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Question: The electric field potential in space has the form \(V(x,y,z) = - 2xy + 3y{z^{ - 1}}\). The electric...

The electric field potential in space has the form V(x,y,z)=−2xy+3yz−1V(x,y,z) = - 2xy + 3y{z^{ - 1}}. The electric field intensity E⃗\vec E magnitude at the point (−1,1,2)( - 1,1,2) is
(A) 286units2\sqrt {86} units
(B) 2163units2\sqrt {163} units
(C) 163units\sqrt {163} units
(D) 86units\sqrt {86} units

Explanation

Solution

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:
E⃗=−dVdr\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-
E⃗=−dVdr\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 E⃗\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+3yz−1V(x,y,z) = - 2xy + 3y{z^{ - 1}}
There will be different values of E⃗\vec Ein all the different axes. The resultant of all these values will be the net Electric Field Intensity at the given point.
The value of E⃗\vec Eat each axis is calculated by partially differentiating the V for that axis.
The component ofE⃗\vec Ein the x axis is given by-
E⃗x=−∂V∂x=−∂dx(−2xy+3yz){\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-
E⃗x=−(−2y)=2yi^{\vec E_x} = - \left( { - 2y} \right) = 2y\hat i
Similarly for the y direction-
E⃗y=−∂∂y(−2xy+3yz){\vec E_y} = - \dfrac{\partial }{{\partial y}}\left( { - 2xy + \dfrac{{3y}}{z}} \right)
E⃗y=−(−2x+3z)j^{\vec E_y} = - \left( { - 2x + \dfrac{3}{z}} \right)\hat j
E⃗y=(2x−3z)j^{\vec E_y} = \left( {2x - \dfrac{3}{z}} \right)\hat j
For the z direction-
E⃗z=−∂∂z(−2xy+3yz){\vec E_z} = - \dfrac{\partial }{{\partial z}}\left( { - 2xy + \dfrac{{3y}}{z}} \right)

E⃗z=−(−3yz2)k^{\vec E_z} = - \left( { - \dfrac{{3y}}{{{z^2}}}} \right)\hat k
E⃗z=(3yz2)k^{\vec E_z} = \left( {\dfrac{{3y}}{{{z^2}}}} \right)\hat k
For point (−1,1,2)\left( { - 1,1,2} \right)the values or Ex,EyandEz{E_x},{E_y}and{E_z}are given by-
E⃗x=2yi^=2×1=2i^{\vec E_x} = 2y\hat i = 2 \times 1 = 2\hat i
E⃗y=(2x−3z)j^=(2×(−1)−32)j^{\vec E_y} = \left( {2x - \dfrac{3}{z}} \right)\hat j = \left( {2 \times ( - 1) - \dfrac{3}{2}} \right)\hat j
E⃗y=−(2+32)j^=−72j^{\vec E_y} = - \left( {2 + \dfrac{3}{2}} \right)\hat j = - \dfrac{7}{2}\hat j
E⃗z=(3yz2)k^=(3×12×2)=(34)k^{\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(−1,1,2)\left( { - 1,1,2} \right),
Enet=(Exi^)2+(Eyj^)2+(Ezk^)2{E_{net}} = \sqrt {{{({E_x}\hat i)}^2} + {{({E_y}\hat j)}^2} + {{({E_z}\hat k)}^2}}
Enet=(2)2+(−72)2+(34)2{E_{net}} = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - \dfrac{7}{2}} \right)}^2} + {{\left( {\dfrac{3}{4}} \right)}^2}}
Enet=4+494+916{E_{net}} = \sqrt {4 + \dfrac{{49}}{4} + \dfrac{9}{{16}}}
Enet=64+196+916{E_{net}} = \sqrt {\dfrac{{64 + 196 + 9}}{{16}}}
Enet=26916=14269{E_{net}} = \sqrt {\dfrac{{269}}{{16}}} = \dfrac{1}{4}\sqrt {269}
The net electric field at that point is 14269\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.