Related Concepts: 2.01 Fundamental Postulates of Electrostatics & Gauss_s Law Applications | 2.03 Conductors, Dielectrics & Polarization Charge Densities | 1.02 Vector Operators (Gradient, Divergence, Curl & Laplacian) | 2.05 Electrostatic Energy, Work Done & Solved PYQ Numericals

2.02 Electric Potential, Equipotential Contours & Dipole Derivations

Core Idea

Electric potential $V$ is a scalar field — potential energy per unit charge. Working with $V$ instead of $\vec{E}$ is a massive simplification: you add scalars instead of vectors, and then recover the field by one differentiation, $\vec{E}=-\nabla V$. The dipole potential derivation built on this idea is one of the two or three most frequently examined derivations in the whole course (2016, 2017, 2019, 2021, 2023, 2024).

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1. Electric Potential and the Relation $\vec{E}=-\nabla V$ [PYQ: 2016, 2017, 2021]

Abstract

The electric potential $V$ at a point is the work done by an external agent, per unit positive charge, in bringing a test charge from a reference point (usually infinity, where $V=0$) to that point, without acceleration. $V=-{\int }_{\infty }^P\vec{E}\cdot d\vec{l}\text{(V or J/C)}$

Abstract

The electric field intensity $\vec{E}$ at a point is the force per unit positive test charge placed at that point: $\vec{E}={\lim }_{q\to 0}\frac{\vec{F}}{q}\text{(V/m or N/C)}$ The limit $q\to 0$ ensures the test charge does not disturb the source distribution it is measuring.

Potential difference between two points: $V_{AB}=V_B-V_A=-{\int }_A^B\vec{E}\cdot d\vec{l}$

1.1 Proof: Work Done Moving a Unit Charge = Potential Difference [PYQ: 2017]

PYQ — 2017 (08 marks)

Define electric potential. Show that in an electric field, work done in moving a unit charge from one point to another is equal to the electric potential difference between those two points.

Proof

Step 1. A charge $q$ in a field $\vec{E}$ experiences a force ${\vec{F}}_E=q\vec{E}$.

Step 2. To move it without acceleration, an external agent must apply an equal and opposite force: ${\vec{F}}_{\text{ext}}=-q\vec{E}$

Step 3. The differential work done by the external agent over a displacement $d\vec{l}$ is: $dW={\vec{F}}_{\text{ext}}\cdot d\vec{l}=-q\vec{E}\cdot d\vec{l}$

Step 4. Integrate from $A$ to $B$: $W_{A\to B}=-q{\int }_A^B\vec{E}\cdot d\vec{l}$

Step 5. For a unit positive charge ($q=1$ C): $W_{A\to B}=-{\int }_A^B\vec{E}\cdot d\vec{l}=V_B-V_A=V_{AB}■$

Step 6 — Path independence. Because $\nabla \times \vec{E}=0$ (second postulate), ${\oint }_C\vec{E}\cdot d\vec{l}=0$, so the integral depends only on the endpoints, never the path. This is what makes $V$ a single-valued function of position and hence a legitimate scalar field.

1.2 Deriving $\vec{E}=-\nabla V$ for a Point Charge [PYQ: 2016, 2021]

PYQ — 2016 (10 marks), 2021 (06 marks)

Define electric field intensity and electric potential. Also derive the relation between them when caused by a point charge.

Derivation

Step 1. The field of a point charge $q$ at distance $R$ is $\vec{E}={\hat{a}}_R\frac{q}{4\pi {ϵ}_0{R}^2}$.

Step 2. Bring a unit charge from $\infty$ to $R$ along a radial path, so $d\vec{l}={\hat{a}}_{R}d{R}^{\prime }$: $V=-{\int }_{\infty }^R\frac{q}{4\pi {ϵ}_0{R}^{\prime 2}}d{R}^{\prime }=-\frac{q}{4\pi {ϵ}_0}{\left[-\frac{1}{R^{\prime }}\right]}_{\infty }^R=\frac{q}{4\pi {ϵ}_{0}R}$

Step 3. Now take the gradient. Since $V$ depends only on $R$, in spherical coordinates only the radial term survives: $\nabla V={\hat{a}}_R\frac{\partial V}{\partial R}={\hat{a}}_R\frac{\partial }{\partial R}\left(\frac{q}{4\pi {ϵ}_{0}R}\right)=-{\hat{a}}_R\frac{q}{4\pi {ϵ}_0{R}^2}$

Step 4. The bracketed quantity is exactly $\vec{E}$: $\nabla V=-\vec{E}⟹\boxed{\vec{E}=-\nabla V}■$

Why the minus sign is physically necessary

$\nabla V$ points toward increasing potential. But a positive charge released in a field naturally accelerates toward lower potential energy. The field must therefore point “downhill” in $V$, which is exactly what the minus sign encodes.

The identity that licenses all of this

$\vec{E}$ can be written as a gradient only because $\nabla \times \vec{E}=0$ and $\nabla \times (\nabla V)\equiv 0$ — see 1.03 Integral Theorems (Divergence Theorem & Stokes_s Theorem) & Identity Proofs, Identity I. Cite this in a “derive the relation” answer.

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2. Electric Field Intensity $\vec{E}$ vs. Electric Flux Density $\vec{D}$ [PYQ: 2022, 2025]

PYQ — 2022 (05 marks), 2025 (07 marks)

2022: Differentiate between electric field intensity and electric flux density, emphasizing their physical significance. 2025: Distinguish between $\vec{E}$ and $\vec{D}$ with respect to definition, unit, governing relation, and physical significance.

The 2025 phrasing names the exact four rows the examiner wants. Reproduce this table:

Basis	Electric Field Intensity $\vec{E}$	Electric Flux Density $\vec{D}$
Definition	Force experienced per unit positive test charge at a point	Electric flux passing per unit area normal to the flux, arising from free charge only
Unit	volt per metre (V/m), equivalently N/C	coulomb per square metre (C/m²)
Governing relation	$\vec{E}=-\nabla V$; $\vec{E}=\vec{F}/q$; $\nabla \times \vec{E}=0$	$\vec{D}=ϵ\vec{E}={ϵ}_0\vec{E}+\vec{P}$; $\nabla \cdot \vec{D}={\rho }_v$
Physical significance	Describes the force effect of the field on charges; the quantity that appears in energy and work calculations	Describes the source–flux relationship; measures the flux due to free charge and is what Gauss’s law counts
Medium dependence	Depends on the medium — for the same free charge, $\vec{E}$ falls by a factor ${ϵ}_r$ inside a dielectric	Independent of the medium — determined solely by the free charge distribution
Behaviour at a dielectric boundary	Its tangential component is continuous: $E_{1t}=E_{2t}$	Its normal component is continuous when ${\rho }_s=0$: $D_{1n}=D_{2n}$
Related to polarization?	No — it is the total field acting on a charge	Yes — explicitly contains $\vec{P}$

The single sentence that earns the "physical significance" mark

”$\vec{D}$ is a source-related quantity determined only by free charge and unaffected by the medium; $\vec{E}$ is a force-related quantity that does depend on the medium, since the dielectric’s polarization partially cancels the applied field.”

Why we need two vectors at all

Inside a dielectric the bound polarization charges create their own field opposing the applied one. If we only had $\vec{E}$, Gauss’s law would have to count both free and bound charge, which we usually cannot measure. Defining $\vec{D}={ϵ}_0\vec{E}+\vec{P}$ absorbs the bound charge into the definition, so $\nabla \cdot \vec{D}={\rho }_v$ involves free charge only. See 2.03 Conductors, Dielectrics & Polarization Charge Densities.

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3. Equipotential Lines & Field-Line Sketches [PYQ: 2024, 2025]

Abstract

An equipotential line (in 2D) or equipotential surface (in 3D) is the locus of all points having the same electric potential. No work is done in moving a charge along it, since $dW=-qdV=0$.

3.1 Three properties always worth stating

1. Orthogonality. Field lines and equipotentials meet at exactly $90^{\circ }$ everywhere. Proof: along an equipotential $dV=\nabla V\cdot d\vec{l}=0$, so $\nabla V\perp d\vec{l}$, and $\vec{E}=-\nabla V$ is therefore perpendicular to the equipotential.
2. Direction. Field lines point from high $V$ to low $V$ — the direction of maximum decrease of potential.
3. Zero work. Moving a charge anywhere along an equipotential requires no work. Equipotentials never cross each other (a point cannot have two potentials).

3.2 Sketching the Two Required Cases [PYQ: 2024, 2025]

PYQ — 2024 (07 marks), 2025 (10 marks)

2024: Define equipotential line. Draw the electric field lines and the equipotential lines of a uniform charge sphere. 2025: Make a two-dimensional sketch of the electric field lines and the equipotential lines of a uniform charge sphere and a dipole. Ensure the lines are distinguishable.

[FIGURE: Two-panel 2D sketch. Left — uniformly charged sphere: straight radial field lines with arrowheads pointing outward from the sphere; concentric dashed circles as equipotentials, spaced increasingly far apart with radius. Right — electric dipole: curved solid field lines emerging from $+q$ and terminating on $-q$, symmetric about the dipole axis; dashed equipotential contours forming closed ovals around each charge, with the perpendicular bisector plane being the $V=0$ equipotential (a straight dashed line). — source: David K. Cheng, Ch. 3, Fig. 3-12]

Uniform charge sphere (radius $a$, total charge $Q$):

Feature	Field lines $\vec{E}$	Equipotentials $V$
Shape	Straight, radial, pointing outward (for $+Q$)	Concentric circles (spheres in 3D)
Outside ($R>a$)	$E=\frac{Q}{4\pi {ϵ}_0{R}^2}$	$V=\frac{Q}{4\pi {ϵ}_{0}R}$
Spacing	Lines spread apart as $R$ grows (field weakens)	Circles get further apart as $R$ grows

Electric dipole:

Feature	Field lines $\vec{E}$	Equipotentials $V$
Shape	Curved loops leaving $+q$, entering $-q$	Closed ovals around each charge
Symmetry plane	Field is purely perpendicular to the bisecting plane ($\theta =90^{\circ }$)	The bisecting plane is the $V=0$ equipotential — a straight line in the 2D sketch
Far field	$E\propto 1/R^3$	$V\propto 1/R^2$

Marks the examiner is looking for in the sketch

1. Arrowheads on the field lines (direction from $+$ to $-$).
2. Different line styles — the 2025 paper explicitly says “ensure the lines are distinguishable”: use solid for $\vec{E}$, dashed for equipotentials, and add a legend.
3. Visible perpendicularity at every crossing.
4. The $V=0$ line through the dipole’s midplane — this is the detail most students omit.
5. Field lines start and end on charges; equipotentials are closed and never cross.

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4. Master Derivation: The Electric Dipole [PYQ: 2016, 2017, 2019, 2021, 2023, 2024]

The most-repeated derivation in Chapter 2

Asked in six separate years. The potential derivation ($V$) is asked in 2017, 2019, 2021, 2024; the field derivation ($\vec{E}$ from the dipole moment) in 2016, 2023.

Abstract

An electric dipole is a pair of equal and opposite point charges, $+q$ and $-q$, separated by a distance $d$ that is small compared with the distance to the observation point.

The electric dipole moment is the vector $\vec{p}=q\vec{d}\text{(C}\cdot \text{m)}$ whose magnitude is $qd$ and whose direction is from the negative charge toward the positive charge.

[FIGURE: Electric dipole with $+q$ at $(0,0,d/2)$ and $-q$ at $(0,0,-d/2)$; observation point $P$ at distance $R$ from the origin making angle $\theta$ with the $z$-axis; distances $R_{+}$ and $R_{-}$ marked, with the far-field approximation showing $R_{-}-R_{+}\approx d\cos \theta$ — source: David K. Cheng, Ch. 3, Fig. 3-10]

4.1 Derivation of the Dipole Potential $V(R,\theta )$ [PYQ: 2017, 2019, 2021, 2024]

Step-by-step derivation

Step 1 — Exact expression. Place $+q$ at $z=+d/2$ and $-q$ at $z=-d/2$. Potential is a scalar, so simply add: $V=\frac{q}{4\pi {ϵ}_0}\left(\frac{1}{R_{+}}-\frac{1}{R_{-}}\right)=\frac{q}{4\pi {ϵ}_0}\left(\frac{R_{-}-R_{+}}{R_{+}{R}_{-}}\right)$

Step 2 — Far-field approximation ($R\gg d$). By the law of cosines, $R_{\pm }^2=R^2+{\left(\frac{d}{2}\right)}^2\mp Rd\cos \theta \approx R^2\left(1\mp \frac{d}{R}\cos \theta \right)$ neglecting $(d/2R{)}^2$. Taking the inverse square root and applying the binomial expansion $(1+x{)}^n\approx 1+nx$ with $n=-1/2$: $\frac{1}{R_{\pm }}\approx \frac{1}{R}\left(1\pm \frac{d}{2R}\cos \theta \right)$

Step 3 — Subtract. $\frac{1}{R_{+}}-\frac{1}{R_{-}}\approx \frac{1}{R}\left[\left(1+\frac{d\cos \theta }{2R}\right)-\left(1-\frac{d\cos \theta }{2R}\right)\right]=\frac{d\cos \theta }{R^2}$

Equivalently: $R_{-}-R_{+}\approx d\cos \theta$ and $R_{+}{R}_{-}\approx R^2$.

Step 4 — Substitute back. $V=\frac{qd\cos \theta }{4\pi {ϵ}_0{R}^2}$

Step 5 — Express via the dipole moment. With $\vec{p}={\hat{a}}_{z}qd$ and $\vec{p}\cdot {\hat{a}}_R=p\cos \theta$: $\boxed{V=\frac{p\cos \theta }{4\pi {ϵ}_0{R}^2}=\frac{\vec{p}\cdot {\hat{a}}_R}{4\pi {ϵ}_0{R}^2}}\text{(V)}■$

4.2 How $V$ Varies with Distance and Angle [PYQ: 2019]

The 2019 paper explicitly asked you to “explain how electric potential varies with distance and angle of position”:

Dependence	Behaviour	Physical reason
With distance $R$	$V\propto 1/R^2$ — decays faster than a point charge’s $1/R$	At large $R$ the two opposite charges nearly cancel; only their small separation leaves a residual, so the potential falls off one power faster
With angle $\theta$	$V\propto \cos \theta$	Maximum positive at $\theta =0^{\circ }$ (on the $+q$ side of the axis); maximum negative at $\theta =180^{\circ }$; exactly zero at $\theta =90^{\circ }$
At $\theta =90^{\circ }$	$V=0$ everywhere on the bisecting plane	Every point there is equidistant from $+q$ and $-q$, so their contributions cancel exactly

[GRAPH: $V$ versus $\theta$ at fixed $R$ — a cosine curve peaking at $\theta =0$, crossing zero at $\theta =90^{\circ }$, minimum at $\theta =180^{\circ }$. Governing equation $V=p\cos \theta /(4\pi {ϵ}_0{R}^2)$.]

4.3 Derivation of the Dipole Field $\vec{E}(R,\theta )$ [PYQ: 2016, 2023]

Step-by-step derivation

Step 1. Apply $\vec{E}=-\nabla V$ in spherical coordinates. Since $V$ is independent of $\varphi$, the ${\hat{a}}_{\varphi }$ term vanishes: $\vec{E}=-\left[{\hat{a}}_R\frac{\partial V}{\partial R}+{\hat{a}}_{\theta }\frac{1}{R}\frac{\partial V}{\partial \theta }\right]$

Step 2 — Radial component. $E_R=-\frac{\partial }{\partial R}\left(\frac{p\cos \theta }{4\pi {ϵ}_0{R}^2}\right)=-\left(\frac{-2p\cos \theta }{4\pi {ϵ}_0{R}^3}\right)=\frac{2p\cos \theta }{4\pi {ϵ}_0{R}^3}$

Step 3 — Polar component. $E_{\theta }=-\frac{1}{R}\frac{\partial }{\partial \theta }\left(\frac{p\cos \theta }{4\pi {ϵ}_0{R}^2}\right)=-\frac{1}{R}\left(\frac{-p\sin \theta }{4\pi {ϵ}_0{R}^2}\right)=\frac{p\sin \theta }{4\pi {ϵ}_0{R}^3}$

Step 4 — Combine. $\boxed{\vec{E}=\frac{p}{4\pi {ϵ}_0{R}^3}\left(2\cos \theta {\hat{a}}_R+\sin \theta {\hat{a}}_{\theta }\right)}\text{(V/m)}■$

Coordinate-free form (occasionally useful): $\vec{E}=\frac{1}{4\pi {ϵ}_0{R}^3}\left[\frac{3(\vec{p}\cdot \vec{R})}{R^2}\vec{R}-\vec{p}\right]$

4.4 Special Cases Worth Quoting

Position	$\theta$	$\vec{E}$	Note
On the axis	$0^{\circ }$	${\hat{a}}_R\frac{2p}{4\pi {ϵ}_0{R}^3}$	Twice as strong as at the same distance broadside
Broadside (bisecting plane)	$90^{\circ }$	${\hat{a}}_{\theta }\frac{p}{4\pi {ϵ}_0{R}^3}$	Points antiparallel to $\vec{p}$; note $V=0$ here even though $\vec{E}\ne 0$

The classic conceptual trap

On the bisecting plane $V=0$ but $\vec{E}\ne 0$. Zero potential does not mean zero field — $\vec{E}$ depends on the rate of change of $V$ across the surface, not on its value. Examiners like this one.

4.5 Dipole Scaling Summary

Quantity	Point charge	Dipole
Potential $V$	$\propto 1/R$	$\propto 1/R^2$
Field $\vec{E}$	$\propto 1/R^2$	$\propto 1/R^3$
Angular dependence	None (isotropic)	$\cos \theta$ for $V$; $2\cos \theta {\hat{a}}_R+\sin \theta {\hat{a}}_{\theta }$ for $\vec{E}$

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5. PYQ Coverage for This Note

Question (verbatim, condensed)	Marks	Year(s)
Define electric field intensity and electric potential. Also derive the relation between them when caused by a point charge.	10	2016
Define electric potential. Show that work done in moving a unit charge from one point to another equals the potential difference.	08	2017
Define electric potential and state the relation between electric potential and electric field intensity.	06	2021
Differentiate between $\vec{E}$ and $\vec{D}$, emphasizing their physical significance.	05	2022
Distinguish between $\vec{E}$ and $\vec{D}$ with respect to definition, unit, governing relation, and physical significance.	07	2025
Define equipotential line. Draw the electric field lines and equipotential lines of a uniform charge sphere.	07	2024
2D sketch of field lines and equipotential lines of a uniform charge sphere and a dipole.	10	2025
Define electric dipole moment. Deduce $\vec{E}$ of an electric dipole in terms of its dipole moment.	10/13	2016, 2023
Define electric dipole moment. Estimate $V$ at any point P due to an electric dipole ($R\gg d$).	12	2021
Define electric dipole. Derive $V$ due to a dipole and explain how $V$ varies with distance and angle of position.	13	2019
Two equal opposite charges separated by $l$ form a dipole. Derive $V$ at an arbitrary point P.	12	2017, 2024

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6. Exam Hacks & Traps

Key Exam Checkpoints

1. State the approximation explicitly. Write “Since $R\gg d$, we apply the binomial expansion $(1+x{)}^n\approx 1+nx$“. Students who silently jump to $d\cos \theta$ lose presentation marks.
2. Never compute exact $R_{+}$ and $R_{-}$. The whole point is the far-field expansion.
3. $\vec{p}$ points from $-q$ to $+q$. The opposite of the field direction between the charges. Getting this backwards flips the sign of $V$.
4. $V=0$ but $\vec{E}\ne 0$ on the bisecting plane. Know why.
5. For the $\vec{E}$ derivation, use spherical $\nabla$ with the $1/R$ factor on the $\theta$ term. Forgetting the $1/R$ gives $E_{\theta }$ with the wrong power of $R$.
6. In the 2025 sketch question, use two distinguishable line styles and a legend. The question says so directly.
7. For the $\vec{E}$ vs $\vec{D}$ table, use the four headings named in 2025 — definition, unit, governing relation, physical significance. Add medium-dependence as a bonus row.
8. Units: $V$ in volts, $\vec{E}$ in V/m, $\vec{D}$ in C/m², $\vec{p}$ in C·m.

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7. Self-Check

1. Define $V$ and prove that the work per unit charge equals $V_B-V_A$.
2. Why can $\vec{E}$ be written as $-\nabla V$? Which identity licenses it?
3. Give four differences between $\vec{E}$ and $\vec{D}$.
4. Derive $V$ for a dipole and state the approximation used.
5. Why is $V\propto 1/R^2$ for a dipole but $1/R$ for a point charge?
6. Where is $V=0$ for a dipole, and is $\vec{E}$ zero there?

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Next: 2.03 Conductors, Dielectrics & Polarization Charge Densities