Figure (a) shows a nonconducting rod with a uniformly distributed charge +Q. The rod forms a 10/24 of circle with radius R and produces an electric field of magnitude Earc at its center of curvature P. If the arc is collapsed to a point at distance R from P (see Figure (b)), by what factor is the magnitude of the electric field at P multiplied?

Question

austinl · Answer

Figures a and b.

anonymous · Answer

Ok.. whats the problem?..

austinl · Answer

That is the problem lol

anonymous · Answer

i mean.. what part are u having difficulty? :P

austinl · Answer

Well, after I get about a sentence in it loses me. If you know how to do this, could you walk me through it?

anonymous · Answer

well.. first thing is you need to calculate the electric field at P due to the circular shape like thingy

to do that.. u need to consider a small differential element.. and then do some vectors.. and then integrate :D.. have u done smething like that before?

roadjester · Answer

@Mashy "...circular shape like thingy" ??

anonymous · Answer

i am tooo sleeepppy to wakl to through this.. .. maybe roadjester can help!

roadjester · Answer

oh man, E&M it's been a semester

anonymous · Answer

come on roady.. this one's easy peezy!

austinl · Answer

I think so. I did some work on my page. I will post my final answer and can you tell me if it makes any sense? I found $\large{E_{arc}}$

$\Large{\overrightarrow{E}_{arc}=\frac{0.92Q}{4\pi \epsilon_{0} r^2}i}$

roadjester · Answer

i can pull out my old notes but i never liked E&M; particularly later on in diffraction in interference

roadjester · Answer

@austinL are you applying Gauss's Law?

austinl · Answer

I honestly am not sure, I followed along with an example problem and that is where I got. I will show you. Just a sec.

anonymous · Answer

|dw:1391371452011:dw|