2.1 Electric Potential due to Point Charges and Dipoles
The electrostatic force is conservative, so every point near a set of charges has a potential — work per unit charge from infinity. For a point charge $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$, for a short dipole $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$, and for many charges the potentials simply add.
An electron microscope fires electrons at a specimen at a large fraction of the speed of light. To get them moving, it releases them from a hot filament and lets them "fall" through a difference in electrical level of tens of thousands of volts — much as a ball gains speed falling off a cliff. A $1.5$ V torch cell, the $230$ V mains, the $11$ kV of a local substation: each of these numbers measures the same thing, the electric potential difference between two points, which tells you how much energy each coulomb of charge gains or loses in going from one to the other.
In Chapter 1 the electric field told us the force on a charge. This section develops the energy picture. Because the electrostatic force, like gravity, is conservative, we can attach to every point in space a single number, the electrostatic potential $V$: the work needed, per unit charge, to bring a charge there from far away. We find $V$ for a point charge, for an electric dipole and for any collection of charges, where — unlike fields — potentials add as plain numbers with no directions to worry about. A charged spherical shell finishes the section and shows that the potential can be constant in a region where the field is zero.
Work, Conservative Forces and Potential Energy
Lift a book from the floor to a shelf slowly, so that it gains no kinetic energy. You do work $mgh$ against gravity, and that work is stored as gravitational potential energy; let the book fall and it is returned as kinetic energy. The amount does not depend on the route you take to the shelf, because gravity is a conservative force (Class 11 Physics, Chapter 5 — Work, Energy and Power).
Now replace gravity by the electric field of a fixed charge $Q$, and the book by a small positive test charge $q$. To move $q$ slowly towards a positive $Q$ we must push it against the repulsion, applying an external force $\mathbf F_\text{ext}=-q\mathbf E$ that just balances the electric force at every point. Like gravity, the Coulomb force is a central force — it acts along the line to a fixed point and depends only on the distance — and every central force of this kind is conservative. Superposition then makes the field of any set of fixed charges conservative.
The work done by the electrostatic field of a set of stationary charges on a charge $q$ that moves from a point $R$ to a point $P$ depends only on the end points $R$ and $P$, not on the path joining them. Equivalently, the work done by the field around any closed path is zero.
Hence we can define an electrostatic potential energy $U$ of the charge $q$: the work done by an external agent in moving $q$ slowly (without acceleration) from $R$ to $P$ equals the increase in its potential energy, $$W_{RP}^{\,\text{ext}}=U_P-U_R=-W_{RP}^{\,\text{field}}.$$
Only differences of potential energy have a physical meaning, so we are free to choose where $U=0$. For charges confined to a finite region the natural choice is a point infinitely far away, where the field has died out. Then the potential energy at $P$ is the external work needed to bring $q$ in from infinity to $P$. This work is proportional to $q$ itself — double the test charge and every force, and so every bit of work, doubles. Dividing by $q$ therefore gives a quantity that belongs to the point and to the charges that make the field, not to the test charge.
A convenient energy unit in atomic physics is the electron volt, the energy gained by a charge $e$ moving through a potential difference of $1$ V: $1\ \text{eV}=(1.6\times10^{-19}\ \text{C})(1\ \text{V})=1.6\times10^{-19}\ \text{J}$.
Turned round, Definition 2.1 says what happens to a free charge. If a charge $q$ moves from $A$ to $B$ with only the electric force acting on it, the field does work $W=q(V_A-V_B)$ on it, and by the work–energy theorem $$\Delta K=q\,(V_A-V_B)=-q\,\Delta V .$$ A positive charge speeds up as it moves to lower potential (downhill); a negative charge such as an electron speeds up as it moves to higher potential.
Potential due to a Point Charge
Put a point charge $Q$ at the origin and bring a unit positive charge in from infinity along a radial line to the point $P$ at distance $r$ (by Principle 2.1 any other path gives the same answer). At distance $r'$ the field pushes outwards on the unit charge with force $kQ/r'^2$, so the external agent applies $kQ/r'^2$ inwards. Moving inwards by $|dr'|$ it does work $\dfrac{kQ}{r'^2}|dr'|$. Adding these pieces (uses integration): $$V(r)=-\int_\infty^{r}\frac{kQ}{r'^2}\,dr'=\left[\frac{kQ}{r'}\right]_\infty^{r}=\frac{kQ}{r}.$$ The minus sign appears because $r'$ decreases as we move in, while the external force points inwards.
The electrostatic potential at distance $r$ from a point charge $Q$ in vacuum, taking $V=0$ at infinity, is $$V(r)=\frac{1}{4\pi\varepsilon_0}\,\frac{Q}{r}=\frac{kQ}{r}.$$ It is positive all round a positive charge and negative all round a negative charge, has the same value at every point of a sphere centred on the charge, and falls off as $1/r$ — more slowly than the field $E=kQ/r^2$. For a point charge, $V=E\,r$ at every point. The result also holds outside any spherically symmetric charge distribution of total charge $Q$, with $r$ measured from its centre.
Two features are worth fixing in your mind. First, $V$ is a scalar with a sign: near a negative charge an external agent has to hold back a positive test charge as it is brought in, so the work it does is negative. Second, $V$ and $E$ carry different information. $E$ is the rate at which $V$ changes with distance (Section 2.2 makes this exact), so a field that is large means a potential that changes quickly, not a potential that is large.
Potential due to an Electric Dipole
An electric dipole is a pair of charges $-q$ and $+q$ a distance $2a$ apart, with dipole moment $\mathbf p$ of magnitude $p=q(2a)$ pointing from $-q$ to $+q$ (Section 1.2). Take its centre as origin and a point $P$ at distance $r$, along a line making angle $\theta$ with $\mathbf p$. Let $r_1$ and $r_2$ be the distances of $P$ from $+q$ and $-q$. Potentials add as numbers, so $$V=kq\left(\frac{1}{r_1}-\frac{1}{r_2}\right)=kq\,\frac{r_2-r_1}{r_1r_2}.$$ Far from the dipole ($r\gg a$) the lines from the two charges to $P$ are almost parallel, and simple geometry gives $r_1\approx r-a\cos\theta$ and $r_2\approx r+a\cos\theta$. Then $r_2-r_1\approx2a\cos\theta$ and $r_1r_2\approx r^2$ (the correction $a^2\cos^2\theta$ is negligible beside $r^2$), so $$V\approx kq\,\frac{2a\cos\theta}{r^2}=\frac{kp\cos\theta}{r^2}.$$
At a point at distance $r$ from the centre of a dipole of moment $\mathbf p$, along a direction at angle $\theta$ to $\mathbf p$, and for $r\gg a$, $$V=\frac{1}{4\pi\varepsilon_0}\,\frac{p\cos\theta}{r^2}=\frac{1}{4\pi\varepsilon_0}\,\frac{\mathbf p\cdot\hat{\mathbf r}}{r^2}.$$ On the axis ($\theta=0$ or $180^\circ$): $V=\pm\dfrac{kp}{r^2}$ — positive on the side of $+q$. On the equatorial plane ($\theta=90^\circ$): $V=0$ at every point, even though the field there is not zero. On the axis the exact result, valid for any $r\gt a$, is $V=\dfrac{kp}{r^2-a^2}$.
Compare this with a single charge. A dipole's potential falls as $1/r^2$, not $1/r$, because from far away the two equal and opposite charges almost cancel; only their slight difference in distance survives. It also depends on direction: the potential is not the same everywhere on a sphere round a dipole, as it is round a point charge. Both features match the dipole field of Section 1.2, which falls as $1/r^3$ — one power of $r$ faster than the potential, just as $kq/r^2$ is one power faster than $kq/r$.
Potential due to a System of Charges
Since the forces from several charges add (superposition, Section 1.1), so does the work they require, and dividing by the test charge gives the most useful rule of this chapter.
The potential at a point $P$ due to point charges $q_1,q_2,\dots,q_n$ at distances $r_{1P},r_{2P},\dots,r_{nP}$ from $P$ is the algebraic sum of their separate potentials: $$V_P=\frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_{1P}}+\frac{q_2}{r_{2P}}+\cdots+\frac{q_n}{r_{nP}}\right)=k\sum_{i=1}^{n}\frac{q_i}{r_{iP}} .$$ Each $q_i$ is used with its sign; no directions or components are needed. For a continuous charge distribution, divide it into small elements $dq$ and add: $V_P=k\displaystyle\int\frac{dq}{r}$.
- Find the distance $r_i$ from each charge to the point $P$ (Pythagoras, symmetry). Convert to metres and coulombs.
- Add $kq_i/r_i$ with signs. A useful shortcut: if all charges are at the same distance $r$ (centre of a square, a ring, a regular polygon), $V=k(\sum q_i)/r$.
- Do not conclude $V=0$ from $E=0$, or $E=0$ from $V=0$. Fields add as vectors and can cancel where potentials do not, and vice versa.
- To find where $V=0$ on a line through two unlike charges, write $\dfrac{|q_1|}{r_1}=\dfrac{|q_2|}{r_2}$ for each region in turn; such points lie between the charges and beyond the smaller one.
A charged spherical shell. In Section 1.3 Gauss's law showed that a thin spherical shell of radius $R$ carrying charge $Q$ produces the field $kQ/r^2$ outside, as if all its charge were at the centre, and zero field inside. Outside, then, its potential is $kQ/r$. Inside, no work is needed to move a test charge around (there is no force), so the potential cannot change: it keeps the value it has at the surface.
For a thin spherical shell of radius $R$ with total charge $Q$ spread uniformly over it, $$V=\frac{kQ}{r}\quad(r\ge R),\qquad V=\frac{kQ}{R}\quad(r\le R).$$ The potential is the same at every point inside the shell and equal to its value on the surface, although the field inside is zero. The same holds for a charged conducting sphere, solid or hollow, because its charge lives on its outer surface (Section 2.3).
1. Using $1/r^2$ instead of $1/r$. Potential of a point charge is $kq/r$; the field is $kq/r^2$. Halving the distance doubles $V$ but quadruples $E$.
2. Adding potentials as vectors, or dropping signs. Potentials are scalars that carry the sign of the charge: $+q$ and $-q$ at the same distance give $V=0$. Never resolve a potential into components.
3. Confusing "$V=0$" with "$E=0$". At the midpoint of a dipole $V=0$ but $E\ne0$; at the centre of a square of equal charges $E=0$ but $V\ne0$. They answer different questions: $V$ is energy per charge, $E$ is force per charge.
4. Forgetting that only differences are physical. "$V=0$ at infinity" is a choice. Work, energy and the motion of charges depend only on $\Delta V$, which every choice of zero gives the same.
5. Mixing up who does the work. $q(V_B-V_A)$ is the work done by an external agent moving $q$ slowly from $A$ to $B$; the field does the negative of this.
Not required by the CBSE syllabus. Electric potential and Gauss's law are listed by JEE Main, NEET and JEE Advanced; none names the solid sphere, which is a standard application of both, set mostly in JEE problems.
For a non-conducting sphere of radius $R$ with charge $Q$ spread uniformly through its volume, Gauss's law gives $E=\dfrac{kQr}{R^3}$ inside and $\dfrac{kQ}{r^2}$ outside. Outside, $V=kQ/r$ as for any spherical distribution. Inside, start from $V(R)=kQ/R$ and add the work per unit charge needed to move inwards against $E$ (uses integration): $$V(r)=\frac{kQ}{R}+\int_r^R\frac{kQr'}{R^3}\,dr'=\frac{kQ\,(3R^2-r^2)}{2R^3}\qquad(r\le R).$$ At the centre $V=\dfrac32\,\dfrac{kQ}{R}$ — the potential is greatest at the centre, where the field is zero. Compare the hollow shell, for which the potential is constant inside. JEE problems often ask for $V_\text{centre}/V_\text{surface}=3/2$.
Exercises
- 1 How much work is needed to carry a charge of $3.0\ \mu\text{C}$ slowly between two points whose potential difference is $12$ V?
- 2 An electron starts from rest and is accelerated through a potential difference of $500$ V. Find its kinetic energy in eV and in joules, and its final speed.
- 3 As a charge of $+2.0\ \mu\text{C}$ moves from $A$ to $B$, the electric field does $+4.0\times10^{-5}$ J of work on it. Find $V_A-V_B$. Which point is at the higher potential?
- 4 Show that $1\ \text{V}=1\ \text{N m/C}$, and find the dimensional formula of potential.
- 5 A proton moves freely from a point at $100$ V to a point at $40$ V, starting from rest. Find the kinetic energy it gains (in eV and J) and its final speed.
- 6 An alpha particle (charge $+2e$) is accelerated from rest through $1.0$ kV. What kinetic energy does it gain, in keV and in joules?
- 7 In the definition of potential the test charge is brought from infinity "slowly". Why is this condition needed, and what would change without it?
- 8 Can the potential be zero at a point where the field is not zero? Can the field be zero where the potential is not? Give an example of each.
- 1 Find the potential at $30$ cm from a point charge of $6.0$ nC.
- 2 At what distance from a $2.0\ \mu\text{C}$ point charge is the potential $1.0$ kV?
- 3 At a certain point near a point charge the potential is $600$ V and the field is $200$ N/C. How far is the point from the charge, and what is the charge?
- 4 In a hydrogen atom the electron is about $0.53\ \text{Å}$ from the proton. Find the potential due to the proton at that distance, and the potential energy $U=qV$ of the electron in eV.
- 5 A short dipole has $p=5.0\times10^{-9}$ C m. Find the potential $30$ cm from its centre (a) on its axis, (b) on its equatorial line, (c) along a line at $120^\circ$ to $\mathbf p$.
- 6 For a short dipole, find the ratio of the potential on the axis to the potential along a line at $60^\circ$ to the axis, both at the same distance $r$.
- 7 The distance from a short dipole is doubled. By what factor does the potential change? Compare with a point charge, and explain the difference.
- 8 A test charge $q_0$ is carried once round a circle of radius $r$ centred on a point charge $Q$. How much work is done? What if the circle is replaced by any closed loop?
- 1 Charges $+2.0$ nC at $x=0$ and $-3.0$ nC at $x=10$ cm. Find the potential at $x=5.0$ cm.
- 2 For the charges of part 1, find the potential at $x=20$ cm.
- 3 For the charges of part 1, find every point on the $x$-axis where $V=0$.
- 4 Charges $+1.0$, $+2.0$ and $+3.0\ \mu\text{C}$ sit at the corners of an equilateral triangle of side $30$ cm. Find the potential at its centroid.
- 5 Six charges of $+1.0$ nC sit at the corners of a regular hexagon of side $10$ cm. Find the potential and the field at its centre.
- 6 A thin spherical shell of radius $12$ cm carries $6.0$ nC. Find the potential at $6.0$ cm, $12$ cm and $24$ cm from its centre.
- 7 An isolated metal sphere of radius $5.0$ cm is at a potential of $900$ V. What charge does it carry?
- 8 Two concentric thin shells have radii $10$ cm and $20$ cm and charges $+4.0$ nC and $-2.0$ nC. Find the potential at the centre, at $15$ cm and at $30$ cm from the centre.
- 1 Modelling. A thousand identical spherical water droplets, each charged to a potential of $10$ V, merge into one big drop. Find the potential of the big drop. (Charge is conserved; volume is conserved.)
- 2 Modelling. An electron microscope accelerates electrons from rest through $50$ kV. Find their kinetic energy in joules and estimate their speed with $K=\tfrac12mv^2$. Why should you distrust the result?
- 3 Modelling. Air breaks down (sparks) when the field exceeds about $3.0\times10^{6}$ V/m. What is the greatest potential to which an isolated metal sphere of radius $15$ cm can be charged in air, and what charge does it then carry?
- 4 Sketch graphs of $E$ and $V$ against $r$ for a charged thin spherical shell of radius $R$, from $r=0$ to $r=3R$. Mark the values at $r=R$.
- 5 Charges $+1.0$ nC are fixed at $(\pm3.0\ \text{cm},\,0)$. Find the potential at the origin and at $(0,\,4.0\ \text{cm})$. Where on the $y$-axis is the potential greatest, and what is the field there?
- 6 A charge of $3.0$ nC is spread uniformly round a thin ring of radius $3.0$ cm. Find the potential at the centre of the ring and at a point on its axis $4.0$ cm from the centre. Hint: every bit of the ring is the same distance from such a point.
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.
E1. Board · 1 mark Assertion (A): The electrostatic potential at the midpoint of an electric dipole is zero, although the electric field there is not zero. Reason (R): Potential is a scalar quantity, while the electric field is a vector quantity. Choose: (A) both A and R true and R is the correct explanation of A; (B) both true but R is not the correct explanation; (C) A true, R false; (D) A false, R true.
E2. Board · 3 marks Derive an expression for the potential at a point at distance $r$ from the centre of a short electric dipole of moment $\mathbf p$, along a line making angle $\theta$ with $\mathbf p$. Hence write the potential on the axial line and on the equatorial line. Find the potential $20$ cm from a dipole of moment $4.0\times10^{-10}$ C m along a line at $60^\circ$ to its axis.
E3. JEE Main (Numerical value.) Point charges $+5\ \mu\text{C}$ and $-2\ \mu\text{C}$ are fixed $6$ cm apart. The potential is zero at two points on the line through them. Find the distance, in cm, of the point that lies outside the segment joining them from the $+5\ \mu\text{C}$ charge.
E4. NEET A thin spherical shell of radius $R$ carries a charge that gives it a potential $V_0$ at its surface. The potential at a distance $R/2$ from its centre is (A) $V_0/2$ (B) $2V_0$ (C) $V_0$ (D) zero.
E5. JEE Advanced (One or more correct.) Two concentric thin spherical shells have radii $R$ and $2R$. The inner shell carries charge $+Q$ and the outer shell carries charge $q$. Which statements are correct? (A) The potential at the common centre is $\dfrac{kQ}{R}+\dfrac{kq}{2R}$. (B) The potential difference between the inner and outer shells does not depend on $q$. (C) If $q=-Q$, the potential is zero at every point outside the outer shell. (D) If $q=-Q$, the potential at the centre is zero.
E6. JEE Main NEET A proton and an alpha particle start from rest and are accelerated through the same potential difference. The ratio $v_p/v_\alpha$ of their final speeds is (A) $1/\sqrt2$ (B) $\sqrt2$ (C) $2$ (D) $1/2$. (Take $m_\alpha=4m_p$, $q_\alpha=2e$.)
Research question. Even on a calm, sunny day the air above us carries a downward electric field of the order of $100$ V/m. How large is the total potential difference between the ground and the upper atmosphere, and why does it not give us shocks?
Why it matters. The "global electric circuit" links thunderstorms all over the planet: they pump charge upwards, and the charge leaks back down through the fair-weather air everywhere else. Atmospheric scientists in Finland, the UK, Brazil and India measure the near-ground field continuously; changes in it are studied in connection with pollution, cosmic rays and solar activity, and the field must be understood by anyone designing lightning protection or high-altitude balloons and aircraft.
What to do.
- Sample balloon data — field strength $E$ (V/m) at heights $h$ (km): $h=0$: $130$; $h=1$: $60$; $h=2$: $35$; $h=5$: $15$; $h=10$: $5$.
- For a uniform field the potential changes by $E\,\Delta h$ over a height $\Delta h$. Using the average field in each layer, estimate the potential difference across each of the four layers.
- Add them to estimate the potential of the air at $10$ km relative to the ground. Sketch $E$ against $h$ and explain why your sum is the area under the graph.
- A person $1.7$ m tall stands in the $130$ V/m field. Estimate the potential difference between the air at head height and the ground — then explain, using what you know about conductors, why the person feels nothing.
Tools: Calculator or spreadsheet; graph paper.
Hand in: A table of layers with $\Delta h$, average $E$ and $\Delta V$; your total; the graph; and a paragraph answering step 4.
Going further. Fair-weather air conducts weakly, so a current of about $2\ \text{pA/m}^2$ flows down everywhere. Multiply by the Earth's surface area ($5.1\times10^{14}\ \text{m}^2$) to estimate the total current, and look up how many thunderstorms are thought to be active at any moment to keep it going.
🧪 Self-check quiz
Q1. Which unit is the same as the volt?
Q2. The potential at distance $r$ from a point charge is $V$. At distance $2r$ it is
Q3. At a point on the equatorial line of a short dipole, at distance $r$ from its centre, the potential is
Q4. Four charges $+q$ sit at the corners of a square of side $a$. The potential at the centre of the square is
Q5. An electron is accelerated from rest through a potential difference of $200$ V. Its kinetic energy is