Electric potential (V)
The electric potential at a point in electric field is defined as amount of work done for bringing a unit positive charge from infinity to that point.
It is also defined as the work done per unit charge due to electric field at a point. The SI unit of electric potential is Joule/coulomb which is volt.
Consider a +Q charge placed at a point 'O' and a unit +ve charge at point 'A' at a distance 'r' from the charge.
The electromagnetic force between both charge is given as
\( \overrightarrow{F} = \frac{1}{4\pi\varepsilon_{0}} \frac{Q}{r^{2}} \)
\( \therefore \overrightarrow{F}=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r^{2}} \) … (i)
This is also the intensity at point A due to charge \( q' \).
If the test charge brings a small distance \( dr \) against the force. Then the work done can be written as:
Now, The total work done for the bringing of unit +ve charge can be obtained by integrating above equations over limits \( \infty \) to \( r \).
So, \( W = \int dw \)
$$ \therefore \quad W = \int\limits_{\infty}^{r} - F \cdot dr $$
$$ \therefore W = \int\limits_{\infty}^{r}\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r^{2}} dr $$
$$ \therefore \quad W = -\frac{1}{4\pi\varepsilon_{0}} Q \int\limits_{\infty}^{r}\frac{1}{r^{2}} dr $$
$$ \therefore W = -\frac{Q}{4\pi\varepsilon_{0}}\int r^{-2}dr $$
$$ \therefore \quad W = -\frac{Q}{4\pi\varepsilon_{0}}\left[\frac{r^{-2+1}}{-2+1}\right]_{\infty}^{r} $$
$$ \therefore W = \frac{-Q}{4\pi\varepsilon_{0}} \left[ \frac{r^{-1}}{-1} \right]_{\infty}^{r} $$
$$ \text{or}\ W=\frac{-Q}{4\pi\varepsilon_{0}}\left(\frac{r^{-1}}{-1}-\frac{\infty^{-1}}{-1}\right) $$
$$ \text{or}\ W=\frac{-Q}{4\pi\varepsilon_{0}}\left(-\frac{1}{r}-0\right) $$
$$ \therefore W = \frac{Q}{4\pi\varepsilon_{0}r} = \frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r} $$
This is the relation for electric potential at a point in the electric field. It is a scalar quantity.
Definition of 1 volt
A point in an electric field is said to have 1 V of potential if 1 joule of work is to be done in bringing 1 coulomb of positive charge from infinity to that point.
Definition of 1 stat coulomb
Electric potential energy
The electric potential energy at a point in electric field is defined as the work done by bringing a charge from infinity to that point.
According to def of potential,
or \( W = \frac{1}{4\pi\varepsilon_{0}} \frac{Q}{r} \times q \)
\( W = \frac{Qq}{4\pi\varepsilon_{0}r} \)
This is the relation for electric potential energy.
Potential difference
Consider a point A and B having distance \( r_{1} \) and \( r_{2} \) from a point charge +Q. The potential due to these charges at point A and B is given as:
\( V_{a} = \frac{1}{4\pi\varepsilon_{0}} \frac{Q}{r_{1}} \) and
\( V_{b} = \frac{1}{4\pi\varepsilon_{0}} \frac{Q}{r_{2}} \)
Now, the difference in potential at point A to B can be determined as
$$ V_{ab} = \frac{1}{4\pi\varepsilon_{0}} \frac{Q}{r_{1}} - \frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r_{2}} $$
This is the potential difference between point \( A' \) to \( B' \) due to given charge +Q.
Potential gradient \(\left(\frac{dV}{dr}\right)\)
The change in potential per unit length is defined as potential gradient. Let potential at a point due to a charge is given as:
\( \therefore V = \frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r} \)
Now, According the def of potential gradient, we can written as:
\( \frac{dv}{dr} = \lim_{\Delta r \to 0} \frac{\Delta v}{\Delta r} \)
Now, differentiating potential with respect to distance (dr), the potential gradient can be obtained as
\( \frac{dv}{dr} = \frac{d\left(\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r}\right)}{dr} \)
\( \therefore\frac{dv}{dr}=\frac{Q}{4\pi\varepsilon_{0}}\times(-1)\times r^{-2} \)
\( \therefore\frac{dv}{dr}=\frac{-Q}{4\pi\varepsilon_{0}r^{2}}=-E\Rightarrow E=-\frac{dv}{dr} \)
Here, the electric field intensity can also be defined as negative of the potential gradient at a point.
Electron volt (eV)
To measure the energy of charge particles in joules will be quite large unit and hence inconvenient. Electron volt is the unit of energy to represent the small energy of charge particles.
1 electron volt of energy is defined as the energy gained by an electron when accelerated through potential difference of 1 volt. Here
Potential difference (pd) = \( \frac{\text{work done}}{\text{charge}} \)
work = p.d. × charge
for one electron volt (eV),
\( p \cdot d = 1\ \text{V} \)
and charge = \( 1.6 \times 10^{-19} C \)
\( W = 1 \times 1.6 \times 10^{-19} \)
\( = 1.6 \times 10^{-19} \) joule
\( = 1 \text{ eV} \)
\( 1\text{eV} = 1.6 \times 10^{-19}\text{J} \)
Potential due to multiple charges
Since, the electric potential is a scalar quantity, the total potential at a point due to multiple charge is the algebraic sum of individual potential at that point.
Let, \( V_{1} \), \( V_{2} \) and \( V_{3} \) are the potential at point A due to charges \( q_{1}, q_{2}, q_{3} \) respectively. Then, the total potential at point A can be written as:
\( V = V_{1} + V_{2} + V_{3} \)
Equipotential surface
The surface on which the potential at every point is same or there is no potential difference between any two points then the surface is called equipotential surface. For example: equipotential surface due to a charge can be represented by a diagram given as:
Fig.1: Showing equipotential surface / Equip potential surfaces.