1.2 The Electric Field, Field Lines and Electric Dipoles
A charge changes the space around it: every other charge placed there feels a force $\mathbf F=q\mathbf E$. Field lines make that field visible, and a pair of equal and opposite charges — an electric dipole — has a field that falls off as $1/r^3$ and turns in a uniform field.
An inkjet printer fires a stream of tiny ink droplets, gives each one a small charge, and then sends it between two metal plates. The plates do not touch the droplets, yet each droplet is pushed sideways by just the right amount to land on its spot on the paper. A photocopier drum, a smoke precipitator in a power station and the electrostatic spray gun that paints a car all work the same way. In each case it is useful to stop asking "which charges are pushing on this droplet?" and ask instead "what is the condition of space at this point, and what force would any charge feel here?" That condition is the electric field.
In Section 1.1 we found the force between charges with Coulomb's law and superposition. This section repackages that force into a field: a vector $\mathbf E$ attached to every point in space, defined as the force per unit positive charge. We find the field of a point charge and of several charges, draw it with electric field lines, and then study the most important charge arrangement after a single charge — the electric dipole, a pair of equal and opposite charges. Dipoles are everywhere: a water molecule is one, and so is any neutral body whose charges have been pushed slightly apart. We find the field of a dipole along its axis and its equatorial line, and the torque that a uniform field exerts on it.
The Electric Field
Place a charge $Q$ somewhere and bring a second, small charge $q_0$ near it. The small charge feels a force, and by Coulomb's law that force is proportional to $q_0$: double $q_0$ and the force doubles. So the ratio $\mathbf F/q_0$ does not depend on the small charge at all — it depends only on $Q$ and on where $q_0$ was placed. That ratio describes what $Q$ has done to the surrounding space.
Put the test charge $q_0$ at distance $r$ from a point charge $Q$. Coulomb's law gives the force $\mathbf F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Qq_0}{r^2}\hat{\mathbf r}$, where $\hat{\mathbf r}$ is the unit vector from $Q$ towards the point. Dividing by $q_0$ gives the field of a point charge, and because forces add as vectors (Principle 1.3), so do fields.
A point charge $Q$ at rest in vacuum produces at distance $r$ the field $$\mathbf E=\frac{1}{4\pi\varepsilon_0}\,\frac{Q}{r^2}\,\hat{\mathbf r},\qquad E=k\frac{|Q|}{r^2},$$ where $\hat{\mathbf r}$ points from $Q$ to the field point. The field points away from a positive charge and towards a negative charge, and it has the same magnitude at every point of a sphere centred on $Q$ (it is spherically symmetric).
For several point charges $q_1,q_2,\dots,q_n$ the field at a point is the vector sum of the fields each charge would produce alone: $$\mathbf E=\mathbf E_1+\mathbf E_2+\cdots+\mathbf E_n=\frac{1}{4\pi\varepsilon_0}\sum_{i=1}^{n}\frac{q_i}{r_i^{2}}\,\hat{\mathbf r}_i,$$ where $r_i$ is the distance from $q_i$ to the point and $\hat{\mathbf r}_i$ points from $q_i$ to the point.
Two features are worth noticing. First, the field of $Q$ does not mention $q_0$: it exists at every point around $Q$ whether or not a test charge is there to feel it. Second, the field is not only a bookkeeping trick. If $Q$ is suddenly moved, a distant charge does not feel the change instantly; the news travels outwards through the field at the speed of light. Changing electric and magnetic fields carry energy and momentum across empty space — that is what light is (Chapter 8). For charges at rest, however, the field is simply Coulomb's law and superposition written per unit charge.
- Draw the charges and the field point $P$ with coordinates. You are finding a property of the point $P$ — there need not be any charge at $P$.
- For each charge find the magnitude $E_i=k|q_i|/r_i^2$ and draw $\mathbf E_i$ as an arrow at $P$: away from $q_i$ if $q_i\gt0$, towards $q_i$ if $q_i\lt0$.
- Resolve each arrow into components, using symmetry to spot components that cancel, and add: $E_x=\sum E_{ix}$, $E_y=\sum E_{iy}$.
- Give $|\mathbf E|=\sqrt{E_x^2+E_y^2}$ and the direction. If a charge $q$ is then placed at $P$, its force is $q\mathbf E$.
Electric Field Lines
A field is a vector at every point — far too many arrows to draw. Michael Faraday found a better picture: draw curves that follow the direction of the field, the way iron filings line up around a magnet. Where the curves crowd together the field is strong; where they spread out it is weak.
Why is the density rule consistent for a point charge? Suppose $N$ lines leave a charge $Q$ evenly in all directions. At distance $r$ they cross a sphere of area $4\pi r^2$, so the number per unit area is $N/(4\pi r^2)\propto1/r^2$ — exactly how $E$ varies. Field lines work as a picture because Coulomb's law is an inverse-square law. (In a flat drawing the lines are a slice through a three-dimensional pattern, so spacing on the page is only a rough guide.)
1. Field lines start on positive charges and end on negative charges. If there is a net charge, the extra lines start (or end) at infinity.
2. In a region with no charge, field lines are continuous smooth curves without breaks.
3. Two field lines never cross. At a crossing point the field would have two directions at once, which is impossible: the field at a point is a single vector.
4. Electrostatic field lines never form closed loops. (This follows from the fact that the electrostatic field does no net work round a closed path — Chapter 2.)
5. The field is strong where the lines are close together and weak where they are far apart. In a uniform electric field — the same magnitude and direction everywhere, as between two large parallel plates — the lines are parallel, straight and equally spaced.
A field line is not the path a charge follows. A charge released from rest starts off along the field line, but once it is moving its inertia carries it off the line wherever the line curves. Only in a uniform field, where the lines are straight, does a charge released from rest move along a field line.
1. Confusing the field with the force. $\mathbf E$ belongs to the point; $\mathbf F=q\mathbf E$ belongs to the charge placed there. A negative charge feels a force opposite to the field lines.
2. Including the charge at $P$ in its own field. The field at $P$ is produced by the other charges. A charge exerts no net force on itself.
3. Adding magnitudes. Fields are vectors: in Example 2, $2.0\times10^4+5.4\times10^3$ N/C is wrong; the correct magnitude is $1.73\times10^4$ N/C.
4. Thinking the field is zero where no line is drawn. Lines are a sample. The field exists everywhere between them; only its strength is shown by their spacing.
The Electric Dipole and Its Field
Many neutral objects are not electrically "quiet". In a water molecule the oxygen atom pulls the shared electrons towards itself, so the oxygen end is slightly negative and the hydrogen end slightly positive. A neutral atom placed in a field has its electron cloud pulled one way and its nucleus the other. In both cases the total charge is zero, yet the positive and negative charges are not centred at the same point. The simplest model of all such objects is a pair of equal and opposite point charges.
The total charge of a dipole is zero, but its field is not: at any point the two charges are at slightly different distances and in slightly different directions, so their fields fail to cancel exactly. We work out the field in the two places that matter most and where the geometry is simplest.
On the axial line. Put the centre $O$ of the dipole at the origin, $-q$ at $x=-a$ and $+q$ at $x=+a$, and take a point $P$ on the axis at $x=r$ with $r\gt a$. Both fields at $P$ lie along the axis. The field of $+q$ (distance $r-a$) points away from it, along $+\hat{\mathbf p}$; the field of $-q$ (distance $r+a$) points towards it, along $-\hat{\mathbf p}$: $$\mathbf E_{+q}=\frac{kq}{(r-a)^2}\,\hat{\mathbf p},\qquad \mathbf E_{-q}=-\frac{kq}{(r+a)^2}\,\hat{\mathbf p}.$$ Adding, $$\mathbf E=kq\left[\frac{1}{(r-a)^2}-\frac{1}{(r+a)^2}\right]\hat{\mathbf p}=kq\,\frac{4ar}{(r^2-a^2)^2}\,\hat{\mathbf p}=\frac{2kpr}{(r^2-a^2)^2}\,\hat{\mathbf p}.$$ On the equatorial line. Now take $P$ at distance $r$ from $O$ on the perpendicular bisector. Each charge is $\sqrt{r^2+a^2}$ from $P$, so both fields have magnitude $\dfrac{kq}{r^2+a^2}$. The components perpendicular to the axis cancel by symmetry; the components along the axis both point from $+q$ towards $-q$, i.e. along $-\hat{\mathbf p}$, and each equals the magnitude times $\cos\alpha=\dfrac{a}{\sqrt{r^2+a^2}}$: $$\mathbf E=-2\cdot\frac{kq}{r^2+a^2}\cdot\frac{a}{\sqrt{r^2+a^2}}\,\hat{\mathbf p}=-\frac{kp}{(r^2+a^2)^{3/2}}\,\hat{\mathbf p}.$$
For a dipole of moment $\mathbf p$ (charges $\pm q$, separation $2a$) and a point at distance $r$ from its centre:
On the axial line ($r\gt a$), along $\mathbf p$: $$\mathbf E_\text{axial}=\frac{1}{4\pi\varepsilon_0}\,\frac{2pr}{(r^2-a^2)^2}\,\hat{\mathbf p}.$$
On the equatorial line, opposite to $\mathbf p$: $$\mathbf E_\text{eq}=-\frac{1}{4\pi\varepsilon_0}\,\frac{p}{(r^2+a^2)^{3/2}}\,\hat{\mathbf p}.$$
Short dipole ($r\gg a$), the forms used most often: $$\mathbf E_\text{axial}\approx\frac{2k\mathbf p}{r^3},\qquad \mathbf E_\text{eq}\approx-\frac{k\mathbf p}{r^3}.$$
Three things to remember. The dipole field falls off as $1/r^3$ — faster than the $1/r^2$ of a single charge, because from far away the two opposite charges look almost as if they were on top of each other and nearly cancel. At the same large distance the axial field is twice the equatorial field and points the opposite way (along $\mathbf p$ on the axis, against $\mathbf p$ on the equator). And far away the field depends on $q$ and $a$ only through the product $p=2qa$: a large charge close together and a small charge far apart look exactly the same.
A Dipole in a Uniform Electric Field
Place a dipole in a uniform field $\mathbf E$, with $\mathbf p$ at angle $\theta$ to $\mathbf E$. The charge $+q$ feels a force $q\mathbf E$ and the charge $-q$ feels $-q\mathbf E$. These forces are equal and opposite, so they add to zero: a uniform field exerts no net force on a dipole, and its centre does not accelerate. But the two forces act along different lines, a perpendicular distance $2a\sin\theta$ apart. They form a couple, whose torque about the centre has magnitude $$\tau=qE\,(2a\sin\theta)=pE\sin\theta.$$ Its direction is the one that turns $\mathbf p$ towards $\mathbf E$, and it is described neatly by a cross product.
A dipole of moment $\mathbf p$ in a uniform electric field $\mathbf E$ experiences zero net force and a torque $$\boldsymbol\tau=\mathbf p\times\mathbf E,\qquad \tau=pE\sin\theta,$$ where $\theta$ is the angle between $\mathbf p$ and $\mathbf E$. The torque tends to rotate $\mathbf p$ into line with $\mathbf E$. It is zero at $\theta=0$ (stable equilibrium: a small twist produces a torque back towards $\theta=0$) and at $\theta=180^\circ$ (unstable equilibrium: a small twist grows), and greatest, $\tau_\text{max}=pE$, at $\theta=90^\circ$.
To turn the dipole slowly from $\theta_1$ to $\theta_2$ an external agent must supply a torque equal and opposite to $\boldsymbol\tau$, and so does work (uses integration) $$W=\int_{\theta_1}^{\theta_2}pE\sin\theta\,d\theta=pE\left(\cos\theta_1-\cos\theta_2\right).$$ Turning it from the stable position $\theta=0$ to $90^\circ$ takes $pE$; turning it right round to $180^\circ$ takes $2pE$. That work is stored as the potential energy of the dipole in the field, which we study properly in Section 2.2.
In a non-uniform field the two forces are no longer equal, so there is a net force as well as a torque. A dipole that has lined up along the field is pulled towards the region where the field is stronger. This is the second half of the explanation of why a rubbed comb picks up bits of paper: the comb's field induces dipoles in the paper, lines them up, and then — because the comb's field is stronger closer to the comb — pulls them in.
Not required by the CBSE syllabus. The field of a dipole and the torque on it are listed by JEE Main, NEET and JEE Advanced (which also lists angular SHM); the field at a general point and the small oscillations are standard applications that no syllabus names.
Field at a general point. At distance $r\gg a$ from a short dipole, along a direction at angle $\theta$ to $\mathbf p$, resolve $\mathbf p$ into $p\cos\theta$ along $\hat{\mathbf r}$ (the point is on the axis of this part) and $p\sin\theta$ perpendicular to it (the point is on the equator of this part). The two results give field components $E_r=\dfrac{2kp\cos\theta}{r^3}$ and $E_\theta=\dfrac{kp\sin\theta}{r^3}$, so $$E=\frac{kp}{r^3}\sqrt{1+3\cos^2\theta},$$ at an angle $\alpha$ to $\hat{\mathbf r}$ with $\tan\alpha=\tfrac12\tan\theta$. Check: $\theta=0$ gives $2kp/r^3$, $\theta=90^\circ$ gives $kp/r^3$. A favourite JEE fact: the field is perpendicular to $\mathbf p$ where $\tan\theta=\sqrt2$ ($\theta\approx54.7^\circ$).
Small oscillations. A dipole with moment of inertia $I$ about its centre, twisted through a small angle $\theta$ from the field, feels a restoring torque $pE\sin\theta\approx pE\,\theta$. This is angular simple harmonic motion (Class 11 Physics, Chapter 13) with period $$T=2\pi\sqrt{\frac{I}{pE}}.$$ Quadrupling the field halves the period. Neither result is required for the board examination.
Exercises
- 1 Find the electric field $30$ cm from a point charge of $+2.0\ \mu\text{C}$.
- 2 A charge of $-3.0$ nC placed at a point $P$ feels a force of $6.0\times10^{-6}$ N towards the east. Find the electric field at $P$.
- 3 At what distance from a $1.0$ nC point charge is the field $100$ N/C?
- 4 The field of a point charge is $400$ N/C at distance $r$. What is it at $2r$ and at $r/2$?
- 5 Find the force on a proton and its acceleration in a uniform field of $3.0\times10^4$ N/C.
- 6 What uniform field (magnitude and direction) would hold an electron at rest against gravity?
- 7 An oil drop of mass $1.96\times10^{-15}$ kg is held stationary between horizontal plates where the field is $4.0\times10^4$ N/C, pointing downwards. Find its charge and the number of excess electrons on it.
- 8 Charges $+4.0$ nC at $x=0$ and $+9.0$ nC at $x=10$ cm. Where on the $x$-axis is the electric field zero?
- 1 Charges $+5.0$ nC at $x=0$ and $-5.0$ nC at $x=10$ cm. Find the field at the midpoint.
- 2 For the same pair, find the field at $x=20$ cm.
- 3 Equal charges $+1.0$ nC sit at the four corners of a square of side $10$ cm. (a) What is the field at the centre? (b) One of the charges is replaced by $-1.0$ nC. Find the field at the centre now.
- 4 Charges of $+2.0$ nC are placed at two corners of an equilateral triangle of side $10$ cm. Find the field at the third corner.
- 5 Charges $+4.0$ nC are fixed at $(-3,\,0)$ and $(3,\,0)$. Find the field at $(0,\,4)$.
- 6 Give two reasons why two electric field lines can never cross.
- 7 Charges $+2q$ and $-q$ are drawn with $10$ field lines per $q$. How many lines leave $+2q$, how many end on $-q$, and what happens to the rest? Sketch the pattern.
- 8 In some region the field lines are straight, parallel to the $x$-axis, and get closer together as $x$ increases. Is the field uniform? Where is it strongest? Could the field lines in a charge-free region be closed loops?
- 1 Charges $\pm3.0$ nC are $2.0$ cm apart. Find the magnitude and direction of the dipole moment.
- 2 For this dipole find the field on the axis $20$ cm from the centre, exactly and with the short-dipole formula.
- 3 Find the field on its equatorial line $20$ cm from the centre, exactly and approximately. State the direction.
- 4 The axial field of a short dipole at $10$ cm is $E$. At what distance on the equatorial line is the field also $E$?
- 5 The field on the axis of a short dipole is $160$ N/C at $5.0$ cm. Find the field at $10$ cm on the axis and at $10$ cm on the equatorial line.
- 6 Show that $\dfrac{2kp}{r^3}$ has the unit N/C.
- 7 A dipole has $2a=1.0$ cm. Beyond what distance on the axis does the short-dipole formula underestimate the exact field by less than $1\%$?
- 8 Estimate the field of a water molecule ($p=6.2\times10^{-30}$ C m) at $2.0$ nm on its axis.
- 1 A dipole of moment $2.0\times10^{-8}$ C m makes $60^\circ$ with a uniform field of $5.0\times10^4$ N/C. Find the torque on it.
- 2 Explain why the net force on a dipole in a uniform field is zero, while the torque need not be.
- 3 At what angles between $\mathbf p$ and $\mathbf E$ is the torque on a dipole half its maximum value?
- 4 A dipole made of charges $\pm1.0\ \mu\text{C}$ experiences a maximum torque of $8.0\times10^{-3}$ N m in a field of $4.0\times10^5$ N/C. Find the separation of the charges.
- 5 A dipole $\mathbf p=(2.0\,\hat{\mathbf i}+3.0\,\hat{\mathbf j})\times10^{-9}$ C m is in a field $\mathbf E=4.0\times10^5\,\hat{\mathbf i}$ N/C. Find the torque vector.
- 6 Describe the positions of stable and unstable equilibrium of a dipole in a uniform field, and explain the difference.
- 7 How much work is needed to turn the dipole of part 1 from its stable position through $90^\circ$? Through $180^\circ$?
- 8 Modelling. A water molecule ($p=6.2\times10^{-30}$ C m) is in a field of $1.0\times10^6$ N/C. Find the greatest torque on it and the work needed to turn it from aligned to anti-aligned. Compare that work with the typical thermal energy of a molecule at room temperature, about $4\times10^{-21}$ J. What do you conclude about how well the field lines up the molecules?
E1. Board · 3 marks Derive an expression for the electric field at a point on the equatorial line of an electric dipole, at distance $r$ from its centre. Hence find the field at $15$ cm on the equatorial line of a dipole made of charges $\pm0.50\ \mu\text{C}$ placed $2.0$ mm apart.
E2. Board · 1 mark Assertion (A): At a large distance $r$, the field on the axis of a short dipole is twice the field at the same distance on its equatorial line. Reason (R): The field of a dipole falls off as $1/r^3$. (A) Both A and R are true and R is the correct explanation of A. (B) Both are true but R is not the correct explanation of A. (C) A is true, R is false. (D) A is false, R is true.
E3. JEE Main A charge $+4q$ is at $x=0$ and a charge $-q$ is at $x=d$. The electric field on the $x$-axis is zero at (A) $x=d/3$ (B) $x=2d/3$ (C) $x=2d$ (D) $x=-d$.
E4. NEET An electric dipole of moment $p$ is in stable equilibrium in a uniform field $E$. The work needed to turn it through $60^\circ$ is (A) $pE$ (B) $pE/2$ (C) $\sqrt3\,pE/2$ (D) $2pE$.
E5. JEE Advanced (One or more correct.) A short dipole $\mathbf p=p\,\hat{\mathbf i}$ is at the origin. Points $A=(r,0)$, $B=(0,r)$ and $C=(0,2r)$ are all far from the dipole compared with its length. Which statements are correct? (A) The field at $B$ is $-\dfrac{kp}{r^3}\hat{\mathbf i}$. (B) The field at $A$ is $\dfrac{2kp}{r^3}\hat{\mathbf i}$. (C) $|\mathbf E_A|:|\mathbf E_C|=16:1$. (D) A positive point charge placed at $B$ feels a force in the direction of $\mathbf p$.
E6. JEE Main NEET An electron starts from rest in a uniform electric field of $2.0\times10^3$ N/C. After moving $1.0$ cm its speed is about (A) $1.3\times10^6$ m/s (B) $2.7\times10^6$ m/s (C) $3.8\times10^6$ m/s (D) $7.0\times10^6$ m/s. ($e/m_e=1.76\times10^{11}$ C/kg)
Research question. How close to a dissolved ion must a water molecule be for the ion's electric field to line it up against the jostling of thermal motion?
Why it matters. When table salt dissolves, each $\text{Na}^+$ and $\text{Cl}^-$ ion gathers a "hydration shell" of water molecules turned to face it. These shells control how fast ions move through water, how batteries and desalination membranes work, and how nerve cells pass sodium and potassium ions through protein channels only a few tenths of a nanometre wide — research pursued by chemists and biophysicists worldwide. A simple dipole-in-a-field estimate already shows why the shell is only a molecule or two thick.
What to do.
- Treat the ion as a point charge $e$ and use the vacuum field $E=ke/r^2$ (this overestimates the field in water — note that as a limitation). Compute $E$ at $r=0.3,\ 0.5,\ 1.0$ and $2.0$ nm.
- For a water molecule ($p=6.2\times10^{-30}$ C m) compute the alignment energy scale $pE$ at each distance.
- Compare each value with the thermal energy $k_BT$ at $300$ K ($k_B=1.38\times10^{-23}$ J/K). Make a table of $pE/k_BT$ against $r$ and plot it on log–log axes. What slope do you expect, and why?
- Decide at what distance $pE\approx k_BT$. Inside that distance the molecules are strongly aligned; outside, they are nearly random. Then explain, using Section 1.1, why the real field in water is weaker than your estimate.
Tools: Calculator or spreadsheet; log–log graph paper or plotting software.
Hand in: Your table, the graph with its measured slope, the crossover distance, and a paragraph on the main limitation of the model.
Going further. A water molecule next to the ion also feels a net force because the ion's field is non-uniform. Which way does it point for an aligned dipole, and how does it change with $r$?
🧪 Self-check quiz
Q1. The SI unit of electric field is
Q2. The field of a point charge is $E$ at distance $r$. At distance $3r$ it is
Q3. Two electric field lines never cross because
Q4. The field of a short dipole at distance $r$ on its axis is $E_0$. At distance $2r$ on its equatorial line the field is
Q5. A dipole is placed in a uniform field with $\mathbf p$ perpendicular to $\mathbf E$. It experiences