Recent content by adamp121

Hi, I've know that accelerated charge generates electromagnetic radiation which eventually should cause the electron to crash into the atom nuclear, until Bohr atom model. Suppose that we have a mass which cause to gravitation field. If this mass will be accelerated, will it radiate...

I've tried it also, where I set the image ring with charge - q'=-2/pi\frac{Rs}{Rr} and R'=\frac{Rs^2}{Rr} as described here - http://en.wikipedia.org/wiki/Method_of_image_charges#Method_of_images_for_spheres

Yes. I know how to replace a grounded sphere near to a single particle, but don't know how to replace a grounded sphere with surrounded ring. I have only the ring’s potential on the Z axis, and it makes it harder.

Can you give me some clue about the calculation of the sphere's potential?

Homework Statement Hi, I'm trying to find the potential of conducting grounded sphere with radius Rs which located in the center of charged ring with Rr (>Rs) with charge density λ, h meters up to the z axis (see the attached images) Rs=4.3[cm] Rr=6.6[cm] h=13.1[cm] λ=1.0[esu/cm]...

Sorry... The upper and the lower charges are negative while the two middles are positive.

Hi, I'm trying to use image charges to find the force that apply on the following particle (on the attachment) , but it seems that I do something wrong - F=2q^2/(2a^2)-2cos(60)q^2/a^2 The gray section have 0 potential, so I've tried to put image charges like it is in the second attachment...

Now it's all clear :-) Thank you again

So, can we say that it impossible to use the divergence for fields which are not defined at r=0 and we need to use stokes' theorem for getting the right answer using Path integral?

Hi, Thank you very much for your response. I still don't understand why when we are calculate \nabla\cdot E we "naively" ignored what happens at r=0. Is it comes up from the divergence definition?

Hi, I'm trying to use stock's theorem with the following magnetic field - B=1/r\hat{\theta} on Cylindrical coordinate. From one side I get - \nabla X B=0 = \mu \int\int J \cdot dA, means that the current density is zero. From the other side I get - \oint B \cdot dl = 2 \pi r \cdot...