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 $.