Calculate the potential at a point due to a charge of $4 \times {10^{ - 7}}C$ located at $0.09m$ away.

Hint: This question can easily be solved if we know the relation between the electric potential due to a charge and the value of the electric field created by it. Also, we need to relate the values of electric field and electric potential. After that, we need to solve the equations, to get the required value of the electric potential. Also, we need to know about the properties of electric potential and charges in order to finally conclude with the solution.

Complete step by step solution:
In the question, the value of charge is given as,\[q = 4 \times {10^{ - 7}}C\]
And the distance where it is kept is given as,$r = 0.09m$
Now, we know the relation between Electric field and electric potential as,
$E = - \dfrac{{dV}}{r}$ ………………… (i)
And the relation between Electric field and charge is given as,
$E = \dfrac{{kq}}{{{r^2}}}$ …………….. (ii)
Where, $'k'$is a constant and its value is given as $9 \times {10^9}N{m^2}{C^{ - 2}}$
Now, we need to equate equation (i) and (ii).
Therefore, by equating the above equations, we will get,
$\dfrac{V}{r} = \dfrac{{kq}}{{{r^2}}}$
$ \Rightarrow V = \dfrac{{kqr}}{{{r^2}}}$,
$ \Rightarrow V = \dfrac{{kq}}{r}$
$ \Rightarrow V = \dfrac{{9 \times {{10}^9} \times 4 \times {{10}^{ - 7}}}}{{0.09}}$
$\therefore V = 4 \times {10^4}V$

Hence, the required value of the electric potential is $4 \times {10^4}V$.

Note: We define electric field as the electric force experienced per unit charge. Mathematically, it can also be represented as,$E = \dfrac{F}{q}$. Electric field is a property which is associated with the space between the charges. We find that the electric field lines flow from positive charge negative to negative charge.
Electric potential is defined as the work done per unit charge in bringing the charge from one reference point to the other point. Mathematically, it can be represented as,$V = \dfrac{W}{q}$.