Question

The electric field potential in space has the form $V(x,y,z)$ = $-2xy+3y{z}^{-1}$,. The electric field intensity $\overrightarrow{E}$ magnitude at the point (-1, 1 , 2) is

• A. $2\sqrt{86}$ units
• B. $2\sqrt{163}$ units
• C. $\sqrt{163}$ units
• D. $\sqrt{86}$ units

Answer 1

The correct option is A $2\sqrt{86}$ units

The electric field potential $=V(x,y,z)=-2xy+3{yz}^{-1}$

then at point $=(-1,1,2)$

$=\sqrt{1^2+1^2+2^2}$

$=\sqrt{1+1+4}=\sqrt{6}$ unit

$\frac{dv}{dx}=E$

or, $\frac{d}{dx}(2xy+3{yz}^{-1})=E$

or, $E_x=-2y+0$

$=-2\times 1=-2$

$\frac{d}{dy}(-2xy+{3yz}^{-1})$

$E_y=-2x+{3z}^{-1}=+2+\frac{3}{2}=\frac{4+3}{2}=\frac{7}{2}$

$E_z=\frac{d}{dz}{(-2xy+{3yz}^{-1})}^2$

$=+3y(-1)\frac{1}{z^2}$

$=-3\times 1\times \frac{1}{4}=-\frac{3}{4}$

$\therefore$ Total $=E_x+E_y+E_z=-2+\frac{7}{4}-\frac{3}{4}=0.75=2\sqrt{86}$ unit.