The electric potential at a point in free space due to a charge ‘Q’ coulomb is $Q\times {{10}^{11}}V$. The electric field at the point is
A. $12\pi {{\in }_{0}}Q\times {{10}^{22}}V{{m}^{-1}}$
B. $4\pi {{\in }_{0}}Q\times {{10}^{22}}V{{m}^{-1}}$
C. $12\pi {{\in }_{0}}Q\times {{10}^{20}}V{{m}^{-1}}$
D. $4\pi {{\in }_{0}}Q\times {{10}^{20}}V{{m}^{-1}}$

Hint: The electric potential is given for a charge ‘Q’. It is based on Coulomb's law of electrostatic forces. We shall apply the given values in Coulomb’s law to find the electric field at the point for the charge ‘Q’.

Complete step by step answer:
The formula is given by Coulomb’s law of electrostatic force which states that, the force of attraction or repulsion between the point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them. The direction of force is always along the line that joins the two charges.
$F=k\dfrac { { Q }^{ 2 } }{ { r }^{ 2 } }$
Where, k is the proportionality constant and
     r is the assumed distance between the charges.
Now, let us substitute the value of ‘k’ in the above-mentioned formula. We get,
$F=\dfrac{1}{4\pi {{\in }_{0}}}\dfrac{{{Q}^{2}}}{{{r}^{2}}}$
Where $\dfrac{1}{4\pi {{\in }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{c}^{-2}}$.
We also know, $V=\dfrac{Q}{4\pi {{\in }_{0}}r}=Q\times {{10}^{11}}$ (from the question)
Which means $4\pi {{\in }_{0}}r={{10}^{-11}}$
On substituting the values in the formula, we get
$E=\dfrac{Q}{4\pi {{\in }_{0}}{{r}^{2}}}$
$E=\dfrac{Q\times 4\pi {{\in }_{0}}}{{{\left( 4\pi {{\in }_{0}}r \right)}^{2}}}$
$E=\dfrac{Q\times 4\pi {{\in }_{0}}}{{{\left( {{10}^{-11}} \right)}^{2}}}$
$E=4\pi {{\in }_{0}}Q\times {{10}^{22}}V{{m}^{-1}}$
Therefore, the correct answer for the given question is option (B).

Note: If the charge is present in a medium of permittivity $\in $, the magnitude of force of charge will be, ${{F}_{m}}=\dfrac{1}{4\pi \in }\dfrac{{{Q}^{2}}}{{{r}^{2}}}$ where Fm is the force of the charge in a medium.