Potential due to Electric Dipole — Revision Notes

⚡ 30-Second Revision

Electric Dipole — Two equal and opposite charges ($+q, -q$) separated by $2a$.
Dipole Moment — $\vec{p} = q(2\vec{a})$, direction from $-q$ to $+q$.
General Potential Formula (for $r \gg a$) — $V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2}$
Axial Line Potential — $V_{axial} = \pm \frac{p}{4\pi\epsilon_0 r^2}$ (for $\theta = 0^circ, 180^circ$)
Equatorial Line Potential — $V_{equatorial} = 0$ (for $\theta = 90^circ$)
Distance Dependence — $V_{dipole} \propto 1/r^2$ (vs. $V_{point charge} \propto 1/r$)
Nature — Potential is a scalar quantity.
Potential Energy of Dipole in External Field — $U = -\vec{p} \cdot \vec{E} = -pE \cos\theta$

2-Minute Revision

For NEET, quickly recall that an electric dipole is a pair of equal and opposite charges. Its strength is quantified by the dipole moment $\vec{p}$, directed from negative to positive charge. The electric potential $V$ created by this dipole at a distant point ($r \gg a$) is given by $V = \frac{p \cos\theta}{4\pi\epsilon_0 r^2}$.

This formula is key. Notice the $1/r^2$ dependence, which is faster than the $1/r$ for a single point charge, and the crucial $\cos\theta$ term, indicating angular dependence. Remember the two special cases: on the axial line ($\theta = 0^circ$ or $180^circ$), potential is maximum in magnitude ($V = \pm \frac{p}{4\pi\epsilon_0 r^2}$), while on the equatorial line ($\theta = 90^circ$), the potential is always zero.

This zero potential on the equatorial line is a frequently tested concept. Don't confuse potential (scalar) with electric field (vector); a zero potential doesn't mean a zero field. Also, be ready for questions involving the potential energy of a dipole in an external electric field, $U = -pE \cos\theta$, and related work done calculations.

5-Minute Revision

To thoroughly revise potential due to an electric dipole for NEET, start by solidifying the definition: an electric dipole is a system of two equal and opposite charges, $+q$ and $-q$, separated by a small distance $2a$.

The electric dipole moment, $\vec{p}$, is a vector of magnitude $q(2a)$, pointing from $-q$ to $+q$. The core formula for electric potential $V$ at a point $(r, \theta)$ far from the dipole ($r \gg a$) is $V = \frac{p \cos\theta}{4\pi\epsilon_0 r^2}$.

This formula is derived using the superposition principle and binomial approximations. Understand that $r$ is the distance from the dipole's center, and $\theta$ is the angle between $\vec{p}$ and the position vector $\vec{r}$.

Crucial special cases: On the axial line ($\theta = 0^circ$ or $180^circ$), $\cos\theta = \pm 1$, so $V_{axial} = \pm \frac{p}{4\pi\epsilon_0 r^2}$. On the equatorial line ($\theta = 90^circ$), $\cos\theta = 0$, leading to $V_{equatorial} = 0$. This zero potential on the equatorial line is a high-yield concept. Remember, even if potential is zero, the electric field is generally not zero there.

Compare and contrast: The potential due to a dipole falls off as $1/r^2$, much faster than a point charge's $1/r$ dependence. This is a common conceptual question. Also, the dipole potential is direction-dependent (anisotropic), unlike a point charge's spherically symmetric potential.

For numerical problems, ensure you correctly identify $p$, $r$, and $\theta$, and use the correct value for $1/(4\pi\epsilon_0)$ (approximately $9 \times 10^9,\text{N m}^2/\text{C}^2$). Practice calculations involving $\cos\theta$ for common angles like $0^circ, 30^circ, 45^circ, 60^circ, 90^circ, 180^circ$.

Finally, briefly review the potential energy of a dipole in an external uniform electric field, $U = -\vec{p} \cdot \vec{E}$, as questions on work done in rotating a dipole are often linked to this topic.

For example, rotating a dipole from parallel ($\theta=0^circ$) to perpendicular ($\theta=90^circ$) to the field requires work $W = U_{final} - U_{initial} = 0 - (-pE) = pE$.

Prelims Revision Notes

Electric Potential due to an Electric Dipole (NEET Focus)

1. Definition of Electric Dipole:

Two equal and opposite point charges, $+q$ and $-q$.
Separated by a small fixed distance, $2a$.

2. Electric Dipole Moment ($\vec{p}$):

Magnitude: $p = q(2a)$.
Direction: From $-q$ to $+q$.
Units: Coulomb-meter (C m).

3. General Formula for Electric Potential (for $r \gg a$):

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

* $V$: Electric potential at point P. * $p$: Magnitude of electric dipole moment. * $r$: Distance of point P from the center of the dipole. * $\theta$: Angle between the dipole moment vector $\vec{p}$ and the position vector $\vec{r}$ of point P. * $\epsilon_0$: Permittivity of free space ($1/(4\pi\epsilon_0) \approx 9 \times 10^9,\text{N m}^2/\text{C}^2$).

Key takeaway — Potential is a scalar quantity.

4. Special Cases:

On the Axial Line:

* Point P lies on the axis of the dipole. * $\theta = 0^circ$ (towards $+q$) or $\theta = 180^circ$ (towards $-q$). * $\cos 0^circ = 1$, $\cos 180^circ = -1$. * $V_{axial} = \pm \frac{p}{4\pi\epsilon_0 r^2}$. * Potential is maximum (magnitude) on the axial line.

On the Equatorial Line (Perpendicular Bisector):

* Point P lies on the line perpendicular to the dipole axis, passing through its center. * $\theta = 90^circ$. * $\cos 90^circ = 0$. * $V_{equatorial} = 0$. * Crucial point: Potential is zero everywhere on the equatorial plane. However, the electric field is NOT zero on the equatorial line.

5. Distance Dependence:

Dipole Potential — $V_{dipole} \propto 1/r^2$ (for $r \gg a$).
Point Charge Potential — $V_{point charge} \propto 1/r$.
NEET Tip — Be careful not to confuse these. Dipole potential falls off faster with distance.

6. Angular Dependence:

Dipole potential is anisotropic, depending on $\cos\theta$.
Point charge potential is spherically symmetric, no angular dependence.

7. Potential Energy of a Dipole in an External Uniform Electric Field (Related Concept):

$U = -\vec{p} \cdot \vec{E} = -pE \cos\theta$.
Work done by external agent to rotate dipole from $\theta_1$ to $\theta_2$: $W = U_2 - U_1 = -pE(\cos\theta_2 - \cos\theta_1)$.

8. Common Mistakes to Avoid:

Confusing potential (scalar) with electric field (vector).
Mixing up distance dependence for point charge vs. dipole.
Incorrectly applying $\cos\theta$ or misremembering values for standard angles.
Assuming zero potential implies zero electric field (especially on equatorial line).

Vyyuha Quick Recall

To remember the dipole potential formula: 'P Cosey on R Squared'.

P — Dipole moment ($p$)
Cosey — $\cos\theta$
R Squared — $r^2$ in the denominator

So, $V = \frac{p \cos\theta}{4\pi\epsilon_0 r^2}$.

For equatorial line: 'Equator is Zero' (potential is zero).