Question

The potential at a point $\left( x,0,0 \right)$ is given by $V=\left( \dfrac{1000}{x}+\dfrac{1500}{{{x}^{2}}}+\dfrac{500}{{{x}^{3}}} \right)$ . Find the field intensity at the point where $x=1m$
(A) ${{E}_{x}}=5500\widehat{i}V{{m}^{-1}}$
(B) ${{E}_{x}}=55\widehat{i}V{{m}^{-1}}$
(C) ${{E}_{x}}=550\widehat{i}V{{m}^{-1}}$
(D) ${{E}_{x}}=5.5\widehat{i}V{{m}^{-1}}$

Answer 1

Hint
In the given question, we have been asked to find the electric field intensity at a given location and the potential at a given distance from the origin along the x-axis is given in terms of the distance. We all know that the potential and the field intensity at any point is related by differential and integral property. Let’s see the detailed solution for a better understanding.
${{E}_{x}}=-\dfrac{dV}{dx}$

Complete step by step answer
As we discussed above, the electric field intensity and the potential are related by the integral and differential property, that is, the electric field intensity is equal to the differential of the electrical potential at any given location and the inverse also stands true; the electric potential at any given location is the integral of the electric field intensity at that point.
Mathematically, we can represent the electric field intensity as $({{E}_{x}})=-\dfrac{dV}{dx}$ where $\dfrac{dV}{dx}$ is the potential gradient, which is the rate of change of electric potential with distance.
We know the expression for the potential as $V=\left( \dfrac{1000}{x}+\dfrac{1500}{{{x}^{2}}}+\dfrac{500}{{{x}^{3}}} \right)$
Differentiating the expression for electric potential, we get
$\begin{align}
  & \dfrac{dV}{dx}=\dfrac{d}{dx}\left( \dfrac{1000}{x}+\dfrac{1500}{{{x}^{2}}}+\dfrac{500}{{{x}^{3}}} \right) \\
 & \Rightarrow \dfrac{dV}{dx}=\left( -\dfrac{1000}{{{x}^{2}}}-\dfrac{3000}{{{x}^{3}}}-\dfrac{1500}{{{x}^{4}}} \right) \\
 & \Rightarrow \dfrac{dV}{dx}=-\left( \dfrac{1000}{{{x}^{2}}}+\dfrac{3000}{{{x}^{3}}}+\dfrac{1500}{{{x}^{4}}} \right) \\
\end{align}$
We can now say that
$\begin{align}
  & {{E}_{x}}=-\dfrac{dV}{dx} \\
 & \Rightarrow {{E}_{x}}=\left( \dfrac{1000}{{{x}^{2}}}+\dfrac{3000}{{{x}^{3}}}+\dfrac{1500}{{{x}^{4}}} \right) \\
\end{align}$
We have obtained the general expression for the electric field at any given value of
We have been given a specific value $x=1m$
Substituting this value in the general expression, we get
$\begin{align}
  & {{E}_{x=1}}=\left( \dfrac{1000}{{{1}^{2}}}+\dfrac{3000}{{{1}^{3}}}+\dfrac{1500}{{{1}^{4}}} \right) \\
 & \Rightarrow {{E}_{x=1}}=5500\widehat{i}V{{m}^{-1}} \\
\end{align}$
Hence we can say that option (A) is the correct answer.

Note
The symbol denotes the direction of the electric field; it tells us that the electric field is directed along the positive x-axis. The potential at any given location is not directed and hence it does not require any direction vector. If the electric potential is a constant value, then its differentiation will give you a zero; hence constant potential does not produce an electric field.