Recent content by JoelKTH

Yes, that is correct. I fixed this now, thanks. Since the distance from the point charge q and the metal plates is at a distance a, does that mean I can define r=a##\hat r## for all the mirrored charges aswell? One problem then is that the first mirror charge -q_1=-q would then be 0 as...

I placed the field vector at the point vector and are not redefining it for each charge. I am thinking now that maybe the field vector to the point charge q should be a/cos(30deg). Does this make sense? Do you recommend me to redefine the field vector for each charge or have r=a/cos(30deg) and...

Right, angle of 30 to each plate. Not sure what you mean by angle to plate A but between every charge there should be 60 degrees, I get that now. So I place the charges 60 degrees from eachother and then find out the electric field. Should V=0 at every 60 degrees after 30degrees(where the...

This is a good tip, thanks. I believe I made some progress. By mirroring I managed to get a somewhat result where the potential V=0 at the x-axis. However, where the other plate is, I don't believe V=0 as it should, due to that it can be considered grounded. Are my thinking so far correct? (1)...

Yes, but are those two requirements not for using the method of images? As you say they might not be needed to solve the actual problem. Judging from the method of images and my knowledge, my first step would be to place a negative charge -q 180° from the point charge. See image attached. Not...

Hi, I'm having some trouble understanding how to solve this problem. I have a few questions: 1. I understand that I need to make an educated guess for the electric potential, where \(V_1\) is given by: V_1 = \frac{q}{4\pi\epsilon_0} \left(\frac{1}{r_1} - \frac{1}{r_2} + \frac{1}{r_3} -...

Then I did dimensionanalysis and its ok. But I noticed that one needs to check the boundary conditions. My professor writes in the solution: To convince ourselves that the charge distribution is correct, lets examine the boundary condition. First, let z = 0. The symmetry in the choice of the...

Here are my calculations and answer:

I was a bit confused when to use what equation. From my calculations i used ##E =\frac{\rho_l}{2\pi\epsilon_0} \frac{(r-r')}{|r-r'|^3}## and got the correct result. I don't quite get why this equation is to be used compared to my first instinct: ##dE =\frac{dq}{4\pi\epsilon_0}...

I managed to get the same answer as my professor, however we defined the charges differently. Am I free to choose both alternative A and B(see figure attached) or is one of them to prefer?

I actually solved it by knowing that V=0 both at the top of the cylinder and the bottom. I used two equations by adding R to the figure.

Hi, I want to solve the problem by method of mirroring and by using the electric field by doing superposition and then adding them up to use in Lorentz law to get the force. I have attached a figure that represents the problem. How do I know from the figure that $-p_l$ is from the...

Yes you are correct. Here is my calculation: ##d\mathbf{l}' = d\mathbf{r} = (r_c)\hat{\mathbf{d}}r_c' + (\varphi)\hat{\mathbf{r}}_c' d\varphi'## Then, ##dr/d\varphi = \hat{\mathbf{r}}_c \frac{dr_c'}{d\varphi} + \hat{\boldsymbol{\varphi}} \frac{r_c' d\varphi'}{d\varphi'}## Since ##r_c = r_0...

##r=r_0ke^{k\varphi}##

Yes, you are correct. I ment what you wrote but I was a bit too quick here in Latex. Not sure what you mean by expressing ##dr## in terms of ##d\varphi##? I have ##\vec {dl}=\hat \varphi r.d\varphi+\hat rdr##. I know that $$d\mathbf{l'} = \left(\mathbf{\hat{r_c}} \frac{d...