Recent content by kirito

thats actually why I even made this post , I got weirded out by the calculation using cylindrical coordinates since there is no component in the theta direction whereas the z results in 0 and the component in the r direction results in something other than zero unless I had to do something to...

this is the field I was provided and this is the charge density that I have reached I tried to use this yet the output was different I also used Cartesian it gave me the same output as the spherical ones

could you elaborate on what should one do to gain knowledge "path to success ", Any advice on what one should include in a post in general would help , I did read the guidelines, but I’m still getting used to the format of such sites. I usually do not include the whole problem fearing that the...

Thank you! It provided clarity and reminded me to always recheck the conditions of any simplification, rather than becoming too familiar with the result and assume the conditions apply intuitively.i overlooked the issues until something seemed off with the solution.

I do not

I tried to solve the question using two different approaches to gain a better understanding of the subject. However, I reached two different results with each approach. I believe I used Gauss's law to find the electric charge distribution and the electric field inside the cavity incorrectly...

I think I am starting to understand what they meant thanks to all of you it seems like they did indeed consider a spherical shell but with a small hollow disk removed and gave me a radius a<<R so I can use an approximation to solve the integral or use superposition and assumed with such...

first and for most thanks to everyone for the quick replies , I think as everyone stated imagining something like this would do the trick and solving using superposition instead of integration would be wiser , the thing is though I am learning the course in a language thats not my native one ...

I am having a bit of a problem understanding what to visualise since in the first part we found the electric field resultant from the disc shell The goal is to calculate the electric field at a point on the z-axis, at a distance z from the center of the disk. now after they cut of the circle...

in addition I can t see how c has the same flux as a and be I tried to rearrange it to get a closed surface yet got stuck seems like there is a simpler way to approach this

I am currently studying a section from \textit{Electricity and Magnetism} by Purcell, pages 81 and 82, and need some clarification on the following concept. Here’s what I understand so far: 1. The integral of a function $ \mathbf{F} $ over a surface \( S \) is equal to the sum of the integrals...

your explanation is much appreciated, thank you

if you don't mind further explaining what this means , not sure if I get it , to set the potential at t+1 from r=h is bit unclear to me

For r=0 i thought its the condition so potential at z=0 is zero For rmax i though i should take it as 0 since at infinite distance we consider potential zero