Question

A uniform electric field having magnitude ${{\text{E}}_0}$ and direction along the positive x-axis exists. If the potential V is zero at x = 0, then what will be its value at x = +x?
\[
  {\text{A}}{\text{. V}}\left( x \right) = {x^2}{{\text{E}}_0} \\
  {\text{B}}{\text{. V}}\left( x \right) = - {x^2}{{\text{E}}_0} \\
  {\text{C}}{\text{. V}}\left( x \right) = - x{{\text{E}}_0} \\
  {\text{D}}{\text{. V}}\left( x \right) = x{{\text{E}}_0} \\
 \]

Answer 1

Hint: Here, we will proceed by defining the term electric field. Then, we will be using the general relation between electric field and the potential and we will integrate it in order to find the required potential as a function of x.
Formula Used: ${\text{E}} = - \dfrac{{d{\text{V}}}}{{dx}}$.

Complete Step-by-Step solution:
 When charge is present in any form, the electric field is an electrical property associated with each point in space. The magnitude of the electric field is represented by the value of E, which is called the strength of the electric field or the intensity of the electric field, or simply the electric field. Knowledge of the value of the electrical field at one point, without any clear knowledge by what the field has generated, is all that is required to decide what will happen to electrical charges at that point.
According to the relation between electric field and potential,
$
  {\text{E}} = - \dfrac{{d{\text{V}}}}{{dx}} \\
  {\text{E}}dx = - d{\text{V}} \\
 $
Here, replacing E with ${{\text{E}}_0}$ in the above equation, we get
${{\text{E}}_0}dx = - d{\text{V }} \to {\text{(1)}}$
The electrical field is in the direction where potential decreases most steeply. The magnitude of the electric field is given by changing the magnitude of potential at the point to the equipotential surface per unit displacement normal.
Given, when x = 0, potential difference = 0
Magnitude of uniform electric field = ${{\text{E}}_0}$
Let the potential at x = +x be V
Now, integrating equation (1) on both sides having limits of x from 0 to x and limits of V from 0 to V
\[\int_0^x {{{\text{E}}_0}dx} = - \int_0^{\text{V}} {d{\text{V}}} \]
It is given that the electric field \[{{\text{E}}_0}\] is uniform (i.e., constant and independent of x) so it can be taken outside of the above integral
 \[
   \Rightarrow {{\text{E}}_0}\int_0^x {dx} = - \int_0^{\text{V}} {d{\text{V}}} \\
   \Rightarrow {{\text{E}}_0}\left[ x \right]_0^x = - \left[ {\text{V}} \right]_0^{\text{V}} \\
   \Rightarrow {{\text{E}}_0}\left[ {x - 0} \right] = - \left[ {{\text{V}} - 0} \right] \\
   \Rightarrow {{\text{E}}_0}\left( x \right) = - {\text{V}} \\
   \Rightarrow {\text{V}} = x{{\text{E}}_0} \\
   \Rightarrow {\text{V}}\left( x \right) = x{{\text{E}}_0} \\
 \]
Therefore, potential which is a function of x is given by \[x{{\text{E}}_0}\] i.e., \[{\text{V}}\left( x \right) = x{{\text{E}}_0}\]
Hence, option D is correct.

Note- For a uniform electric field, the field lines are parallel and uniformly spaced, as the field strength does not change. Uniform fields are generated by creating a potential difference between two conductive plates positioned at a certain distance from each other.