Electric Field

# 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 $q_0$ 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 $r_1$ and $OB$ be $r_2$. At any instant, suppose a charge $q_0$ 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 $r_2$ to $r_1$, $W=-\int_{r_2}^{r_1} \frac{qq_0}{4πε_0x^2}\;dx$ $W=-\frac{qq_0}{4πε_0}\left[\frac{x^{-2+1}}{-2+1}\right]_{r_2}^{r_1}$ $W=\frac{qq_0}{4πε_0}\left[\frac{1}{r_1}-\frac{1}{r_2}\right]$ $W=\frac{qq_0}{4πε_0}\left[\frac{1}{r_1}-\frac{1}{r_2}\right]$ Thus, potential difference between $A$ and $B$ is, $V_A-V_B=\frac{W_{BA}}{q_0}$ $V_A-V_B=\frac{q}{4πε_0}\left[\frac{1}{r_1}-\frac{1}{r_2}\right]$

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