Question

In a region, the potential is represented by $V(x,y,z)=2xyz–4y+3zx$, where $V$ is in $\text{Volts}$ and $x,y$ and $z$ are in meters. The electric field at the point $(1,1,1)$ is

• A. $\sqrt{41}\text{V/m}$
• B. $\sqrt{54}\text{V/m}$
• C. $\sqrt{64}\text{V/m}$
• D. $\sqrt{144}\text{V/m}$

Answer 1

The correct option is B $\sqrt{54}\text{V/m}$

Given:
$V(x,y,z)=2xyz–4y+3zx$
Point $=(1,1,1)$

As we know that,

$E_{x(1,1,1)}=-\frac{\partial V}{\partial x}$

$E_{x(1,1,1)}=-\frac{\partial }{\partial x}(2xyz–4y+3zx)$

$E_{x(1,1,1)}=-(2yz+3z)=-(2+3)$

$E_{x(1,1,1)}=-5\text{V/m}$

$E_{y(1,1,1)}=-\frac{\partial V}{\partial y}$

$E_{y(1,1,1)}=-\frac{\partial }{\partial y}(2xyz–4y+3zx)$

$E_{y(1,1,1)}=-(2xz-4)=-(2-4)$

$E_{y(1,1,1)}=2\text{V/m}$

$E_{z(1,1,1)}=-\frac{\partial }{\partial z}(2xyz–4y+3zx)$

$E_{z(1,1,1)}=-(2xy+3x)=-(2+3)$

$E_{z(1,1,1)}=-5\text{V/m}$


$E=\sqrt{(E_x^2+E_y^2+E_z^2)}=\sqrt{\left[\right(-5{)}^2+2^2+(-5{)}^2]}$

$E=\sqrt{54}\text{V/m}$

Hence, option $\left(B\right)$ is correct.