The electric potential is $v={{x}^{2}}-2x$. What is the electric field strength at x=1?
A. 2
B. Zero
C. 2
D. 4

Hint: The relation between electric field and electric potential gives an answer to this question. Electric field is the negative gradient of electric potential so the component of electric field in any direction is the negative of potential gradient in that direction.

Complete step by step answer:
Electric field strength: Electric field strength at a point is the force that a unit positive test charge experiences when placed at that point.
Electric potential: It is the work done in bringing a unit positive charge from infinite distance to a required point.
Now let us see the relationship between them,
The relation between electric and magnetic field is given by the formula
$E=-\dfrac{dv}{dx}$…... (1)
Where, E – electric field strength and v – electric potential
Therefore, the electric field is the negative derivative of potential with respect to position.
Now let us differentiate given ‘v’ with respect to ‘x’, we get
$E=-\dfrac{d({{x}^{2}}-2x)}{dx}$
On differentiating E is,
     $=-(2x-2)$
Now multiply -ve sign
 $E=-2x+2$….. (2)
Thus, this is the electric field due to potential $v={{x}^{2}}-2x$
Now to field strength at x=1, substitute x=1 in (2) we get
$\begin{align}
  & E=-2(1)+2 \\
 & E=0 \\
\end{align}$
Thus, the electric field at x=1 due to given potential v is zero. Correct option is B.

Note: Students may get confused with the sign in formula while answering these types of questions. The electric field is -ve differentiation of electric potential.
As electric field is the differentiation of potential, we can express potential as the integration of electric field i.e., $v=-\int{E}dx$.
As we find electric fields along ‘x’ we also find along the ‘y’ and ‘z’ axis also by differentiating ‘v’ (electric potential) with respect to ‘y’ and ‘z’ axis, respectively.