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

Options

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)`

Answer

A

Explanation

`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`