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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.
3 years ago

Answers : (6)

Nicho priyatham
625 Points
							
by integration 
is it force per unit length.....?
3 years ago
Nicho priyatham
625 Points
							
is it QS/4e0....?
 
3 years ago
Nemanja Grubor
8 Points
							
@Nicho priyatham Yes,  it is correct. Could you show the steps? Thanks.
3 years ago
Nemanja Grubor
8 Points
							
@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=QEy=Qs/(4e0)
3 years ago
Nemanja Grubor
8 Points
							
@Nicho priyatham Ok, I solved it.
3 years ago
Nicho priyatham
625 Points
							
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 …..?
3 years ago
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