Question

If $\overrightarrow{E}=-(2y^3-3y{z}^2)\hat{x}-(6x{y}^2-3x{z}^2)\hat{y}+\left(6xyz\right)\hat{z}$ is the electric field in a source free region, a valid expression for the electrostatic potential is

• A. $x{y}^3-y{z}^2$
• B. $2x{y}^3-xy{z}^2$
• C. $y^3+xy{z}^2$
• D. $2x{y}^3-3xy{z}^2$

Answer 1

The correct option is D $2x{y}^3-3xy{z}^2$

Ans : (d)

$E=-\frac{\partial V}{\partial x}{\hat{a}}_x-\frac{\partial V}{\partial y}\hat{a}y-\frac{\partial V}{\partial Z}{\hat{a}}_z$

$\frac{\partial V}{\partial x}=2y^2-3y{z}^{2}V=2x{y}^3-3xy{z}^2+K_1$

$\frac{\partial V}{\partial y}=6x{y}^2-3x{z}^{2}V=2x{y}^3-3xy{z}^2+K_2$

$\frac{\partial V}{\partial z}=-6xyzV=-3xy{z}^2+K_3$
Comparing all three for uniqueness solution of potential

$K_1=K_2=0K_3=2x{y}^3$

$\overrightarrow{E}=-(2y^3-3y{z}^2)\hat{x}-(6x{y}^2-3x{z}^2)\hat{y}+\left(6xyz\right)\hat{z}$

as we know that $E=-\nabla V$

$E=-(\frac{\partial }{\partial x}\hat{x}+\frac{\partial }{\partial y}\hat{y}+\frac{\partial }{\partial z}\hat{z})V$ ... (i)
Clearly option (d) satisfies the given relation.