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Electrostatics

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.

Nemanja Grubor

11 Years agoGrade

6 Answers

Nicho priyatham

11 Years ago

by integration

is it force per unit length.....?

Nicho priyatham

Approved Tutor Answer11 Years ago

is it QS/4e₀....?

Nemanja Grubor

11 Years ago

@Nicho priyatham Yes, it is correct. Could you show the steps? Thanks.

Nemanja Grubor

11 Years ago

@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

dE_y=cos(w)(sdl/2pie₀r, where ‘w’ is angle between dE_yand dE

So, E_yis 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

E_y=(s/pie₀)arctg(1) which is indeed E_y=s/(4e₀). Now,

F^‘=Q^‘E_y=Q^‘s/(4e₀)

Nemanja Grubor

11 Years ago

@Nicho priyatham Ok, I solved it.

Nicho priyatham

11 Years ago

dats good nemanja

sorry i was feeling lazy to type the solution

i think u got it now if still any doubt i will definetly say u

can i no what book u r using n wh clg ur in …..?

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