A charge particle `q = -10 mu C` is carried along OP and PQ and then back to O along QO as shown in an electric field `vec E = (x + 2y) hat i + 2x hat j` Find the work done by an external agent in (i) each path and (ii) the round trip.
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Explanation

(i) `W = -q int vec(E) . d vec(l) = - q (int E_x dx + int E_y dy)`
In Path O rarr P : Since the displacement is along the x - axis for the path OP,
`W_(O rarr P) = -q int E_x d x = - (-10 xx 10^-6) int _0 ^2 (x + 2 y ) d x,`
where `y = 0`
or `W_(O rarr P) = + 20 xx 10^-6 J`
In Path P rarr Q : Since the displacement is along y - axis for PQ,
`W_(P rarr Q) = -q int E_(y) d_(y) = -q (2 int_0 ^2 x d y)`
where (x = 2)
= `- 2 (- 10 xx 10^(-6)) int _(0)^(2) 2 d y = 80 xx 10^(-6) J`
In Path Q rarr O : Since the displacement
`d l = d x hat(i) + d y hat(j) "along" Q O,`
`W_(Q rarr O) = -q int (E_x d x + E_y d_y) = -q int_(2)^(0) (x + 2y) d x + 2 x d y `
=`- q [ int _2 ^0 (x + 2 y) d x + 2 int _2 ^0 x d y]`
where the equation of the path (QO) is given as (y = x)
or `W_(Q rarr O) = -q [ int _2 ^0 (x + 2 x) d x + 2 int _2 ^0 y d y]`
= `- q ((3 x ^2)/(2)|._(2)^(0) + y^2|_(2)^(0)) = - (-10 xx 10^(-6)){ (- 3 xx 4)/(2) - 4}`
=` - 100 xx 10^(-6) J`
(ii) Then, the total work done in the round trip is
`W = W_(O rarr P) + W_(P rarr Q) + W _(Q rarr O) = 0`
In the above example, `oint vec(E) . d vec(l) = 0`. Hence, the field is conservation and static.