Electric field vector in a region is given by `E=(3hati+4hatj)V-m^(-1)`. The potential at the origin is zero. Then, the potential at a point (2, 1) m is

Options

A. 7V
B. 8V
C. `-8V`
D. `-7V`

Answer

C

Explanation

As, `E=(-delV_(x))/(delx)hati+(-delV_(y))/(dely)hatj`
`(-delV_(x))/(delx)=3and(-delV_(y))/(dely)=4y`
`rArr" "V_(x)=-3xandV_(y)=-2y^(2)`
So, V = potential function
= `-(3x+2y^(2))`
Potential at point (2, 1) is
`V=-(3xx2+2xx1^(2))=-8V`