Gravitational field in `x-y` plane is given as
`E = (2xhati+3Y^(2)hatj)N//kg`
Fin difference in gravitation potential between two points A and B, where
co-ordinates of A and B are `(2m, 4m) and (6m,0)`.

Answer

B, C

Explanation

We know that, `dV = -E*dr` …(i)
Here `dr = dx hat i +dy hat j`
`:. E.dr = (2xdx+3y^(2)dy)`
Now, Eq. (i) can be written as
`int _(B)^(A) dV = -int_(B)^(A) (2xdx+3y^(2) dy`
or `V_(A) -V_(B) = -[x^(2) +y_(3)]_(6m, 0)^((2m, 4m)`
`= -[{(2)^(2)+(4)^(3)}-(6)^(2)+(0)^(3)]`
`= - 32 J//kg`.