. I'm not sure if the area you are thinking of is correct though. In the area you are...

Oh, haha. Cant believe I forgot that. I would think the upper limit on rho depends on phi because with this shape, you can change theta all you want and rho is not going to change. However, changing phi changes rho, this is because the solid is symmetrical about the z-axis. Right?

Homework Statement Using spherical coordinates, find the volume of the solid that lies within the sphere x2+y2+z2=4, above the xy-plane and below the cone z=√(x2+y2)Homework Equations The Attempt at a Solution This is what I have so far...

update, I proved everything, however I'm not sure If my proof for b is what they're asking for. I said that \int_{-\infty}^{\infty}e^{-x^2}dx\int_{-\infty}^{\infty}e^{-y^2}dy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dy dx and since I had proved that the rhs of the...

Homework Statement (a) we define the improper integral (over the entire plane R2) I=\int\int_{R^2}e^{-(x^2+y^2)}dA=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dy dx=\lim_{a\rightarrow\infty}\int\int_{D_{a}} e^{-(x^2+y^2)} dA where Da is the disk with radius a and center the...

Homework Statement show that the curvature of a plane curve is \kappa=|\frac{d\phi}{ds}| where phi is the angle between T and i; that is, phi is the inclination of the tangent line.Homework Equations The Attempt at a Solution I'm not sure how to start this one out. Any ideas?