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**Hint:**Electric field potential of a point is defined as the energy which is required to bring a unit positive charge from infinity to that point. The electric field intensity of a point is defined as the force that is experienced by a unit positive charge at that point.

Formula used:

$\vec E = - \dfrac{{dV}}{{dr}}$

Where V is the electric field potential at a point

And r is the distance from the point.

E is the electric field intensity.

**Complete step by step solution:**

The Relation between the electric field intensity and electric field potential is given by the relation-

$\vec E = - \dfrac{{dV}}{{dr}}$

This means that Electric field intensity is the derivative of the Electric field potential. The negative sign implies that the direction of $\vec E$ is opposite to that of V.

In the question it is given that,

The electric field potential is related to space as, $V(x,y,z) = - 2xy + 3y{z^{ - 1}}$

There will be different values of $\vec E$in all the different axes. The resultant of all these values will be the net Electric Field Intensity at the given point.

The value of $\vec E$at each axis is calculated by partially differentiating the V for that axis.

The component of$\vec E$in the x axis is given by-

${\vec E_x} = - \dfrac{{\partial V}}{{\partial x}} = - \dfrac{\partial }{{dx}}\left( { - 2xy + \dfrac{{3y}}{z}} \right)$

In partial differentiation with respect to x the variables other than x are treated as constant, thus the equation is-

${\vec E_x} = - \left( { - 2y} \right) = 2y\hat i$

Similarly for the y direction-

${\vec E_y} = - \dfrac{\partial }{{\partial y}}\left( { - 2xy + \dfrac{{3y}}{z}} \right)$

${\vec E_y} = - \left( { - 2x + \dfrac{3}{z}} \right)\hat j$

${\vec E_y} = \left( {2x - \dfrac{3}{z}} \right)\hat j$

For the z direction-

${\vec E_z} = - \dfrac{\partial }{{\partial z}}\left( { - 2xy + \dfrac{{3y}}{z}} \right)$

${\vec E_z} = - \left( { - \dfrac{{3y}}{{{z^2}}}} \right)\hat k$

${\vec E_z} = \left( {\dfrac{{3y}}{{{z^2}}}} \right)\hat k$

For point $\left( { - 1,1,2} \right)$the values or ${E_x},{E_y}and{E_z}$are given by-

${\vec E_x} = 2y\hat i = 2 \times 1 = 2\hat i$

${\vec E_y} = \left( {2x - \dfrac{3}{z}} \right)\hat j = \left( {2 \times ( - 1) - \dfrac{3}{2}} \right)\hat j$

${\vec E_y} = - \left( {2 + \dfrac{3}{2}} \right)\hat j = - \dfrac{7}{2}\hat j$

${\vec E_z} = \left( {\dfrac{{3y}}{{{z^2}}}} \right)\hat k = \left( {\dfrac{{3 \times 1}}{{2 \times 2}}} \right) = \left( {\dfrac{3}{4}} \right)\hat k$

The net electric field at the point$\left( { - 1,1,2} \right)$,

${E_{net}} = \sqrt {{{({E_x}\hat i)}^2} + {{({E_y}\hat j)}^2} + {{({E_z}\hat k)}^2}} $

${E_{net}} = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - \dfrac{7}{2}} \right)}^2} + {{\left( {\dfrac{3}{4}} \right)}^2}} $

${E_{net}} = \sqrt {4 + \dfrac{{49}}{4} + \dfrac{9}{{16}}} $

\[{E_{net}} = \sqrt {\dfrac{{64 + 196 + 9}}{{16}}} \]

${E_{net}} = \sqrt {\dfrac{{269}}{{16}}} = \dfrac{1}{4}\sqrt {269} $

The net electric field at that point is $\dfrac{1}{4}\sqrt {269} $

**No option is the correct answer.**

**Note:**The electric field intensity is vector quantity, the reason why the electric potential is partially differentiated is because it is a scalar quantity. To specify the values associated with the particular directions of Electric field intensity, the partial differentiation is done.

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