Recent content by Sergey Vilov

Yes, you're right, I must have admitted a mistake. Though you can replace e^C_0+e^C_1 with just D, where D is an arbitrary constant(because both C0 and C1 are arbitrary constants). Thus, x=-r+D is the solution.

You should take an indefinite integral(without limits). Thus,x=\frac{C_1}{r},where C1 is a constant(to be determined through the boundary conditions). Then you remember that x=\bigtriangledown\phi, so you have a new differential equation:\frac{d\phi}{dr}=\frac{C_1}{r} whose solution is...

As I wrote above, you can not use the quadratic formula to solve a second order differential equation. This is wrong! You need to study math books in order to learn how to solve this equation, there is no physics here.

Now your equation seems ok apart from the fact that you have already substituted the equation for the gradient of the potential, so you should replace \bigtriangledown\phi with \frac{d\phi}{dr} otherwise it seems that you add a vector(gradient) to a scalar(laplasian) which makes no sense. On...

Yes, your general strategy is right. First,find potential,then E and D. But there is something strange with your equations...what does the gradient of epsilon equal? if \epsilon=\frac{\epsilon_0\alpha r}{R} then \bigtriangledown\epsilon=\frac{\partial}{\partial r}(\frac{\epsilon_0\alpha...

The formula for \bigtriangledown(\epsilon \vec{E}) that you are asking about is general for any divergence,it is a pure mathematical thing(you can find it in any math books devoted to vector analysis or even in Wikipedia: http://en.wikipedia.org/wiki/Divergence). On the other hand, it is indeed...

First, \bigtriangledown(\epsilon\vec{E})=\bigtriangledown\epsilon\vec{E}+\epsilon div\vec{E}=-(\bigtriangledown\epsilon \bigtriangledown\phi)-\epsilon \bigtriangleup\phi=0 Then, \bigtriangledown\epsilon=\frac{\alpha}{R}\vec{r_0} Hence, r\frac{d^2\phi}{dr^2}+\frac{d\phi}{dr}=0

As concerns part A, I would say that a dipole moment is absent because there are several rotation axes C3. You can see this through analysing the character table of the symmetry group. The dipole moment can not exist when there are more than one Cn axes because it can not have more than one...

Usually such simulations are performed in reduced(atomic) units:http://en.wikipedia.org/wiki/Atomic_units

I think you are on the right way, \bigtriangledown(\epsilon\vec{E})=0 because all of the charges between the cylinders are non-free. Then, I would use the formula \bigtriangledown(\epsilon\vec{E})=\bigtriangledown\epsilon\vec{E}+\epsilon div\vec{E} and the relation between electric field and...

I think, in order to calculate the centre of pressure you can use the same formula as for the centre of mass(you can simply look it up in wikipedia https://en.wikipedia.org/wiki/Center_of_mass). Just replace the total mass with the total force and the volume density with the pressure and...

I assume it is not necessary to consider forces f1 and f2. This would be redundant as N1 and N2 are needed. I would choose an axis which would go through the centre point between man's feet perpendicular to the radius and the direction of gravity force(so that torques of N1 N2 and centrifugal...

If you have an abstract sphere of positive(negative) charges, imagine how electric field lines go. Their general property is that they should start at positive charges and end at negative charges. Imagine them going from each elementary charge of the sphere. As there is no negative(positive)...

I think you solution is correct. As concerns the standing wave, note that the spatial wave function should be multiplied by exp(-iwt) with w=E/hbar in order to get the solution to the Shrodinger equation. This will give you a standing wave.

Could you say exactly what the book states? Actually, in electrostatics the electric field is always zero in conductors. The usual explanation is that conductors are always in the external electric field created by all of the other charges in the world. This field render the conductor to...