Question

The potential at point P due to a single point charge \[q_1\] at distance \[r_{1P}\] is expressed as:

पर्याय

• \[V_1 = \frac{1}{4\pi\varepsilon_0}\cdot q_1 r_{1P}\]

• \[V_1 = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1}{r_{1P}^2}\]

• \[V_1 = \frac{1}{2\pi\varepsilon_0}\cdot\frac{q_1}{r_{1P}}\]

• \[V_1 = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1}{r_{1P}}\]

Answer 1

The potential due to a point charge varies inversely with the first power of distance: \[V_1 = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1}{r_{1P}}\]. The constant \[\frac{1}{4\pi\varepsilon_0}\] is Coulomb's constant, and the result is a signed scalar depending on the sign of the charge.