Chapter 2 · Electrostatic Potential and Capacitance

2.2 Equipotential Surfaces and Electrostatic Potential Energy

Points at the same potential form equipotential surfaces, and the field always crosses them at right angles, pointing towards lower potential with strength $E=-\dfrac{dV}{dr}$. The same potential tells us the energy stored in a set of charges, $U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}$ per pair, and in a dipole turned in a field, $U=-\mathbf p\cdot\mathbf E$.

A hiking map shows a mountain with contour lines: every point on a contour is at the same height. Where the contours crowd together the slope is steep, and water poured on the ground runs straight downhill, crossing every contour at right angles. Electric potential can be mapped in exactly the same way. Doctors do it every day: the electrodes of an electrocardiogram (ECG) sample the potential at a few points on the skin, and the pattern of those potentials — set up by the heart acting as a changing electric dipole — reveals the direction and strength of each heartbeat.

In this section we draw the "contour lines" of electrostatics, the equipotential surfaces, and prove that the field is everywhere perpendicular to them and strongest where they are closest. That gives the key relation $E=-dV/dr$, which lets us get the field from the potential and the other way round. Then we use potential to count energy: the potential energy of a system of charges (the work done to assemble it), the energy of a charge or a pair of charges placed in an external field, and the energy of a dipole as it turns in a uniform field.

Equipotential Surfaces

📘 Definition 2.2 — Equipotential surface
An equipotential surface is a surface on which the potential has the same value at every point. In a two-dimensional drawing it appears as an equipotential line. Equipotentials are usually drawn at equal steps of potential (every $10$ V, say), like the contours of a map drawn every $10$ m of height.

For a single point charge, $V=kq/r$ is the same at every point of a sphere centred on the charge: the equipotentials are concentric spheres. Because $V\propto1/r$, spheres drawn at equal steps of $V$ get farther apart as $r$ increases — the field weakens. For a uniform field, say along $+x$, the potential changes only with $x$, so the equipotentials are planes perpendicular to the field, equally spaced. Two equal like charges produce a figure-of-eight equipotential through the neutral point between them; a dipole produces a plane of zero potential (its equatorial plane) with closed surfaces of positive potential round $+q$ and of negative potential round $-q$. Interactive 1 draws all of these.

📐 Result 2.6 — Field and equipotentials are perpendicular

(i) No work is done by the electric field (or against it) in moving a charge along an equipotential surface. (ii) The electric field at every point is perpendicular to the equipotential surface through that point, and points towards decreasing potential. (iii) Two different equipotential surfaces never intersect.

Why? (i) Between two points of the same surface $\Delta V=0$, so the work $q\Delta V$ is zero. (ii) Move a charge $q$ a short distance $\delta l$ along the surface. The field does work $qE\,\delta l\cos\theta$, where $\theta$ is the angle between $\mathbf E$ and the displacement; this must be zero for every direction of $\delta l$ in the surface, so $\cos\theta=0$ and $\mathbf E$ is normal to the surface. A positive charge moves along $\mathbf E$ of its own accord and loses potential energy, so $\mathbf E$ points from higher to lower $V$. (iii) If two surfaces with different potentials met, the meeting point would have two values of potential, and the field there would have two different normal directions — both impossible.

The Field as the Slope of the Potential

Take two close equipotential surfaces, at $V$ and $V+\delta V$, a perpendicular distance $\delta l$ apart. Carry a unit positive charge from the higher surface to the lower one along the normal, the direction of $\mathbf E$. The field does work $E\,\delta l$ on it, and this equals the drop in potential, $-\delta V$. Hence $E=-\dfrac{\delta V}{\delta l}$, and in the limit:

📐 Result 2.7 — Relation between field and potential

The electric field at a point points in the direction in which the potential decreases fastest — perpendicular to the equipotential surface through the point — and its magnitude is the rate of decrease of $V$ with distance in that direction: $$E=-\frac{dV}{dl}\qquad\left(\text{so } |\mathbf E|=\left|\frac{dV}{dl}\right|\ \text{measured normal to the equipotential}\right).$$ Along a line (say the $x$-axis) the component of $\mathbf E$ in that direction is $E_x=-\dfrac{dV}{dx}$. Conversely, $V_A-V_B=\displaystyle\int_A^B\mathbf E\cdot d\mathbf l$ along any path; in a uniform field this gives $V_A-V_B=E\,d$, where $d$ is the distance from $A$ to $B$ measured along the field.

Two consequences are worth stating. Since $E=|\delta V|/\delta l$, equipotentials drawn at equal steps of $V$ are closer together where the field is stronger, exactly as crowded contours mark a steep slope. And the unit of field may equally be written V/m: $1\ \text{N/C}=1\ \text{V/m}$. Notice also that $E=0$ in a region means $V$ is constant there, not zero — the inside of a charged shell in Section 2.1 is the standard example.

✏️ Example 1 — Equipotentials between charged plates
Two large parallel metal plates are $2.0$ cm apart and are kept at $0$ V and $400$ V. (a) Find the field between them. (b) How far apart are the equipotentials at $100$ V, $200$ V and $300$ V, and what shape are they? (c) How much work is needed to carry a charge of $+3.0$ nC from the $100$ V surface to the $300$ V surface? (d) And along the $200$ V surface for $5.0$ mm?
Concept Between large parallel plates the field is uniform, so $V$ changes steadily with distance: $E=\dfrac{\Delta V}{d}$, pointing from the $400$ V plate towards the $0$ V plate.
(a) Field $E=\dfrac{400\ \text{V}}{0.020\ \text{m}}=2.0\times10^{4}$ V/m.
(b) Equipotentials Planes parallel to the plates, $\Delta x=\dfrac{\Delta V}{E}=\dfrac{100}{2.0\times10^4}=5.0\times10^{-3}$ m $=5.0$ mm apart — equally spaced, because the field is uniform.
(c), (d) Work $W=q\,\Delta V=(3.0\times10^{-9})(300-100)=6.0\times10^{-7}$ J, by any path. Along an equipotential $\Delta V=0$, so $W=0$ however far the charge moves.
✏️ Example 2 — A path at an angle to a uniform field
A uniform field of $500$ V/m points along $+x$. Point $A$ is at the origin and point $B$ is at $(3.0\ \text{cm},\ 4.0\ \text{cm})$. (a) Find $V_A-V_B$. (b) How much work does the field do on an electron that moves from $A$ to $B$, in eV?
Idea Go from $A$ to $B$ by any convenient path: $3.0$ cm along $\mathbf E$, then $4.0$ cm perpendicular to it. Only the part along the field changes the potential; the perpendicular part runs along an equipotential.
(a) $V_A-V_B=E\,\Delta x=(500)(0.030)=15$ V. $A$ is higher: $\mathbf E$ points from $A$'s plane towards $B$'s.
(b) $W^\text{field}=q(V_A-V_B)=(-e)(15\ \text{V})=-15$ eV. The field does negative work — an electron is pushed against $\mathbf E$, so moving it along $+x$ needs work from outside.
Check Using $\mathbf E\cdot\Delta\mathbf l$ directly: $(500,0)\cdot(0.030,0.040)=15$ V ✓. The full distance $AB=5.0$ cm is irrelevant.
✏️ Example 3 — Field from a potential graph and from a formula
(a) Along the $x$-axis the potential is $50$ V for $0\le x\le2$ cm, falls steadily to $-30$ V at $x=6$ cm and stays at $-30$ V beyond. Find $E_x$ in each region. (b) In another region $V=4x^2$ (with $V$ in volts and $x$ in metres). Find the field at $x=0.50$ m and at $x=-1.0$ m.
Rule $E_x=-\dfrac{dV}{dx}$: minus the slope of the $V$–$x$ graph.
(a) Graph For $0\le x\le2$ cm and $x\ge6$ cm the graph is flat: $E_x=0$ (but $V\ne0$). For $2\le x\le6$ cm the slope is $\dfrac{-30-50}{0.04}=-2000$ V/m, so $E_x=+2.0\times10^{3}$ V/m, along $+x$ — towards lower potential ✓.
(b) Formula $E_x=-\dfrac{d}{dx}(4x^2)=-8x$. At $x=0.50$ m: $E_x=-4.0$ V/m (magnitude $4.0$ V/m, along $-x$). At $x=-1.0$ m: $E_x=+8.0$ V/m (along $+x$). In both cases $\mathbf E$ points towards $x=0$, where $V$ is lowest.
Interactive 1 — Potential Map: Equipotentials and Field

Colour shows the potential ($q=1$ nC, grid square $1$ cm): red where $V\gt0$, blue where $V\lt0$, deeper colour for larger $|V|$. Curves are equipotentials at $\pm25,\ 50,\ 100,\ 200,\ 400,\ 800$ V (dashed: $V=0$); small arrows show the direction of $\mathbf E$. Drag the charges or the probe ◆. Press Play to release the probe as a free positive charge drifting along $\mathbf E$: it crosses every equipotential at right angles, always to lower $V$.

potential $V$ at probe—
field $|\mathbf E|$ at probe—
direction of $\mathbf E$ from $+x$—
$|\Delta V|/\Delta l$ across 1 mm, normal—

Potential Energy of a System of Charges

A set of charges held at fixed positions stores energy, just as a stretched spring does. To find how much, imagine assembling the system one charge at a time, starting with all of them infinitely far apart, and add up the work done by an external agent.

  1. Bring $q_1$ from infinity to its place. No other charges are present yet, so there is no field and no work: $W_1=0$.
  2. Bring $q_2$ to its place, a distance $r_{12}$ from $q_1$. It moves through the field of $q_1$, from potential $0$ to $V_1=kq_1/r_{12}$, so $W_2=q_2V_1=\dfrac{kq_1q_2}{r_{12}}$.
  3. Bring $q_3$ into the field of both: $W_3=q_3\left(\dfrac{kq_1}{r_{13}}+\dfrac{kq_2}{r_{23}}\right)$.

The total is a sum of one term for each pair. The order of assembly does not matter (each pair is counted once whichever arrives first), and because the force is conservative the result does not depend on the routes either.

📘 Definition 2.3 — Electrostatic potential energy of a system of charges
The electrostatic potential energy of a system of point charges is the total work done by an external agent in assembling them, slowly, from infinite separation into their present positions. It is the energy that would be released if they were allowed to fly apart to infinity.
📐 Result 2.8 — Potential energy of a system of point charges

For two point charges a distance $r_{12}$ apart, and for three: $$U=\frac{1}{4\pi\varepsilon_0}\,\frac{q_1q_2}{r_{12}},\qquad U=\frac{1}{4\pi\varepsilon_0}\left(\frac{q_1q_2}{r_{12}}+\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}}\right).$$ For $n$ charges, add $kq_iq_j/r_{ij}$ over every pair, each pair once. Signs are kept: $U\gt0$ for like charges (work was needed to push them together; they will fly apart if released), $U\lt0$ for unlike charges (they are bound; work must be done to separate them).

✏️ Example 4 — Assembling three charges
Charges $+1.0\ \mu\text{C}$, $+2.0\ \mu\text{C}$ and $-3.0\ \mu\text{C}$ are placed at the corners of an equilateral triangle of side $10$ cm. (a) Find the potential energy of the system. (b) How much work is needed to separate them completely? (c) How much work is needed to bring a fourth charge of $+1.0\ \mu\text{C}$ from infinity to the centroid of the triangle?
Setup All three pairs are $0.10$ m apart, so $U=\dfrac{k}{0.10}\,(q_1q_2+q_1q_3+q_2q_3)$ with charges in $\mu\text{C}$ giving $10^{-12}\ \text{C}^2$.
(a) Energy $q_1q_2+q_1q_3+q_2q_3=(1)(2)+(1)(-3)+(2)(-3)=2-3-6=-7$. $$U=\frac{9.0\times10^9}{0.10}\,(-7\times10^{-12})=-0.63\ \text{J}.$$
(b) Separation Taking them to infinity brings $U$ to $0$, so the agent must supply $0-(-0.63)=+0.63$ J. The system is bound.
(c) Fourth charge All corners are equidistant from the centroid, so $V_\text{centroid}=\dfrac{k(1+2-3)\ \mu\text{C}}{r}=0$ and $W=q_4V=0$ — even though the field at the centroid is not zero.
✏️ Example 5 — How close can an alpha particle get to a nucleus?
An alpha particle (charge $+2e$) with kinetic energy $5.0$ MeV is fired straight at a gold nucleus (charge $+79e$), which stays at rest. (a) Find the distance of closest approach. (b) What is the kinetic energy of the alpha particle when it is twice that distance from the nucleus?
Concept Energy is conserved: kinetic energy far away becomes electrostatic potential energy of the alpha–nucleus pair. At closest approach the particle is momentarily at rest.
(a) Equation $K=\dfrac{k(2e)(79e)}{d}\Rightarrow d=\dfrac{k(158)e^2}{K}$. With $K=5.0\times10^6\times1.6\times10^{-19}=8.0\times10^{-13}$ J: $$d=\frac{(9.0\times10^9)(158)(1.6\times10^{-19})^2}{8.0\times10^{-13}}\approx4.6\times10^{-14}\ \text{m}.$$
(b) At $2d$ $U\propto1/r$, so at $2d$ the potential energy is half its maximum, $2.5$ MeV, and the kinetic energy is $5.0-2.5=2.5$ MeV.
Meaning $46$ fm is several times the size of a gold nucleus. Measurements like this, in Rutherford's laboratory, gave the first upper limit on the size of the nucleus (Chapter 12).

Potential Energy in an External Field

Often the charges that make a field are far away and fixed — the plates of a capacitor, a charged cloud, the electrodes of an instrument — and we care only about the energy of a charge placed in that external field. If the external potential at the position $\mathbf r$ is $V(\mathbf r)$, then by Definition 2.1 the work needed to bring $q$ from infinity to $\mathbf r$ is $qV(\mathbf r)$, and that is its potential energy. For two charges in an external field we must add their energy of interaction with each other: $$U=q_1V(\mathbf r_1)+q_2V(\mathbf r_2)+\frac{kq_1q_2}{r_{12}}.$$ A dipole is the most important case. In a uniform field the charge $+q$ sits a distance $2a\cos\theta$ farther along the field than $-q$, so its potential is lower by $E\,(2a\cos\theta)$; the energy is $qV_+-qV_-=q(V_+-V_-)=-qE(2a\cos\theta)$. Equivalently, it is the work $pE(\cos\theta_1-\cos\theta_2)$ needed to turn the dipole (Section 1.2), measured from $\theta=90^\circ$.

📐 Result 2.9 — Potential energy of a dipole in a uniform field

A dipole of moment $\mathbf p$ in a uniform field $\mathbf E$, at angle $\theta$ between $\mathbf p$ and $\mathbf E$, has potential energy $$U(\theta)=-pE\cos\theta=-\mathbf p\cdot\mathbf E,$$ taking $U=0$ when $\mathbf p\perp\mathbf E$. It is least, $-pE$, at $\theta=0$ (stable equilibrium) and greatest, $+pE$, at $\theta=180^\circ$ (unstable equilibrium). The work needed to turn the dipole slowly from $\theta_1$ to $\theta_2$ is $W=U(\theta_2)-U(\theta_1)=pE(\cos\theta_1-\cos\theta_2)$.

✏️ Example 6 — Energy of a dipole as it turns
A dipole of moment $4.0\times10^{-9}$ C m is placed in a uniform field of $2.0\times10^5$ N/C. (a) Find its potential energy at $\theta=0^\circ$, $90^\circ$ and $180^\circ$. (b) How much work is needed to turn it from $60^\circ$ to $120^\circ$? (c) It is released from rest at $\theta=90^\circ$. What is its greatest kinetic energy of rotation, and where does it have it?
Setup $pE=(4.0\times10^{-9})(2.0\times10^5)=8.0\times10^{-4}$ J.
(a) $U(0^\circ)=-8.0\times10^{-4}$ J, $\ U(90^\circ)=0$, $\ U(180^\circ)=+8.0\times10^{-4}$ J.
(b) $W=pE(\cos60^\circ-\cos120^\circ)=8.0\times10^{-4}\,(0.5+0.5)=8.0\times10^{-4}$ J.
(c) It swings towards $\theta=0$, losing potential energy $U(90^\circ)-U(0^\circ)=8.0\times10^{-4}$ J, which is its greatest kinetic energy, reached as it passes $\theta=0$. (It then overshoots and oscillates about $\theta=0$.)
⚠️ Common mistakes with equipotentials and energy

1. "$E=0$ so $V=0$." $E=-dV/dl$ is a slope. Zero slope means constant $V$, which can be any value.

2. Getting the direction of $\mathbf E$ backwards. $\mathbf E$ points from high to low potential. Positive charges move that way of their own accord; electrons go the other way.

3. Counting pairs twice. Three charges have three pairs, four charges six. Do not write $q_1q_2/r_{12}$ and $q_2q_1/r_{21}$.

4. Using the distance instead of the distance along the field. In a uniform field $\Delta V=E\,d$ only when $d$ is measured parallel to $\mathbf E$ (Example 2).

5. Mixing up $U$ of a charge and $V$. $V$ belongs to the point; $U=qV$ belongs to the charge placed there, and changes sign with $q$.

Interactive 2 — Same Work by Every Path

Three test charges ($q_0=+1.0$ nC) are carried slowly from $A$ to $B$ near a fixed charge $Q$ (grid square $1$ cm). Path 1 goes out radially then round an equipotential; path 2 goes round first, then radially; path 3 wanders. The bars show the external work done so far, added up step by step as $-q_0\mathbf E\cdot\Delta\mathbf l$. Watch the bars grow differently but finish level: $W=q_0(V_B-V_A)$. On the equipotential sections the bars do not move at all.

+5
$V_A$, $V_B$—
path 1, 2, 3 work so far (J)—
$q_0(V_B-V_A)$—
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — the field as the gradient of $V$ in three dimensions

Not required by the CBSE syllabus. Electric field and potential are listed by JEE Main, NEET and JEE Advanced; none names partial derivatives, and this three-dimensional form appears mainly in JEE problems.

When $V$ depends on $x$, $y$ and $z$, Result 2.7 applies along each axis separately. Each component of $\mathbf E$ is minus the rate of change of $V$ along that axis with the other two coordinates held fixed (a partial derivative, written $\partial$): $$E_x=-\frac{\partial V}{\partial x},\qquad E_y=-\frac{\partial V}{\partial y},\qquad E_z=-\frac{\partial V}{\partial z}.$$ For example, if $V=3x^2y-2z$ (volts, metres), then $E_x=-6xy$, $E_y=-3x^2$, $E_z=+2$; at $(1,\,2,\,0)$ m, $\mathbf E=(-12,\,-3,\,2)$ V/m and $|\mathbf E|=\sqrt{157}\approx12.5$ V/m. Going the other way, for a uniform field $\mathbf E=(E_x,E_y)$ and a displacement $(\Delta x,\Delta y)$, $\Delta V=-(E_x\Delta x+E_y\Delta y)$. JEE problems use both forms; the board examination uses only the one-dimensional $E=-dV/dr$.

Exercises

I
Equipotentials and $E=-dV/dl$.
  • 1 Sketch, with field lines, the equipotential surfaces of (a) a uniform field, (b) a positive point charge, (c) an electric dipole. In (b), why are equipotentials drawn at equal steps of $V$ not equally spaced?
  • 2 Explain why two equipotential surfaces can never cross, and why field lines always cross equipotentials at right angles.
  • 3 Two large parallel plates $4.0$ mm apart are connected to a $120$ V supply. Find the field between them.
  • 4 A uniform field of $2.0\times10^3$ V/m points along $+x$. Find $V_A-V_B$ for $A=(0,0)$ and $B=(5.0\ \text{cm},\ 12\ \text{cm})$. Which point is at the higher potential?
  • 5 Plane equipotentials, drawn every $10$ V, are $2.0$ cm apart. Find the size and direction of the field.
  • 6 Along the $x$-axis $V=6x-3x^2$ ($V$ in volts, $x$ in metres). Find $E_x$ at $x=0$ and at $x=2.0$ m, and the point where $E_x=0$.
  • 7 The potential has the same value, $4.0$ V, everywhere inside a certain region. What can you say about the field there?
  • 8 How much work is done in carrying a charge of $5.0\ \mu\text{C}$ a distance of $3.0$ cm along an equipotential surface at $200$ V?
1 (a) Equally spaced planes perpendicular to the parallel field lines. (b) Concentric spheres crossed by radial lines. (c) Closed surfaces round each charge, squashed towards the equatorial plane, which is the $V=0$ surface; field lines run from $+q$ to $-q$ crossing them at right angles. In (b), $E=|\Delta V|/\Delta r$ falls as $1/r^2$, so a fixed step $\Delta V$ needs a larger $\Delta r$ farther out.
2 A point on two surfaces would have two different potentials — impossible, since $V$ has one value at each point. The field does no work on a charge moving along an equipotential, so $\mathbf E\cdot\delta\mathbf l=0$ for every displacement in the surface: $\mathbf E$ is perpendicular to it.
3 $E=\dfrac{V}{d}=\dfrac{120}{4.0\times10^{-3}}=3.0\times10^{4}$ V/m.
4 Only the $x$-displacement counts: $V_A-V_B=E\,\Delta x=(2.0\times10^3)(0.050)=100$ V. $A$ is higher.
5 $E=\dfrac{10\ \text{V}}{0.020\ \text{m}}=500$ V/m, perpendicular to the planes, pointing towards the lower-potential planes.
6 $E_x=-\dfrac{dV}{dx}=-(6-6x)=6x-6$. At $x=0$: $-6.0$ V/m (along $-x$). At $x=2.0$ m: $+6.0$ V/m. $E_x=0$ at $x=1.0$ m, where $V$ is greatest ($3.0$ V).
7 $E=-dV/dl=0$ everywhere in the region; the value $4.0$ V itself tells you nothing about the field.
8 Zero: $\Delta V=0$ along an equipotential.
II
Potential energy of systems of charges. $k=9.0\times10^9\ \text{N m}^2\,\text{C}^{-2}$, $e=1.6\times10^{-19}$ C, $m_e=9.1\times10^{-31}$ kg.
  • 1 Find the potential energy of two charges of $+3.0\ \mu\text{C}$ held $30$ cm apart.
  • 2 Find the potential energy, in J and in eV, of an electron and a proton $0.53\ \text{Å}$ apart.
  • 3 Two charges of $+2.0\ \mu\text{C}$ are $20$ cm apart. How much work is needed to push them to $10$ cm apart?
  • 4 Three charges of $+1.0\ \mu\text{C}$ are placed at the corners of an equilateral triangle of side $10$ cm. Find the potential energy of the system.
  • 5 Four charges of $+1.0\ \mu\text{C}$ are placed at the corners of a square of side $10$ cm. Find the potential energy of the system.
  • 6 Repeat part 5 with charges $+q,-q,+q,-q$ ($q=1.0\ \mu\text{C}$) taken in order round the square. Interpret the sign.
  • 7 A proton with kinetic energy $1.0$ MeV moves straight towards another proton held fixed. Find the distance of closest approach.
  • 8 Two electrons are released from rest $1.0\times10^{-10}$ m apart. Find the speed of each when they are very far apart.
Check: potential energy of up to four point charges
total $U$—
pairs counted—
work to separate completely—
1 $U=\dfrac{(9.0\times10^9)(3.0\times10^{-6})^2}{0.30}=0.27$ J.
2 $U=-\dfrac{ke^2}{r}=-\dfrac{(9.0\times10^9)(1.6\times10^{-19})^2}{5.3\times10^{-11}}\approx-4.3\times10^{-18}$ J $\approx-27$ eV.
3 $kq^2=(9.0\times10^9)(4.0\times10^{-12})=0.036$ J m. $W=\Delta U=0.036\left(\dfrac{1}{0.10}-\dfrac{1}{0.20}\right)=0.36-0.18=0.18$ J.
4 Three pairs, all at $a$: $U=\dfrac{3kq^2}{a}=\dfrac{3(9.0\times10^9)(1.0\times10^{-12})}{0.10}=0.27$ J.
5 Six pairs: four sides at $a$ and two diagonals at $a\sqrt2$. $U=\dfrac{kq^2}{a}\left(4+\dfrac{2}{\sqrt2}\right)=(0.090\ \text{J})(5.41)\approx0.49$ J.
6 The four sides join unlike charges, the two diagonals like charges: $U=\dfrac{kq^2}{a}\left(-4+\sqrt2\right)=(0.090)(-2.59)\approx-0.23$ J. Negative: the arrangement is bound; $0.23$ J must be supplied to pull it apart.
7 $K=\dfrac{ke^2}{d}\Rightarrow d=\dfrac{(9.0\times10^9)(1.6\times10^{-19})^2}{1.0\times10^6\times1.6\times10^{-19}}=1.44\times10^{-15}\ \text{m}\approx1.4$ fm.
8 $U=\dfrac{ke^2}{r}=\dfrac{2.30\times10^{-28}}{1.0\times10^{-10}}=2.30\times10^{-18}$ J is shared equally (equal masses, equal and opposite momenta): each gets $1.15\times10^{-18}$ J, so $v=\sqrt{\dfrac{2(1.15\times10^{-18})}{9.1\times10^{-31}}}\approx1.6\times10^{6}$ m/s.
III
Energy in an external field.
  • 1 A charge of $-2.0\ \mu\text{C}$ is placed at a point where the external potential is $300$ V. What is its potential energy?
  • 2 A dipole of moment $2.0\times10^{-8}$ C m is in a uniform field of $1.0\times10^4$ N/C. Find its potential energy when $\mathbf p$ is along $\mathbf E$, perpendicular to $\mathbf E$, and opposite to $\mathbf E$.
  • 3 How much work is needed to turn the dipole of part 2 from $\theta=0^\circ$ to $\theta=60^\circ$?
  • 4 Use $U(\theta)=-pE\cos\theta$ to explain which orientation of a dipole in a uniform field is stable and which is unstable.
  • 5 Charges $+4.0$ nC and $-4.0$ nC, $3.0$ cm apart, are placed in an external field at points where the external potential is $+100$ V and $-50$ V respectively. Find the total potential energy of the pair (include their interaction).
  • 6 Modelling. A water molecule has $p=6.2\times10^{-30}$ C m. Compare the largest change in its energy as it turns in a field of $1.0\times10^6$ V/m, $2pE$, with the typical thermal energy $k_BT$ at $300$ K ($k_B=1.38\times10^{-23}$ J/K). What does this suggest about how well the field can line the molecules up?
  • 7 The dipole of part 2 is released from rest at $\theta=90^\circ$. What is its greatest rotational kinetic energy?
1 $U=qV=(-2.0\times10^{-6})(300)=-6.0\times10^{-4}$ J.
2 $pE=2.0\times10^{-4}$ J: $U=-2.0\times10^{-4}$ J (along), $0$ (perpendicular), $+2.0\times10^{-4}$ J (opposite).
3 $W=pE(\cos0^\circ-\cos60^\circ)=(2.0\times10^{-4})(0.5)=1.0\times10^{-4}$ J.
4 $U$ is least at $\theta=0$: any twist raises $U$, and the dipole is pushed back — stable. $U$ is greatest at $\theta=180^\circ$: any twist lowers $U$, and the dipole runs away from that position — unstable. (Equilibrium in both, since $dU/d\theta=pE\sin\theta=0$ — the torque is zero.)
5 $U=q_1V_1+q_2V_2+\dfrac{kq_1q_2}{r}=(4.0\times10^{-9})(100)+(-4.0\times10^{-9})(-50)+\dfrac{(9.0\times10^9)(-16\times10^{-18})}{0.030}$ $=4.0\times10^{-7}+2.0\times10^{-7}-4.8\times10^{-6}=-4.2\times10^{-6}$ J.
6 $2pE=2(6.2\times10^{-30})(1.0\times10^6)=1.24\times10^{-23}$ J; $k_BT=4.14\times10^{-21}$ J, about $330$ times larger. Thermal jostling wins easily: the field only slightly biases the random orientations, so the alignment (and the polarisation it produces, Section 2.3) is small and grows in proportion to $E$.
7 $K_\text{max}=U(90^\circ)-U(0^\circ)=pE=2.0\times10^{-4}$ J, as it swings through $\theta=0$.
IV
Thinking with potential energy.
  • 1 Modelling. In a sodium chloride crystal a $\text{Na}^+$ ion and its nearest $\text{Cl}^-$ ion are $0.28$ nm apart. Treating them as point charges $\pm e$, find their potential energy in eV.
  • 2 Modelling. A lightning flash carries $20$ C through a potential difference of about $1.0\times10^8$ V. Estimate the energy released. For how long could it run a $100$ W lamp?
  • 3 In an electron gun, electrons leave a cathode at rest and reach an anode $1000$ V higher. Find their speed at the anode. Does the distance between the electrodes matter?
  • 4 A point charge produces a potential of $900$ V at $10$ cm. Find the radii of the equipotentials at $900$ V, $600$ V and $300$ V, and explain what the spacing tells you about the field.
  • 5 Charges $+q$, $-q$ and $+q$ lie on a straight line, each a distance $a$ from the next. Find the potential energy of the system for $q=1.0\ \mu\text{C}$, $a=5.0$ cm.
  • 6 Explain, using the idea of a conservative field, why the work done in carrying a charge round any closed path in an electrostatic field is zero, and why this would not be true for a frictional force.
1 $U=-\dfrac{ke^2}{r}=-\dfrac{2.30\times10^{-28}}{2.8\times10^{-10}}\approx-8.2\times10^{-19}$ J $\approx-5.1$ eV. (The measured energy to separate the ions is close to this; the rest of the crystal adds a correction.)
2 $U\approx QV=(20)(1.0\times10^8)=2.0\times10^{9}$ J. A $100$ W lamp would run for $2.0\times10^7$ s — about $230$ days. (The potential difference falls as the cloud discharges, so this is an overestimate, perhaps by a factor of two.)
3 $K=1000$ eV $=1.6\times10^{-16}$ J; $v=\sqrt{\dfrac{2(1.6\times10^{-16})}{9.1\times10^{-31}}}\approx1.9\times10^{7}$ m/s. The distance does not matter — only the potential difference.
4 $V\propto1/r$ and $V=900$ V at $10$ cm, so $r=10\times\dfrac{900}{V}$ cm: $10$ cm, $15$ cm and $30$ cm. The gaps for equal $300$ V steps grow ($5$ cm, then $15$ cm): the field weakens with distance.
5 Pairs: $(+,-)$ at $a$, $(-,+)$ at $a$, $(+,+)$ at $2a$. $U=\dfrac{kq^2}{a}\left(-1-1+\dfrac12\right)=-1.5\,\dfrac{kq^2}{a}=-1.5\times\dfrac{9.0\times10^{-3}}{0.050}=-0.27$ J.
6 In a conservative field the work between two points depends only on the end points, $W=q(V_B-V_A)$. Round a closed path $B=A$, so $W=0$. Friction always opposes the motion, so it does negative work on every part of the path and the total round a loop is never zero; no potential energy can be defined for it.

Original problems in the style of the CBSE board paper, JEE Main, JEE Advanced and NEET. Try each one against the clock (about 2 minutes for an MCQ, 5 minutes for a long answer) before opening the solution.

🎯 Exam corner — equipotentials and potential energy BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks Show that the electric field at any point is perpendicular to the equipotential surface through that point. Draw the equipotential surfaces for a uniform electric field.

Argument Move a charge $q$ by $\delta\mathbf l$ along the surface. The work done by the field is $q\,\mathbf E\cdot\delta\mathbf l=qE\,\delta l\cos\theta$. But $\Delta V=0$ on the surface, so the work $=-q\Delta V=0$. Since $q$, $E$, $\delta l\ne0$, $\cos\theta=0$: $\mathbf E\perp\delta\mathbf l$ for every direction in the surface, i.e. $\mathbf E$ is normal to it.
Diagram Equally spaced parallel planes (lines in the drawing) perpendicular to the parallel, equally spaced field lines, with $V$ decreasing in the direction of $\mathbf E$.
Exam tip Label the direction of decreasing $V$ on your diagram — it is often the mark examiners look for.

E2. Board · 3 marks Derive an expression for the potential energy of an electric dipole of moment $\mathbf p$ in a uniform field $\mathbf E$. Hence find the work needed to turn a dipole of moment $5.0\times10^{-9}$ C m from its position of stable equilibrium to its position of unstable equilibrium in a field of $4.0\times10^5$ N/C.

Torque At angle $\theta$ the field exerts a torque $pE\sin\theta$ tending to reduce $\theta$. To turn the dipole slowly through $d\theta$ an external torque does work $dW=pE\sin\theta\,d\theta$.
Work From $\theta_1$ to $\theta_2$ (uses integration): $W=\displaystyle\int_{\theta_1}^{\theta_2}pE\sin\theta\,d\theta=pE(\cos\theta_1-\cos\theta_2)$. This is stored as potential energy: $U(\theta_2)-U(\theta_1)$.
Choice of zero Take $U(90^\circ)=0$ and $\theta_1=90^\circ$: $U(\theta)=-pE\cos\theta=-\mathbf p\cdot\mathbf E$.
Number Stable $\theta=0$ to unstable $\theta=180^\circ$: $W=pE(1-(-1))=2pE=2(5.0\times10^{-9})(4.0\times10^5)=4.0\times10^{-3}$ J. Exam tip: state the zero of energy you chose; the answer for $W$ does not depend on it.

E3. JEE Main Charges $q$, $2q$ and $-3q$ are placed at the corners of an equilateral triangle of side $a$. The electrostatic potential energy of the system is (A) $-7kq^2/a$ (B) $+7kq^2/a$ (C) $-11kq^2/a$ (D) zero.

Pairs $U=\dfrac{k}{a}\left[(q)(2q)+(q)(-3q)+(2q)(-3q)\right]=\dfrac{kq^2}{a}(2-3-6)$.
Answer (A) $-7kq^2/a$. Exam tip: (D) is the trap — the charges add to zero, which makes the potential at the centroid zero, not the energy.

E4. NEET Three plane equipotential surfaces at $10$ V, $20$ V and $30$ V are perpendicular to the $x$-axis and $10$ cm apart, with the potential increasing along $+x$. The electric field is (A) $100$ V/m along $+x$ (B) $100$ V/m along $-x$ (C) $200$ V/m along $-x$ (D) $10$ V/m along $-x$.

Concept $E=|\Delta V|/\Delta l$ normal to the surfaces, pointing towards lower potential.
Value $E=\dfrac{10\ \text{V}}{0.10\ \text{m}}=100$ V/m; $V$ rises along $+x$, so $\mathbf E$ points along $-x$.
Answer (B). Exam tip: always convert cm to m — (D) is what you get by forgetting.

E5. JEE Advanced (Numerical value.) Two small particles, $A$ of mass $1.0$ g and $B$ of mass $2.0$ g, each carry a charge of $+1.0\ \mu\text{C}$. They are released from rest $3.0$ cm apart on a smooth horizontal insulating surface. Find the final speed of $A$, in m/s, when they are very far apart.

Concepts No external horizontal force: momentum is conserved, $m_Av_A=m_Bv_B\Rightarrow v_B=v_A/2$. Energy: the initial potential energy becomes kinetic energy.
Energy $U=\dfrac{kq^2}{r}=\dfrac{(9.0\times10^9)(1.0\times10^{-12})}{0.030}=0.30$ J $=\tfrac12m_Av_A^2+\tfrac12m_B\left(\tfrac{v_A}{2}\right)^2=\tfrac34m_Av_A^2$.
Answer $v_A^2=\dfrac{4(0.30)}{3(1.0\times10^{-3})}=400$, so $v_A=\mathbf{20}$ m/s (and $v_B=10$ m/s). Exam tip: when both bodies are free, never give all of $U$ to one of them — share it using momentum conservation.

E6. Board · Case study In a cathode-ray tube, an electron beam passes between two parallel deflecting plates $1.0$ cm apart, held at $+60$ V and $-60$ V. (a) Find the field between the plates and the spacing of the equipotential surfaces drawn every $20$ V. (2 marks) (b) How much work does the field do on an electron that moves from the $-60$ V plate to the $+60$ V plate? (1 mark) (c) How much work is done on an electron moving parallel to the plates along the $0$ V surface? (1 mark)

(a) $\Delta V=120$ V over $0.010$ m: $E=1.2\times10^{4}$ V/m, from the $+60$ V plate to the $-60$ V plate. Equipotentials every $20$ V are planes parallel to the plates, $\dfrac{20}{1.2\times10^4}\approx1.7\times10^{-3}$ m $\approx1.7$ mm apart.
(b) $W^\text{field}=q(V_A-V_B)=(-e)(-60-60)=+120$ eV $=1.92\times10^{-17}$ J $\approx1.9\times10^{-17}$ J (the electron is pulled towards the positive plate).
(c) Zero: the path lies on an equipotential. Exam tip: in case studies, quote the general rule ("no work along an equipotential") — it earns the mark even before the arithmetic.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. An ECG records potential differences between electrodes on the right arm, left arm and left leg. How can three such readings reveal the direction of the heart's electric dipole — and why must they always satisfy a simple sum rule?

Why it matters. In 1903 Willem Einthoven in Leiden built the first practical electrocardiograph and modelled the heart as a dipole at the centre of an equilateral triangle formed by the three limbs. The direction of the dipole at the peak of each beat — the "electrical axis" — is still read off every ECG worldwide; an axis that has swung far from normal can point to an enlarged heart chamber or a blocked conduction pathway.

What to do.

  1. Lead I is $V_\text{LA}-V_\text{RA}$, lead II is $V_\text{LL}-V_\text{RA}$ and lead III is $V_\text{LL}-V_\text{LA}$. Using only the fact that potential has one value at each point, show that $\text{II}=\text{I}+\text{III}$ (Einthoven's law). Which result of this section is this "sum round a loop" rule?
  2. Sample peak readings: lead I $=0.80$ mV, lead III $=0.60$ mV. Predict lead II.
  3. In Einthoven's model each lead reads the component of a dipole-like vector $\mathbf M$ along one side of the triangle: $\text{I}=M\cos\alpha$, $\text{II}=M\cos(\alpha-60^\circ)$, $\text{III}=M\cos(\alpha-120^\circ)$, with the angle $\alpha$ measured from the horizontal, positive downwards. Use leads I and III to find $M$ and $\alpha$, then check lead II.
  4. The normal axis lies between about $-30^\circ$ and $+90^\circ$. Is this heart's axis normal?

Tools: Calculator; a protractor and graph paper for a vector diagram.

Hand in: The proof of Einthoven's law, your values of $M$ and $\alpha$ with working, the check on lead II, and a labelled triangle diagram showing $\mathbf M$.

Going further. Real hearts are not point dipoles and the body is not a uniform conductor. Find out why modern 12-lead ECGs add chest electrodes, and what extra information they give.

Instructor note: $(V_\text{LA}-V_\text{RA})+(V_\text{LL}-V_\text{LA})=V_\text{LL}-V_\text{RA}$, the electrostatic form of "zero work round a closed path" (Principle 2.1). Lead II $=1.40$ mV. From I and III: $M\cos\alpha=0.80$ and $-0.40+0.866\,M\sin\alpha=0.60$, so $M\sin\alpha=1.155$, $\alpha\approx55^\circ$, $M\approx1.4$ mV; then $M\cos(\alpha-60^\circ)\approx1.40$ mV ✓. The axis is normal.

🧪 Self-check quiz

0/5 correct

Q1. The angle between the electric field and an equipotential surface at any point is

Q2. The equipotential surfaces of an isolated point charge are

Q3. Along the $x$-axis $V=5x^2$ (volts, metres). At $x=2$ m the field is

Q4. Two electrons held close together are released. As they fly apart, the electrostatic potential energy of the pair

Q5. A dipole in a uniform field has its lowest potential energy when the angle between $\mathbf p$ and $\mathbf E$ is

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