Question #5ef26

1 Answer
May 12, 2016

(3) -12hati Vm^-1

Explanation:

Electric field vecE is the gradient of potential V and is written as
vecE=-grad V

In vector form it can be expressed as
vecE=-(hati(del)/(delx)+hatj(del)/(dely)+hatk( del)/(delz))V ......(1)

It is given that Electric Potential at any point (x,y,z) is
V=3x^2
As the given function of Potential V is independent of y and z, in the equation (1) derivative with respect to y and z terms vanish. (1) reduces to
vecE=-hati(delV)/(delx)
Inserting values of V, we get
vecE=-hati(del(3x^2))/(delx)
or vecE=-6xhat i .....(2)

To find the electric field at the desired point (2,0,1), inserting x=2 in (2) above, (distance in m)
vecE=-(6xx2)hat i=-12hati Vm^-1
assuming electric potential function is given in Volt