Question

Electric potential is given by $V=6x-8x{y}^2-8y+6yz-4z^2$ then electric force acting on $2\text{C}$ charge placed at origin will be

• A. $10\text{N}$
• B. $20\text{N}$
• C. $8\text{N}$
• D. $6\text{N}$

Answer 1

The correct option is B $20\text{N}$

Given,
electric potential, $V=6x-8x{y}^2-8y+6yz-4z^2$

charge placed at origin, $q=2\text{C}$

Electric field along $\text{x}-$direction,
$E_x=-\frac{\partial V}{\partial x}=-(6-8y^2)$

At origin $(x,y,z)=(0,0,0)$, electric field vector
${\overrightarrow{E}}_x=E_x\hat{i}=-6\hat{i}$

Similarly, at origin $(x,y,z)=(0,0,0)$, the electric field vector along $\text{y}-$direction,
$E_y=-\frac{\partial V}{\partial y}=-(-16xy-8+6z)$
${\overrightarrow{E}}_y=E_y\hat{j}=8\hat{j}$

And,
$E_z=-\frac{\partial V}{\partial z}=-(6y-8z)$
${\overrightarrow{E}}_z=E_z\hat{k}=0\hat{k}$

So, the net electric field at origin,
$\overrightarrow{E}=E_x\hat{i}+E_y\hat{j}+E_z\hat{k}$

Substituting the values obtained, we get

$\overrightarrow{E}=-6\hat{i}+8\hat{j}+0\hat{k}$

The force acting on charge placed at origin,
$\overrightarrow{F}=q\overrightarrow{E}$

$\Rightarrow \overrightarrow{F}=-12\hat{i}+16\hat{j}+0\hat{k}$

Therefore magnitude of the electric force at origin,
$F=|\overrightarrow{F}|$
$\Rightarrow F=\sqrt{{(-12)}^2+{\left(16\right)}^2}=20\text{N}$

Hence, option (b) is the correct answer.