Electric Potential for Point Charge and Dipole, Potential Difference, Equipotential Surfaces

Electric Potential due to a Point Charge

In electromagnetism, electric potential is a scalar quantity that represents the amount of electric potential energy per unit charge at a specific point in an electric field. It is defined as the work done per unit charge in moving a test charge from infinity to that point against the electric field. The SI unit of electric potential is the Volt (V), where 1 Volt equals 1 Joule per Coulomb.

Consider a point charge 'Q' placed at the origin. We want to find the electric potential 'V' at a point 'P' located at a distance 'r' from the charge. The electric field 'E' at point P due to charge Q is given by Coulomb's law:

$E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$

where $\epsilon_0$ is the permittivity of free space. The direction of the electric field is radially outward if Q is positive and radially inward if Q is negative.

The electric potential at point P is the work done by an external agent to bring a unit positive test charge from infinity to point P. This work done is equal to the negative of the work done by the electric field. The potential V at a distance r from a point charge Q is given by:

$V = \frac{W}{q_0} = \int_{\infty}^{r} \vec{E} \cdot d\vec{l}$

Since $\vec{E}$ and $d\vec{l}$ are in opposite directions (if we move from infinity towards the charge), $\vec{E} \cdot d\vec{l} = -E dr$.

$V = \int_{\infty}^{r} (-E dr) = -\int_{\infty}^{r} \frac{1}{4\pi\epsilon_0} \frac{Q}{r'^2} dr'$

$V = -\frac{Q}{4\pi\epsilon_0} \int_{\infty}^{r} \frac{1}{r'^2} dr'$

$V = -\frac{Q}{4\pi\epsilon_0} \left[-\frac{1}{r'}\right]_{\infty}^{r}$

$V = -\frac{Q}{4\pi\epsilon_0} \left(-\frac{1}{r} - (-\frac{1}{\infty})\right)$

Since $\frac{1}{\infty} = 0$,

$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$

This is the electric potential at a distance 'r' from a point charge 'Q'.

  • If Q is positive, V is positive.
  • If Q is negative, V is negative.
  • The potential decreases as the distance 'r' increases.
  • The potential is inversely proportional to the distance from the charge.

For a system of 'n' point charges $q_1, q_2, ..., q_n$ located at distances $r_1, r_2, ..., r_n$ respectively from a point P, the total electric potential at P is the algebraic sum of the potentials due to each charge:

$V_{total} = V_1 + V_2 + ... + V_n = \sum_{i=1}^{n} \frac{1}{4\pi\epsilon_0} \frac{q_i}{r_i}$

Electric Potential due to an Electric Dipole

An electric dipole consists of two equal and opposite point charges ($+q$ and $-q$) separated by a small distance (2a). The product of the charge and the distance between them is called the dipole moment (p), which is a vector quantity directed from the negative charge to the positive charge.

$p = q \times (2a)$

We will consider two special cases for the electric potential due to a dipole:

1. Potential at an Axial Point (on the dipole axis)

Let the dipole be placed along the x-axis with the center at the origin. The charges are at $(-a, 0)$ and $(a, 0)$. Consider a point P on the axis at a distance 'r' from the center of the dipole. The distance of P from $+q$ is $(r-a)$ and from $-q$ is $(r+a)$.

The potential at P due to $+q$ is $V_1 = \frac{1}{4\pi\epsilon_0} \frac{q}{(r-a)}$.

The potential at P due to $-q$ is $V_2 = \frac{1}{4\pi\epsilon_0} \frac{-q}{(r+a)}$.

The total potential at P is the algebraic sum:

$V_{axial} = V_1 + V_2 = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{(r-a)} - \frac{q}{(r+a)} \right)$

$V_{axial} = \frac{q}{4\pi\epsilon_0} \left( \frac{(r+a) - (r-a)}{(r-a)(r+a)} \right)$

$V_{axial} = \frac{q}{4\pi\epsilon_0} \left( \frac{2a}{r^2 - a^2} \right)$

Since $p = q \times 2a$, we have:

$V_{axial} = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2 - a^2}$

For a short dipole, $a \ll r$, so $r^2 - a^2 \approx r^2$.

$V_{axial} \approx \frac{1}{4\pi\epsilon_0} \frac{p}{r^2}$

2. Potential at a Broadside-on Point (on the perpendicular bisector)

Consider a point P on the perpendicular bisector of the dipole at a distance 'r' from the center. The distance of P from both $+q$ and $-q$ is $\sqrt{r^2 + a^2}$.

The potential at P due to $+q$ is $V_1 = \frac{1}{4\pi\epsilon_0} \frac{q}{\sqrt{r^2 + a^2}}$.

The potential at P due to $-q$ is $V_2 = \frac{1}{4\pi\epsilon_0} \frac{-q}{\sqrt{r^2 + a^2}}$.

The total potential at P is:

$V_{broadside} = V_1 + V_2 = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{\sqrt{r^2 + a^2}} - \frac{q}{\sqrt{r^2 + a^2}} \right) = 0$

So, the electric potential at any point on the perpendicular bisector of a dipole is zero.

3. Potential at an Arbitrary Point

Let P be a point at a distance 'r' from the center of the dipole, and let $\theta$ be the angle between the position vector of P and the dipole moment vector.

The potential at an arbitrary point P is given by:

$V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2 - a^2 \sin^2\theta}$

For a short dipole ($a \ll r$), $a^2 \sin^2\theta$ is negligible compared to $r^2$.

$V \approx \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2}$

Note that $\cos\theta = \frac{\vec{p} \cdot \vec{r}}{pr}$, where $\vec{p}$ is the dipole moment vector and $\vec{r}$ is the position vector of point P.

$V = \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \vec{r}}{r^3}$ (for a short dipole)

This formula confirms the previous results:

  • If $\theta = 0$ (axial point), $V = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2}$.
  • If $\theta = 90^\circ$ (broadside-on point), $V = 0$.
  • If $\theta = 180^\circ$ (axial point on the other side), $V = -\frac{1}{4\pi\epsilon_0} \frac{p}{r^2}$.

Potential Difference

Potential difference between two points in an electric field is defined as the work done per unit charge in moving a test charge from one point to the other against the electric field. It is the difference in electric potential between two points.

Let $V_A$ be the electric potential at point A and $V_B$ be the electric potential at point B. The potential difference between A and B is given by:

$\Delta V = V_B - V_A$

This potential difference is equal to the work done by an external agent to move a unit positive charge from A to B, or the negative of the work done by the electric field to move a unit positive charge from A to B.

$V_B - V_A = W_{ext, A \to B} = -W_{field, A \to B}$

Mathematically, the potential difference can be expressed as an integral of the electric field:

$V_B - V_A = -\int_{A}^{B} \vec{E} \cdot d\vec{l}$

If the electric field is uniform (like between the plates of a parallel plate capacitor), and the distance between points A and B is 'd', then the potential difference is:

$\Delta V = E \times d$

Key takeaway: Potential difference is the energy per unit charge required to move a charge between two points. It drives the flow of charge (current). A positive charge moves from a region of higher potential to lower potential, while a negative charge moves from lower to higher potential.

Example: Consider a 12V battery. This means the potential difference between its terminals is 12 Volts. If you connect a device across it, charge will flow, and for every Coulomb of charge that flows, 12 Joules of energy are transferred.

Equipotential Surfaces

An equipotential surface is a surface on which the electric potential is the same at every point. In other words, the potential difference between any two points on an equipotential surface is zero.

Properties of Equipotential Surfaces:

  1. No Work Done: No work is done in moving a charge between any two points on an equipotential surface. This is because $W = q \Delta V$, and $\Delta V = 0$ on an equipotential surface.
  2. Perpendicular to Electric Field: Equipotential surfaces are always perpendicular to the electric field lines. This can be understood from the relation $dV = -\vec{E} \cdot d\vec{l}$. If $dV = 0$ (on an equipotential surface), then $\vec{E} \cdot d\vec{l} = 0$, which means $\vec{E}$ is perpendicular to $d\vec{l}$ (the displacement along the surface).
  3. Unique for each potential value: For a given charge distribution, there is a unique equipotential surface for each value of potential.
  4. Do not intersect: Two equipotential surfaces cannot intersect. If they did, the intersection point would have two different potentials, which is impossible.
  5. Examples of Equipotential Surfaces:

    1. Point Charge: For a single point charge, the equipotential surfaces are concentric spheres centered at the charge. The potential is constant on each sphere ($V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$). Electric field lines (radially outward or inward) are perpendicular to these spherical surfaces.

    2. Electric Dipole: The equipotential surfaces for a dipole are more complex. They are not simple spheres. However, the perpendicular bisector is an equipotential surface with V=0.

    3. Uniform Electric Field: In a uniform electric field (e.g., between parallel plates), the equipotential surfaces are planes perpendicular to the electric field lines. For instance, if the field is along the x-axis, the equipotential surfaces are planes of constant x (e.g., x = constant).

    4. Infinite Line Charge: For an infinite line charge, the equipotential surfaces are concentric cylinders with the line charge as their axis.

    Mnemonic for Equipotential Surfaces: Think of contour lines on a topographical map. Each contour line represents a constant altitude (potential). The steepest slope (electric field) is always perpendicular to the contour lines.

    Relationship between Electric Field and Potential: The electric field can be expressed as the negative gradient of the electric potential:

    $\vec{E} = -\nabla V$

    In Cartesian coordinates, this is:

    $\vec{E} = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right)$

    This equation shows that the electric field points in the direction of the steepest decrease in electric potential. The magnitude of the electric field is equal to the rate of change of potential with distance in that direction.

    Summary Table: Point Charge vs. Dipole Potential

    Property Point Charge (Q) Electric Dipole (p)
    Potential Formula (at distance r) $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$ $V \approx \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2}$ (for short dipole)
    Dependence on Distance $V \propto \frac{1}{r}$ $V \propto \frac{1}{r^2}$
    Dependence on Angle None (spherically symmetric) Depends on $\cos\theta$
    Potential at Perpendicular Bisector $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$ $V = 0$

    Understanding electric potential, potential difference, and equipotential surfaces is crucial for analyzing electric fields and their effects on charges. These concepts form the foundation for understanding capacitance, energy stored in electric fields, and the behavior of electrical circuits.