Recent content by brainslush

Thanks for the help

Of course... \nabla \circ (\epsilon \textbf{E}) = (\nabla \epsilon) \circ \textbf{E} + \epsilon (\nabla \circ \textbf{E}) \Rightarrow \frac{\frac{\partial}{\partial \rho} \epsilon (\rho)}{\epsilon (\rho)} = -\frac{\frac{\partial}{\partial \rho} (\rho E(\rho))}{\rho E(\rho)} =...

\nabla \circ \textbf{D} = \epsilon (\nabla \circ \textbf{E}) + (\nabla \circ \epsilon) \textbf{E} = 0 Now this gives the diferential eq. (we only care about the radial direction) \frac{\frac{\partial}{\partial \rho} \epsilon (r)}{\epsilon (r)} = -\frac{\frac{\partial}{\partial r}...

Homework Statement Consider a long cylindrical coaxial capacitor with an inner conductor of radius a, and outer conductor of radius b, and a dielectric with a relative electric permittivity or dielectric ε(r), varying with the cylindrical radius. The capacitor is charged to the voltage V...

After deeper consideration the only reasonable posibility is a surface integral. In that case I do following: One knows that r^{\rightarrow}(\theta,z) = (a*cos(\theta), a*sin(\theta), z) then one gets \left\|\frac{\partial r}{\partial \theta} \times \frac{\partial r}{\partial z}\right\| = a...

Sorry that's all we got. That's the reason why I'm not sure what to do. Oh my fault. Yes it should be three integrals. It looks more like a surface integral to me but this means my calculations up there are incorect

Homework Statement Find the integral of the function x^2 on a cylinder (excluding button and top) x^2 + y^2 = a^2, 0 <= z <= 1 Homework Equations \int\int\int x^{2} dx dy dz x = a * cos \Theta y = a * sin \Theta z = zThe Attempt at a Solution I'm not quite sure what to do but I give it a try...

Homework Statement Use Green’s theorem to find the integral \oint_{\gamma} \frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy along two different curves γ: first where γ is the simple closed curve which goes along x = −y2 + 4 and x = 2, and second where γ is the square with vertices (−1, 0), (1, 0)...

Our professor just told us that r = a * exp(b*x) and i found the mistake... A is dependent on r and therefore I need to consider this in my calculations... dR = \rho d(\frac{L}{A}) d(\frac{L}{A}) = \frac{dL}{A} - \frac{L*dA}{A^2} dA = 2*\pi*r*dr , dL = dx dr = a*b*e^{bx} dx dR =\rho...

Of course, A is a function of r but r is also a function of x. Therefore the substitution yields what I have written above. Or am I wrong?

Homework Statement The radius r of a wire of length L increases according to r = a * exp(bx^2), x is the distance from one end to the other end of the wire. What is the resistance of the wire?Homework Equations R =\frac{L * \rho}{A}The Attempt at a Solution dR =\frac{dx * \rho}{A} A(r) = \pi *...

Homework Statement Assume two perpendicular planes one with charge density \sigma and the other with a charge density 2\sigma. Find the electric field. //| //| //| //| \sigma | //| _ |__________________ //|////// 2\sigma/////////////// //| Homework Equations The Attempt at a Solution I...

Upss my bad, totally messed up the first solution

Homework Statement Proof that there exist more than one solution to following equation \frac{dx}{dt} = \sqrt[3]{x^{2}} , x(0) = 0Homework Equations The Attempt at a Solution Well, I need a confirmation to my attempt of solution. The one is quite forward: \Rightarrow x=(1/3(t+c))^{3} Pluging...

Upps, I mixed up radiation, conductivity... So dQ/dt = e*\sigma*4\Pi*r^{2}(T^{4}_{sphere}-T^{4}_{vacuum}) but what about the radiation screen? It is a thin shell. Can I neglect it?