The electric potential V at any point ( x,y, z) in space is given by `V = 4 x^(2) V` . The electric field `E ` (in `( V )/( m ))` at the point ( 1,0,2) is

Options

A. 8 along negative X-axis
B. 8 along positive X-axis
C. 16 along negative X-axis
D. 16 along positive X-axis

Answer

A

Explanation

`vec(E) = -vec(nabla)V`
where `vec(nabla) = hat(i)(del)/(del x) + hat(j)(del)/(del y) = hat(k)(del)/(del z)`
`:. Vec(E) = -[hat(i)(del V)/(del x) + hat(j)(del V)/(del y) + hat(k)(del V)/(del z)]`
Here, `V = 4x^(2) :. vec(E) = -8x hat(i)`
The electric field at point (1,0,2) is
`vec(E)_((1,0,2)) = -8hat(i) V m^(-1)`
So electric field is along the negative X-axis.