Electric Potential (V)

Electric field intensity \overrightarrow{E} can be calculated by using either Coulomb’s law/Gauss’s law . when the charge distribution is symmetric another way of obtaining \overrightarrow{E} is from the electric scalar potential V

Assume a test charge Q_{t} at A in an Electric field, let points A and B are located at r_{A}and r_{B} units from the origin O,from Coulomb’s law the force acting on a test charge Q_{t} is \overrightarrow{F}= Q_{t}\overrightarrow{E}

The work done in moving a point charge Q_{t} along a differential length \overrightarrow{dl} is dW is given by dW = -\overrightarrow{F}.\overrightarrow{dl}

dW = -Q_{t}\overrightarrow{E}.\overrightarrow{dl}

so the total work done in moving a point charge Q_{t} from A to B is W=-Q_{t}\int_{A}^{B}\overrightarrow{E}.\overrightarrow{dl}

the direction of work done is always opposite to the direction of displacement.

where A is the initial point and B is the final point. Dividing the work done by the charge Q_{t} gives the potential energy per unit charge denoted by V_{AB},this is also known as potential difference between  the two points A and B.

Thus V_{AB} = \frac{W}{Q_{t}}= -\int_{A}^{B}\overrightarrow{E}.\overrightarrow{dl}

if we take B as initial point and A as final point , then V_{BA} = \frac{W}{Q_{t}}= -\int_{B}^{A}\overrightarrow{E}.\overrightarrow{dl}----Equation(1)

To derive the expression for V in terms of charge Q and distance r , we can use the concept of Electric field intensity \overrightarrow{E} produced by a charge Q, which is placed at a distance r

i.e, \overrightarrow{E} = \frac{Q}{4\pi \epsilon _{o}r^{2}}\overrightarrow{a_{r}}

from Equation(1) V_{BA} = -\int_{B}^{A}\overrightarrow{E}.\overrightarrow{dl}

V_{AB}= -\int_{r_{A}}^{r_{B}}\frac{Q}{4\pi \epsilon _{o}r^{2}}\overrightarrow{a_{r}}.dr \overrightarrow{a_{r}}  since \overrightarrow{dl}=dr.\overrightarrow{a_{r}}

V_{AB}= -\int_{r_{A}}^{r_{B}}\frac{Q}{4\pi \epsilon _{o}r^{2}}dr

V_{AB}= -\frac{Q}{4\pi \epsilon _{o}}\left _{r_{A}}^{r_{B}}

V_{AB}= -\frac{Q}{4\pi \epsilon _{o}}


similarly, V_{BA}=V_{A}-V_{B} 

where V_{A} and V_{B} are the scalar potentials at the points A and B respectively. If A is  located at \infty with respect to origin ,with zero potential V_{A} =0 and B is located at a distance r with respect to origin. then the work done in moving a charge from  A (infinity) to B is given by 

V_{AB} = \frac{Q}{4\pi \epsilon _{o}r_{B}}  here r_{B} = r 

\therefore V = \frac{Q}{4\pi \epsilon _{o}r} volts.

hence the potential at any point is the potential difference between that point and a chosen point at which the potential is zero. In other words assuming Zero potential at infinity .

The potential at a distance r from a point charge is the work done per unit charge by an external agent in transferring a test charge from infinity to that point.

i.e, V = -\int_{\infty }^{r}\overrightarrow{E}.\overrightarrow{dl}

So a point charge Q_{1} located at a point P with position vector \overline{r_{1}} then the potential at another point Q with a position vector \overline{r} is 

V_{at \overline{r}} = \frac{Q_{1}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{1}}\right |}

As like \overrightarrow{E} superposition principle is applicable to V also that is for n point charges Q_{1},Q_{2},Q_{3},Q_{4},........Q_{n} located at points with position vectors  \overline{r_{1}},\overline{r_{2}},\overline{r_{3}},.......\overline{r_{n}}

then the potential at \overline{r} is 

V_{at \overline{r}} = \frac{Q_{1}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{1}}\right |}+\frac{Q_{2}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{2}}\right |}+\frac{Q_{3}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{3}}\right |}+........+\frac{Q_{n}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{n}}\right |}


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Author: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.