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.

Electric Potential at a point figure (1)

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

Electric Potential at a point figure (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.

Potential Difference Between Two Points figure (1)

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

Potential Difference Between Two Points figure (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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