Electric field?

please solve the question and explain?

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1 Answer
Sep 10, 2017

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")]