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in Physics by (87.2k points)
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For an infinite line of charge having linear charge density `lambda` lying along x-axis the work done in moving a charge `q` form `C` to `A` along `CA` is
image
A. `(qlambda)/(2piepsilon_(0))"In"2`
B. `(qlambda)/(2piepsilon_(0))"In"sqrt(2)`
C. `(qlambda)/(4piepsilon_(0))"In"2`
D. `(qlambda)/(2piepsilon_(0))"In"(0.5)`

1 Answer

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Best answer
Correct Answer - A
image
`E=(lambda)/(2piepsilon_(0)y)` for a small displacement
`vec(dr)=-(dyhat(j)+dxhat(i))dw=vec(E).vec(dr)`
`(lambda)/(2piepsilon_(0)y)hat(j)(-dyhat(j)-dxhat(i))=-(lambda)/(2piepsilon_(0))(dy)/(y)`
`W=-(lambdaq)/(2piepsilon_(0))("in"y)_(2a)^(a)=-(lambdaq)/(2piepsilon_(0))"in"(0.5)=(lambdaq)/(2piepsilon_(0))"in"2`

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