Question

Let us assume that the electrostatic potential at particular points in space are given by $V={{x}^{2}}-2x$. The electrostatic field strength at $x=1$ will be given as,
$\begin{align}
  & A.0 \\
 & B.-2 \\
 & C.2 \\
 & D.4 \\
\end{align}$

Answer 1

Hint: The electrostatic field strength can be found by taking the negative of the derivative of the electrostatic potential with respect to the distance at space. Substitute the condition in it and take the derivative of this. Substitute the value $x$ in this equation. This will help you in answering this question.

Complete step by step answer:
The potential somewhere in space has been mentioned in the question as,
$V={{x}^{2}}-2x$
The position at which the value of the electric field strength should be calculated has been given as,
$x=1$
As we all know,
The electrostatic field strength can be found by taking the negative of the derivative of the electrostatic potential with respect to the distance at space. This can be mentioned in the form of equation as,
$E=-\dfrac{dV}{dx}$
Substituting the potential in it and find the derivative will give,
$E=-\dfrac{d\left( {{x}^{2}}-2x \right)}{dx}=-2x+2$
Now it has been mentioned that the field strength should be found at a position which is given as,
$x=1$
This can be substituted for the variable in the equation as,
$E=-2x+2=-2\times 1+2=0$
Hence the value of electric field strength at the position $x=1$ in space will be zero.

So, the correct answer is “Option A”.

Note: An electric potential is defined as the measure of work required to move a unit of electric charge from a specific point taken as reference to a particular position in an electric field without developing an acceleration. The reference point is generally taken as the Earth or a point at infinity. Even any point can be used in context of this. Electric field is defined as the electric force occurring per unit charge.