Question

The electric potential at a point is $V=-5x+3y+\sqrt{15}z$ .Find the magnitude of the electric field intensity.

Answer 1

Hint: The electric field at a point is defined as the gradient of the electric potential at that point. The variation of potential at any point in the space is given as $V=-5x+3y+\sqrt{15}z$. Hence by taking the gradient of the above electric potential we will be able to determine the electric field at that point.
Formula used:
$\begin{array}{rl}& Gradient(V)=\bar{\mathrm{\nabla }}.V\\ & E=\bar{\mathrm{\nabla }}.V=-({\displaystyle \frac{\delta V}{\delta x}}+{\displaystyle \frac{\delta V}{\delta y}}+{\displaystyle \frac{\delta V}{\delta z}})\end{array}$

Complete answer:
The potential due to some charge configuration is given by $V=-5x+3y+\sqrt{15}z$. From this expression we can conclude that the charge distribution has its influence along the entire space i.e. it has components along all the axes. For a point in space having a particular value of potential the electric field is defined as the negative gradient of electric potential. This can be mathematically represented as,
$E=\bar{\mathrm{\nabla }}.V=-({\displaystyle \frac{\delta V}{\delta x}}+{\displaystyle \frac{\delta V}{\delta y}}+{\displaystyle \frac{\delta V}{\delta z}})$
Substituting for the potential at appoint due to the charge distribution in the above equation we get,
$\begin{array}{rl}& E=-({\displaystyle \frac{\delta V}{\delta x}}+{\displaystyle \frac{\delta V}{\delta y}}+{\displaystyle \frac{\delta V}{\delta z}})\\ & E=-({\displaystyle \frac{\delta (-5x+3y+\sqrt{15}z)}{\delta x}}+{\displaystyle \frac{\delta (-5x+3y+\sqrt{15}z)}{\delta y}}+{\displaystyle \frac{\delta (-5x+3y+\sqrt{15}z)}{\delta z}})\end{array}$
Since the derivative of a function with respect to it running variable is non zero, the above equation becomes,
$\begin{array}{rl}& E=-({\displaystyle \frac{\delta (-5x+3y+\sqrt{15}z)}{\delta x}}+{\displaystyle \frac{\delta (-5x+3y+\sqrt{15}z)}{\delta y}}+{\displaystyle \frac{\delta (-5x+3y+\sqrt{15}z)}{\delta z}})\\ & \Rightarrow E=-({\displaystyle \frac{-\delta 5x}{\delta x}}+{\displaystyle \frac{\delta 3y}{\delta x}}+{\displaystyle \frac{\delta \sqrt{15}z}{\delta x}}+{\displaystyle \frac{-\delta 5x}{\delta y}}+{\displaystyle \frac{\delta 3y}{\delta y}}+{\displaystyle \frac{\delta \sqrt{15}z}{\delta y}}+{\displaystyle \frac{-\delta 5x}{\delta z}}+{\displaystyle \frac{\delta 3y}{\delta z}}+{\displaystyle \frac{\delta \sqrt{15}z}{\delta z}})\\ & \Rightarrow E=-({\displaystyle \frac{-5\delta x}{\delta x}}+0+{\displaystyle \frac{3\delta y}{\delta y}}+{\displaystyle \frac{\sqrt{15}\delta z}{\delta z}})\\ & \Rightarrow E=-(-5+3+\sqrt{15})\\ & \Rightarrow E=5-3-\sqrt{15}\\ & \Rightarrow E=2-\sqrt{15}\end{array}$
Therefore the electric field at a point in space of influence of the charge configuration is equal to $(2-\sqrt{15})NC$ .

Note:
If we analyze the above answer, i.e. the value of the electric field is negative. Hence we can conclude that the field is generated by some negative charge distribution. Hence we can say that the potential due to the system keeps on decreasing as we move towards the negative charge distribution. It is to be noted that the potential is always relative to the negative and the positive charge.