IIT JEE , AIEEE Forum - Physics , Electricity

mukulss

the diagram shows 3 infinitely long uniform line charges placed on the X,Y,Z axes......
the work done in moving a unit positive chrge from (1,1,1)
(0,1,1) is equal to:

here @ is the linear charge density
please reply

mukulss

hello???

master_purav

Calculate potential at both points due to all three wires and find potential difference.

That would be the work done...

mukulss

@ purav,
thnx 4 it.
since my this portion is weak will u please help me by telling me ki
(1,1,1) and (1,1,0) par electric field kaise calculate hoga...

feynmann

We know that potential due to a single wire = @ / 2 pi

o ln ( r ) + const

So we have potential at ( x, y , z ) V =

1 / 4pi

{ @ 1 ln ( y^ 2 + z^2 ) + @ 2 ln ( z^2 + x^2 ) + @3 ln ( x^2 + y^2 ) } + const

where @ s are corresponding linear charge density of x , y , z axes respectively .

Now calculate the potential diff . ( const term cancels out )

ans is 5@ /(4 pi

o) ln ( 1/2 )

mukulss

@ feymann,
that is good but my actual doubt is:
in bansal notes i read:
2 k landa ln(R/x)

whare R is the distance from the wire at which the potential iis assumed to be o.
how does this: y^ 2 + z^2 fit here...
plaese xpalin

feynmann

That , what u have found in notes is absolutely right. Here the const term does the work for R . It does not come into picture as we are only interested in potential difference . ( You may try the same problem by ln ( R/ x ) also , all the ln ( R ) terms will cancel ) .

Now regarding ur second doubt see sqrt ( y^2 + z^2 ) is the distance of the pt ( x, y, z ) from the x axis etc .

Hope u have got it now .

mukulss

@ fey sir,
thanks a lot ...........

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