Question

The SI unit of electric potential is
A. \[V{m^{ - 1}}\]
B. C
C. \[N{C^{ - 1}}\]
D. V

Answer 1

Hint: In this question, we will first see the definition of electrical potential. The electrical potential is defined from electrical potential energy. After writing the expression for these terms, we will find the SI unit using these expressions.

Complete Step-by-Step solution:
Imagine a charge ‘q’ fixed at the origin of a coordinate system. We make another charge ${q_0}$, which we call ‘test charge’ and move it from ${{\text{r}}_a}{\text{ to }}{{\text{r}}_b}$ under the influence of the force due to q. The change in potential energy $\vartriangle U$ of this two- charge system is given as:
$\vartriangle U = \dfrac{1}{{4\pi { \in _0}}}{q_0}q\left( {\dfrac{1}{{{r_b}}} - \dfrac{1}{{{r_a}}}} \right)$----------- (1)
Electric potential difference($\vartriangle V$)
It is defined as the electrical potential energy difference per unit test charge.
$\vartriangle V = \dfrac{{\vartriangle U}}{{{q_0}}}$
${V_b} - {V_a} = \dfrac{{{U_b} - {U_a}}}{{{q_0}}}$
Taking ${U_a}$ = 0 for infinite initial separation.
The electrical potential at a point is written as:
$V = \dfrac{U}{{{q_0}}} = \dfrac{1}{{4\pi { \in _0}}}q\left( {\dfrac{1}{{{r_b}}}} \right)$ . --------(2)
The SI unit of potential that follows from equation 2 is the joule per coulomb. This combination is given the name of volt(V).
 1Volt = 1 joule /coulomb.
So, option D is correct.

Note- You should remember the formula for finding electrical potential at a point. A potential of zero at a point does not necessarily mean that the electric force is zero at that point. If the electrical potential is zero at a point, no net work is done by the electric force as the test charge moves from infinity to that point.