Question

Electric field?

please solve the question and explain?

Answer 1

I get
(A) and (C)

Explanation:

Given electric field intensity at a point `(x,y)` as `vecI=(12xy^3-4x)hati+18x^2y^2hatj`

Let's examine the case of a two-dimensional vector field whose
`"*"`scalar curl is `=∂/(∂x)F_2−∂/(∂y)F_1`

For the given field we have

`∂/(∂x)(18x^2y^2)−∂/(∂y)(12xy^3-4x)`
`=36xy^2−(36xy^2-0)`
`=0`

As scalar curl `=0`, the function represents a conservative electric field.

We know that potential in an two dimensional electric field is expressed as

`vecE=-[hatidel/(delx)+hatjdel/(dely)]V(x,y)`

For the given electric field above

`-(delf)/(delx)=(12xy^3-4x)` ....(1) and `-(delf)/(dely)=18x^2y^2` .....(2)

From equation (1) using partial integration

`-f(x,y)=int(12xy^3-4x)dx`
`=>-f(x,y)=(12xxx^2/2y^3-4xxx^2/2+C(y))`
where `C(y)` is a constant of integration dependent on `y`.
`=>-f(x,y)=(6x^2y^3-2x^2+C(y))`

Differentiating this with respect to `y` and setting equal to equation (2) we get

`-(delf(x,y))/(dely)=-del/(dely)(6x^2y^3+2x^2+C(y))=-18x^2y^2`
`=>(6x^2 (3y^2)+d/dyC(y))=18x^2y^2`
`=>d/dyC(y)=0`
`=>C(y)` is actually a constant independent of both `x and y`.

The potential function becomes

`f(x,y)=(6x^2y^3-2x^2+c)`

Given that at origin electric potential is zero. Therefore `c=0` and potential function reduces to

`f(x,y)=6x^2y^3-2x^2` ......(3)

Therefore, electric potential at point `(1,1)` is found from equation (3) as

`-f(1,1)=-4V`

.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

A two dimensional vector field `vecF` given as

`vecF=F_1hati+F_2hatj`

is conservative if partial derivative

`∂/(∂x)F_2−∂/(∂y)F_1=0`

The LHS is also called `"*"`scalar curl.
......................................

To determine whether a three dimensional vector field is conservative we find a function `f` such that `vecF=gradf`.
Therefore if curl `vecI = 0`, then `vecI` is conservative.

Where curl `vecI` is given as
`gradxxvecI=[(hati,hatj,hatk),(del/(delx),del/(dely),del/(delz)),(hati" part",hatj" part",hatk" part")]`