Electric field?

please solve the question and explain?

enter image source here

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")]#