Question #5ef26

1 Answer
May 12, 2016

Answer:

(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