I think you would need calculus to do this properly, specifically that
Force = \frac{d(mv)}{dt}
and the force if the force is just gravitational from a single body,
Force = \frac{Gm_1 m_2}{r^2}
G - Newton's Gravitational constant
m1 and m2 are the masses of the bodies being...
5.8Ghz is still a lot lower than the frequency of visible light ~10^15Hz, and people don't spend too much time worrying about light bulbs causing cancer. I am not so sure about exactly how they would cause cancer though - best ask a biochemist.
You don't need to do all of this. Once a solution has been found that satisfies the boundary conditions, you know that this is the only solution by the uniqueness theorem. This means that we can use trial solutions to find a solution to a problem using Laplace's equation.
Water is occasionally used as a dielectric. There are some large coaxial cables under Imperial College in London that use water as a dielectric, but only for a very short period of time, before it starts to conduct.
Think about the structure of a conductor. An electric field line that is not perpendicular to a conducting plane will have a component parallel to it. This electric field then acts on the electrons in the conductor and they will re-arrange themselves so that there is no longer any force acting...
It is better to use this one as the general solution,
V(r,\phi)=a_{0} + b_{0}lnr + c_{0}r cos\phi + d_{0}\frac{1}{r}cos\phi
As r\rightarrow\infty, we know the solution must be -E_{0}r cos\phi so c_{0} = -E_0.
The only other term that will be able to satisfy the boundary condition for...
The first law of thermodynamics states that
U = dQ + dW (depending on how you define it)
The dW can be replaced with fdL as a unit of work and when the Helmholtz free energy is differentiated with respect to L at a constant temperature, you get the result for the tension.
The system of...
A magnetic dipole is generated by a small current loop (the electron). Try working out the current that the electron generates then crossing that with the area of your little current loop.
Remember:
Current amount of charge per unit time
As you only have one electron, you just have to...
However, you do know what the electric field is at infinity, and this will give you another boundary condition. You also need to think about where your solution needs values, is the potential infinite or finite valued at the origin?