Question

12. If potential function in an electric field is given
as V=-4x²+3y2 +2z, then electric field at point
(1,0,2) is
a) 81-6j+2k c) -8i+2K
b) 8i+2K
d) 8i-2k​

Answer 1

Since the electric field is only in the x axis and z axis, we consider electric potential also in those axis only.

Answer 2

Answer:

The electric field is $8\hat{i}-2\hat{k}$ i.e.option (d).

Explanation:

The relation between potential and electric field is given as,

$E=-\frac{dV}{dr}$   (1)

Where,

E=electric field at a point

dV=electric potential over a point

dr=distance at which the electric field is to be found

From the question we have,

$V= -4x^2\hat{i}+3y^2\hat{j} +2z\hat{k}$     (2)

$r=1\hat{k}+0\hat{j}+2\hat{k}$               (3)

By substituting the values of equation (2) in equation (1) we get;

$E=-(\frac{-4x^2}{dx}+\frac{3y^2}{dy}+\frac{2z}{dz})$

$E=8x-6y-2$      (4)

By using equation (3) in equation (4) we get;

$E=8\times 1\hat{i}-6\times 0\hat{j}-2\hat{k}$

$E=8\hat{i}-2\hat{k}$

Hence, the electric field is $8\hat{i}-2\hat{k}$ i.e.option (d).