Question

The electric field at a point due to a point charge is $30N/C$ and the electric potential at that point is $15J/C$ respectively. What is the distance of the point from the charge and the magnitude of the charge?

Answer 1

Hint: We can solve this problem with the concept of electrostatics. Electric field is the area around a charge which exert a force on the other charge in its field. This field is created by the charge itself. The force exerted on the other charge may be an attractive or repulsive force according to the nature of the charge present. Same nature of charges repel each other and the opposite charges attract each other.

Complete answer:
Electric field at a point due to a point charge is $E = \dfrac{1}{{4\pi \varepsilon }}\dfrac{q}{{{r^2}}}$ or $\dfrac{V}{r}$, where $q$ is the magnitude of charge, $r$ is the distance from the given point to the point charge and $V$ is electric potential at that point. The dimensional formula for the electric field is $ML{T^{ - 3}}{A^{ - 1}}$.

Electric potential is defined as a work needed for moving a charge particle from infinite to a specific point inside the electric field without producing an acceleration. It is symbolized as $V$ and has a dimension $M{L^{ - 2}}{T^{ - 3}}{A^{ - 1}}$. The electric potential at the point in a static field is given as below: $V = \dfrac{1}{{4\pi \varepsilon }}\dfrac{q}{r}$.

So for a point charge $E = \dfrac{V}{r}$, $V = 15J/C, E = 30N/C$
Then the distance of point from the point charge is $r = \dfrac{{15}}{{30}} = 0.5m$
$
  V = \dfrac{q}{{4\pi \varepsilon r}} \\
  \Rightarrow q = 4\pi \varepsilon rV \\
  \Rightarrow q = \dfrac{1}{{9 \times {{10}^9}}} \times 0.5 \times 15 = 0.83 \times {10^{ - 9}}C \\
  \Rightarrow q = 0.83nC \\
$

The distance of the point from the charge and the magnitude of the charge is respectively $0.5m,0.83nC$.

Note:
We know that electric field exists only if there is electric potential difference, thus their relationship can be generally expressed as $E = - \dfrac{{dV}}{{dr}}$. from this relation we can say that electric field is the negative space derivative of electric potential. Negative shows that the work done is against the direction of the field.