An electric field is given by `E=(y hati+x hatj)NC^(-1)`. The work done in moving a 1C charge from `r_(A)=(2hati + 2hatj)m" to "r_(B)=(4hati+2hatj)m` is

Options

A. 2y
B. 3y
C. zero
D. infinity

Answer

A

Explanation

Here. Displacement would be,
`Deltar=(4-2)hati+(2-2)hatj=2hati`
Now, `DeltaV=E.Deltar=(yhati+xhatj).2hati=2y`
`therefore" Work done, "W=q(DeltaV)=q.2y`
`" "=1xx2y=2yJ" "(because" charge = 1C")`