Correct Answer - Option 4 : 2xy
3 – 3xyz
2
Concept:
The electric field is related to the potential as:
\(\vec E = - ∇ V\)
\(∇ =\frac{\partial}{\partial x}\hat a_x+\frac{\partial}{\partial y}\hat a_y+\frac{\partial}{\partial z}\hat a_z\)
Calculation:
We can verify all options with the Maxwell equation \(\vec E = - ∇ V\)
Option 1: V = xy3 – yz2
\( - ∇ V=-(\frac{\partial}{\partial x} (xy^3 – yz^2)\hat a_x+\frac{\partial}{\partial y}(xy^3 – yz^2)\hat a_y+\frac{\partial}{\partial z}(xy^3 – yz^2)\hat a_z)\)
= -y3 ax - (3xy2 - z2) ay - 2yz az
Since this is not equal to the given Electric field, this cannot be a valid expression for the given electrostatic potential.
Option 2: V = 2xy3 – xyz2
\( - ∇ V=-(\frac{\partial}{\partial x} (2xy^3 – xyz^2)\hat a_x+\frac{\partial}{\partial y}(2xy^3 – xyz^2)\hat a_y+\frac{\partial}{\partial z}(2xy^3 – xyz^2)\hat a_z)\)
= -[(2y3 - yz2) ax + (6xy2 - xz2) ay - 2xyz az]
Since this is not equal to the given Electric field, this cannot be a valid expression for the given electrostatic potential.
Option 3: V = y3 + xyz2
\( - ∇ V=-(\frac{\partial}{\partial x} (y^3 + xyz^2)\hat a_x+\frac{\partial}{\partial y}(y^3 + xyz^2)\hat a_y+\frac{\partial}{\partial z}(y^3 + xyz^2)\hat a_z)\)
= -[yz2 ax + (3y2 + xz2) ay + 2xyz az]
Since this is not equal to the given Electric field, this cannot be a valid expression for the given electrostatic potential.
Option 4: V = 2xy3 – 3xyz2
\( - ∇ V=-(\frac{\partial}{\partial x} (2xy^3 – 3xyz^2)\hat a_x+\frac{\partial}{\partial y}(2xy^3 – 3xyz^2)\hat a_y+\frac{\partial}{\partial z}(2xy^3 – 3xyz^2)\hat a_z)\)
= -[(2y3 - 3yz2)ax + (6xy2 - 3xz2) ay - 6xyz az]
= -(2y3 - 3yz2)ax - (6xy2 - 3xz2) ay + 6xyz az
Since E = - ∇ V for this case, this this is a valid expression for the given electrostatic potential.