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

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.

{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.