Electrostatic potential due to a point charge $q$ at distance $r$ in vacuum is:
- $\dfrac{q r}{4\pi\varepsilon_0}$
- $\dfrac{q}{4\pi\varepsilon_0 r^3}$
- $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$
- $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$
Show answer and solution
Correct answer: C. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$
Potential V is the work per unit charge to bring a unit positive charge from infinity to the point: $V = q/(4\pi\varepsilon_0 r)$.
The $1/r^2$ form is the field E, not V.