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The electric field potential in space has the form $V(x,y,z) = - 2xy + 3y{z^{ - 1}}$. The electric field intensity $\vec E$ magnitude at the point $( - 1,1,2)$ is
(A) $2\sqrt {86} units$
(B) $2\sqrt {163} units$
(C) $\sqrt {163} units$
(D) $\sqrt {86} units$

Last updated date: 13th Jun 2024
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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.