# Electric Potential

## Electric Potential at a Point

​Electric potential at a point is defined as the amount of work done in moving a unit charge from infinity to that point. i.e. $V=\frac{W_{∞A}}{q_0}$ Let’s find out the electric potential at a point A which lies inside the region of electric field produced by a source charge q which is placed at point O. Let the distance between OA be r. At any instant, suppose a charge q₀ is at point P which is x distance apart from the source charge.

​The force experienced by the test charge is $F=\frac{qq_0}{4πε_0x^2}$

​If the test charge is displaced through infinitesimally small distance dx, then, small amount of work done is, $dW=-Fdx$ There is negative sign because work is done against electrostatic force. $∴dW=-\frac{qq_0}{4πε_0x²}dx$ Integrating within limits from infinite to r, $W=-\int_∞^r \frac{qq_0}{4πε_0x²}dx$ $W=-\frac{qq_0}{4πε_0}\left[\frac{x^{-2+1}}{-2+1}\right]_∞^r$ $W=\frac{qq_0}{4πε_0}\left[\frac{1}{r}-\frac{1}{∞}\right]$ $W=\frac{qq_0}{4πε_0r}$ Thus, the electric potential at A is, $V=\frac{W_{∞A}}{q_0}$ $V=\frac{q}{4πε_0r}$

## Potential Difference Between Two Points

​Potential Difference between two points is defined as the amount of work done in moving a unit charge from one point to that another point. i.e. $V_A-V_B=\frac{W_{BA}}{q_0}$ Let’s find out the electric potential between two points A and B. Place a source charge at a point O. Let the distance between OA be r₁ and OB be r₂. At any instant, suppose a charge q₀ is at point P which is x distance apart from the source charge.

​​The force experienced by the test charge is $F=\frac{qq_0}{4πε_0x^2}$

​If the test charge is displaced through infinitesimally small distance dx, then, small amount of work done is, $dW=-Fdx$ $∴dW=-\frac{qq_0}{4πε_0x^2}dx$ Integrating within limits from r₂ to r₁, $W=-\int_{r₂}^{r₁} \frac{qq_0}{4πε_0x^2}dx$ $W=-\frac{qq_0}{4πε_0}\left[\frac{x^{-2+1}}{-2+1}\right]_{r₂}^{r₁}$ $W=\frac{qq_0}{4πε_0}\left[\frac{1}{r₁}-\frac{1}{r₂}\right]$ $W=\frac{qq_0}{4πε_0}\left[\frac{1}{r₁}-\frac{1}{r₂}\right]$ Thus, potential difference between A and B is, $V_A-V_B=\frac{W_{BA}}{q_0}$ $V_A-V_B=\frac{q}{4πε_0}\left[\frac{1}{r₁}-\frac{1}{r₂}\right]$