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How to solve the following problem:Very large rod, with constant longitudinal charge density Q’, is set in vacuum parallel to very large conductive ribbon with width ‘a’ , where height between them is ‘a/2’ . If the surface charge density of a ribbon is ‘s‘, calculate longitudinal force on a rod.Thanks for reply.
by integration is it force per unit length.....?
is it QS/4e0....?
@Nicho priyatham Yes, it is correct. Could you show the steps? Thanks.
@Nicho priyatham Here is what I did, but get wrong result. If we define electric field on y-axis (on the right side of ribbon) as dEy=cos(w)(sdl/2pie0r, where ‘w’ is angle between dEy and dE So, Ey is the resultant field of a ribbon. When I integrate it, I get Ey=s/e0pi which is not correct.Also in my solution it says Ey=(s/pie0)arctg(1) which is indeed Ey=s/(4e0). Now, F‘=Q‘Ey=Q‘s/(4e0)
@Nicho priyatham Ok, I solved it.
dats good nemanjasorry i was feeling lazy to type the solution i think u got it now if still any doubt i will definetly say ucan i no what book u r using n wh clg ur in …..?
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