Question

Electric potential is:
A. A scalar quantity
B. A vector quantity
C. Neither scalar nor vector
D. Sometimes scalar and sometimes vector

Answer 1

Hint: Here, in this question, we will first define the electrostatic potential. Now, to find whether the electrostatic potential is scalar or vector we will use the force formula according to Coulomb’s law. Here, the electrostatic potential will depend on charge and distance.

Complete step by step answer:
An electric potential is defined as the amount of work done to move a charge from the reference point to a specific point without any acceleration in an electric field.
Now, we will calculate the formula of electric potential due to a point charge to know whether the electric potential is scalar or vector.
Now, consider a positive charge $+q$ in an electric field. Let $P$ be the point that is at a distance $r$ from the charge where the electric potential is to be calculated which is shown below.

Now, the work is done to move the charge from its position to the point $P$, which is as given below
$dW=F.dx$
Now, the total work done can be calculated by integrating the above equation as shown below
$\int dW=\int F.dx$
$W=\int F.dx$
Where $F$ is the force acting on the charge and $dx$ is the infinitesimal small distance.
Now, according to Coulomb’s law, the force is given by
$F={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{q}{r^2}}$
Therefore, the work done will become
$W=\int \left({\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{q}{r^2}}\right).dx$
As we are working against the force acting on the charge, then the force will be positive.
Now, the work done to move a charge at a $dx$ infinitesimal distance, hence, we will calculate the work done by integrating between the limits $r$ to $\mathrm{\infty }$ as given below
$W=\underset{r}{\overset{\mathrm{\infty }}{\int }}\left({\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{q}{r^2}}\right).dx$
Now, using the formula, $\int {\displaystyle \frac{1}{x^n}}={\displaystyle \frac{x^{n+1}}{n+1}}$ , we get
$W=-{\left({\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{Q}{r}}\right)}_r^{\mathrm{\infty }}$
$\Rightarrow W=-\left({\displaystyle \frac{Q}{4\pi {\epsilon }_0}}\left({\displaystyle \frac{1}{\mathrm{\infty }}}-{\displaystyle \frac{1}{r}}\right)\right)$
$\Rightarrow W=-\left(-{\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{Q}{r}}\right)$
$\Rightarrow W={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{Q}{r}}$
Now, the electrostatic potential is taken as work done, hence, the electric potential is given by
$U={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{Q}{r}}$
Therefore, we can say that the electrostatic potential is dependent on charge and distance of charge from the point where the electrostatic potential is to be calculated. As the charge and the distance are scalar quantities. Therefore, the electrostatic potential is the scalar quantity.

So, the correct answer is “Option A”.

Note:
We know that both the point charge and unit charge are positive. So, we have to work against the force to move the charge. Therefore, the electric potential will be positive.
Now, if the point charge is negative. Therefore, the work down will be along with the force. Hence, the electrostatic potential will be negative.