Question

A potential field is given by V = 3x^2y - yz. Which of the following is not true?

(a) At point (1, 0, -1) V and E vanish.

(b) x^2y = 1 is an equipotential line on the xy-plane.

(c) The equipotential surface V = - 8 passes through point P + 2, -1, 4)

(d) The electric field at P is 12ax - 8ay - az V/m.

(e) A unit normal to the equipotential surface V = - 8 at P is -0,83ax + 0,55ay+ 0,07az.

Answer 1

Gives

$V=3x^2y-yz$

$V_{(1,0,1)}=0$

$E=-\nabla V$

$$E_{1,0,1}=-(6(0)a_x+(3\times1+1)a_y-0a_z)$$

$E=-4a_y$

Option (a) wrong(incorrect)

$P=x^2y-1$

Equpotential line equation

$$\nabla.P=\nabla.(x^2y-1)=2xya_x+x^2a_y\rightarrow(2)$$

Equation (2) is lies xy-plane

Option (b) is correct option

Electric field

$V=3x^2y-yz$

$$V_{(2,-1,4)}=3\times4(-1)-(-1)\times4=-8V$$

Option (c) is correct

$$E=-\nabla V=-(6xya_x+(3x^2-z)a_y-ya_z)$$

$E=12a_x-8a_y-a_z$

Option (d) correct

$E=12ax-8ay-az$

$\hat{V}=\frac{\nabla V}{ |\nabla V|}$

$\hat{V}=\frac{6axy a_x+(3x^2-z)a_y+a_z(-y)}{14.5}$

$$\hat{V}_{(2,-1,4)}=\frac{-12}{14.4}a_x+\frac{8}{14.4}a_y+\frac{1}{14.4}a_z$$

$\hat{V}=-0.83a_x+0.55a_y+0.07a_z$

Option (e) is correct