Question

The electric field at a point due to a point charge is $20\,N/C$ and electric potential at that point is $10J/C$ . Calculate the distance of the point from the charge and the magnitude of the charge.

Answer 1

Hint : In this solution, we will use the formula of electric field and electric potential due to a point charge at a distance. We will then take the ratio of these formulae to find the distance of the point from the charge and then its magnitude.

Formula used: In this solution, we will use the following formulae,
Electric field due to a point charge $E = \dfrac{{kq}}{{{r^2}}}$ where $q$ is the magnitude of the charge and $r$ is the distance of the point from the charge
Electric potential due to a point charge $V = \dfrac{{kq}}{r}$

Complete step by step answer
We’ve been given that the electric field at a point due to a point charge is $20\,N/C$ and electric potential at that point is $10J/C$ . We know that the electric field and potential due to a point charge at a distance $r$ can be calculated respectively as
$\Rightarrow E = \dfrac{{kq}}{{{r^2}}}$
And
$\Rightarrow V = \dfrac{{kq}}{r}$
Taking the ratio of both these quantities, we can write
$\Rightarrow \dfrac{E}{V} = \dfrac{{\dfrac{{kq}}{{{r^2}}}}}{{\dfrac{{kq}}{r}}} = \dfrac{1}{r}$
Alternatively, we can write the distance as
$\Rightarrow r = \dfrac{V}{E}$
$\Rightarrow r = \dfrac{{10}}{{20}} = 0.5\,m$
Thus the distance of the point from the charge is $0.5\,m$
Now that we know the distance of the charge and between the point, we can calculate the magnitude of charge using the relation of electric potential as
$\Rightarrow V = \dfrac{{kq}}{r}$
Substituting the value of $k = 9 \times {10^9}$ , and $r = 0.5$ , we get
$\Rightarrow 10 = \dfrac{{9 \times {{10}^9} \times q}}{{0.5}}$
$\therefore q = 0.55 \times {10^{ - 9}}\,C$
Hence the value of the charge is $q = 0.55 \times {10^{ - 9}}\,C$ .

Note
When taking the ratio of electric field and the electric potential, we must be careful that both these quantities are calculated at a point which is at the same distance from the point charge itself. We can alternatively calculate the distance using the formula for the electric field as
$\Rightarrow E = \dfrac{{kq}}{{{r^2}}}$
Substituting the value of $k = 9 \times {10^9}$ , and $r = 0.5$ , we get
$\Rightarrow 20 = \dfrac{{9 \times {{10}^9} \times q}}{{{{0.5}^2}}}$
Which gives us
$\Rightarrow q = 0.55 \times {10^{ - 9}}\,C$.