: Find the relation between ‘E’ and ‘V’:

Hint: Electric field is a conceptual field that is assumed to be produced from a positive charge and terminate on a negative charge. This concept doesn’t have physical significance but is used to understand many physical phenomena through intuition.
Formula used:
 $dV = -Edr$

Complete answer:
In the equation, $dV = -Edr$, ‘V’ is the potential which is defined as the work done in bringing a unit charge from infinity to a particular point. ‘E’ stands for the electric field which is defined as the region in which a charge could feel the influence of any other charge. ‘r’ is the distance moved in the direction of ‘E’.
In the figure, we could see that potential is decreasing as we move away from the centre. The change in potential is according to rectangular hyperbola which says that the product of two quantities. We observe that the product of potential and distance is constant (equal to 6).
$\implies Vd = 6 = constant$
$\implies V = \dfrac{constant}{d}$
One can observe that this behavior is shown by a point charge ‘q’ as $V = \dfrac{Kq}{d} = \dfrac{constant}{d}$
Hence we can directly say that ‘E’ and distance are related as $E = \dfrac{Kq}{d^2} = \dfrac{constant}{d^2}$
Or we could use $dV = -Edr$
$ V = \dfrac{constant}{d}$
$\implies dV = -\dfrac{constant}{d^2}\delta d = -E \delta d$
$\implies E = \dfrac{constant}{d^2}$
Now, from $V = \dfrac{Kq}{d} = \dfrac{constant}{d}$, putting $d = \dfrac{constant}{V}$ in $E = \dfrac{constant}{d^2}$, we get;
$E = \dfrac{V^2}{constant}$
As $constant = 6$
Therefore $V^2 = 6E$, which is the required relation between ‘V’ and ‘E’.

Note:
Students should always remember that electric potential is a scalar quantity whereas the electric field is a vector quantity. The direction of electric field lines is always along the direction of decreasing electric potential. This is because electric fields always travel from high to low potential, which could be understood by $dV = -Edr,\ V = -\int E.dr$.