Question

In a region, the potential is represented by \[V\left( {x,y,z} \right) = 6x - 8xy - 8y + 6yz\], where V is in volts and x, y, z are in meters. The electric force experienced by a charge of \[2\] coulomb situated at point \[\left( {1,1,1} \right)\] is:
A. \[6\sqrt 5 {\rm{ N}}\]
B. \[30{\rm{ N}}\]
C. \[24{\rm{ N}}\]
D. \[4\sqrt {35} {\rm{ N}}\]

Answer 1

Hint: Electric field in any direction is equal to the negative of the partial derivative of potential in that direction. We will also use the relation of electric force at a point which is equal to the product of the electric field and charge at that point.

Complete step by step answer:
Given:
The potential at any point (x, y, z) is represented as \[V\left( {x,y,z} \right) = 6x - 8xy - 8y + 6yz\].
The value of the charge is \[q = 2{\rm{ C}}\].

We have to evaluate the electric force by charge q when it is situated at a point which is expressed as \[\left( {1,1,1} \right)\].

Let us write the expression for the electric field in the x-direction.
\[{E_x} = - \dfrac{{\partial V\left( {x,y,z} \right)}}{{\partial x}}\]

Substitute \[6x - 8xy - 8y + 6yz\] for \[V\left( {x,y,z} \right)\] in the above expression.
\[\begin{array}{c}
{E_x} = - \dfrac{{\partial \left( {6x - 8xy - 8y + 6yz} \right)}}{{\partial x}}\\
 = - 6 + 8y
\end{array}\]

Write the expression for the electric field in the y-direction.
\[{E_y} = - \dfrac{{\partial V\left( {x,y,z} \right)}}{{\partial y}}\]

Substitute \[6x - 8xy - 8y + 6yz\] for \[V\left( {x,y,z} \right)\] in the above expression.
\[\begin{array}{c}
{E_y} = - \dfrac{{\partial \left( {6x - 8xy - 8y + 6yz} \right)}}{{\partial y}}\\
 = 8x + 8 - 6z
\end{array}\]

Write the expression for the electric field in the z-direction.
\[{E_z} = - \dfrac{{\partial V\left( {x,y,z} \right)}}{{\partial z}}\]

Substitute \[6x - 8xy - 8y + 6yz\] for \[V\left( {x,y,z} \right)\] in the above expression.
\[\begin{array}{c}
{E_z} = - \dfrac{{\partial \left( {6x - 8xy - 8y + 6yz} \right)}}{{\partial z}}\\
 = 6y
\end{array}\]

Let us write the expression for the electric field in vector form.
\[\overrightarrow E = {E_x}\hat i + {E_y}\hat j + {E_z}\hat k\]

Substitute \[\left( { - 6 + 8y} \right)\] for \[{E_x}\], \[\left( {8x + 8 - 6z} \right)\] for \[{E_y}\] and \[6y\] for \[{E_z}\] in the above expression.
\[\overrightarrow E = \left( { - 6 + 8y} \right)\hat i + \left( {8x + 8 - 6z} \right)\hat j + \left( {6y} \right)\hat k\]

Substitute \[1\] for x, y and z in the above equation.
 \[\begin{array}{c}
\overrightarrow E = \left( { - 6 + 8 \cdot 1} \right)\hat i + \left( {8 \cdot 1 + 8 - 6 \cdot 1} \right)\hat j + \left( {6 \cdot 1} \right)\hat k\\
 = 2\hat i + 10\hat j + 6\hat k
\end{array}\]

We can calculate the magnitude of the electric field as below.
\[\begin{array}{c}
\left| {\overrightarrow E } \right| = \sqrt {{2^2} + {{10}^2} + {{\left( { - 6} \right)}^2}} {\rm{ N}}{{\rm{C}}^{ - 1}}\\
 = \sqrt {140} {\rm{ N}}{{\rm{C}}^{ - 1}}\\
 = 2\sqrt {35} \,{\rm{N}}{{\rm{C}}^{ - 1}}
\end{array}\]

We know that electric force is equal to the product of the electric field and charge.
\[F = qE\]

Substitute \[2\sqrt {35} \,{\rm{N}}{{\rm{C}}^{ - 1}}\] for E and \[2{\rm{ C}}\] for q in the above equation.
\[\begin{array}{c}
F = \left( {2{\rm{ C}}} \right)\left( {2\sqrt {35} \,{\rm{N}}{{\rm{C}}^{ - 1}}} \right)\\
 = 4\sqrt {35} \,{\rm{N}}
\end{array}\]

Therefore, the electric force experienced by a charge of \[2\] coulomb situated at the point \[\left( {1,1,1} \right)\] is \[4\sqrt {35} \,{\rm{N}}\]

So, the correct answer is “Option D”.

Note:
While writing the expressions of the electric field in respective directions, do not forget to add a negative sign. Also, it would be better if we remember how to calculate the magnitude of a vector for such kinds of problems.