Part 3 · Chapter 22

Electric Fields

A charge does not reach across empty space to grab another — it fills the space around itself with a field, and any second charge simply responds to the field already present where it sits. This single idea, that the field is the messenger, reshapes all of electromagnetism

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 55 min
i What you'll learn
  • The electric field as force per unit charge, \(\vec{E} = \vec{F}/q_{0}\), measured in newtons per coulomb.
  • How to read electric field lines — their direction, their density, and the charges they begin and end on.
  • The field of a point charge, \(E = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{q}{r^{2}}\), and how to add several by superposition.
  • The field of an electric dipole on its axis, \(E = \dfrac{1}{2\pi\varepsilon_{0}}\dfrac{p}{z^{3}}\), with dipole moment \(p = qd\).
  • The field of a ring and a charged disk on the central axis, built by integration.
  • The force on a point charge in a field, \(\vec{F} = q\vec{E}\) — the heart of the Millikan experiment and the ink-jet printer.
  • The torque \(\vec{\tau} = \vec{p}\times\vec{E}\) and the energy \(U = -\vec{p}\cdot\vec{E}\) of a dipole in a field.
Section 22-1

What Is Physics?

The previous chapter left us with a puzzle hidden inside Coulomb's law. Two charges attract or repel across empty space — but how? Neither one touches the other, and nothing visible passes between them. How does one charge even "know" that the other is there, and how far away it is?

Physics answers with the idea of a field. We say that any charge fills the space around itself with an electric field, a vector quantity defined at every point. A second charge brought into that space does not interact with the first charge directly; it simply feels the field that is already present at its own location. The field is the local middleman. This shift — from action-at-a-distance to a field that lives in space and carries the interaction — is one of the deepest ideas in physics, and the rest of electromagnetism is built on it.

Section 22-2

The Electric Field

To measure the field that a charged object produces at some point \(P\), place a tiny positive test charge \(q_{0}\) there and measure the electrostatic force \(\vec{F}\) on it. The electric field at \(P\) is the force per unit charge:

🧭
Definition of the electric field
E = F / q₀  (a vector, in newtons per coulomb)

The field \(\vec{E}\) points in the direction of the force on a positive test charge, and its SI unit is the newton per coulomb (N/C). The test charge must be small enough that it does not disturb the charges it is probing. Crucially, the field belongs to the source, not the test charge — it exists at \(P\) whether or not any \(q_{0}\) is there to feel it.

Table 22-1 · Some electric fields
Location or situationValue (N/C)
At the surface of a uranium nucleus3 × 10²¹
In a hydrogen atom, at r = 5.29 × 10⁻¹¹ m5 × 10¹¹
Electrical breakdown of air3 × 10⁶
Near a charged photocopier drum10⁵
Near a charged comb10³
In the lower atmosphere10²
Inside copper household wiring10⁻²
The field is real, not a bookkeeping trick. Once a charge wiggles, the change in its field spreads outward at the speed of light, carrying energy and momentum with it — that is exactly what a radio wave or a beam of light is. Treating the field as a physical thing that exists in empty space is what makes those later chapters possible.
Section 22-3

Electric Field Lines

Michael Faraday gave us a way to see a field: draw field lines. The rules are simple, and the whole behaviour of a charge distribution can often be read off a sketch of them.

➡️
Rules for field lines
Lines start on positive charge, end on negative charge; E is tangent to the line; density shows strength.

At any point the electric field vector \(\vec{E}\) is tangent to the field line through that point. Where lines are crowded the field is strong; where they spread out it is weak. Lines never cross — the field has one well-defined direction at each point. For a single positive charge the lines point radially outward; for a negative charge, radially inward.

Reading a two-charge picture. For two equal positive charges the lines bend away from each other, leaving a sparse region between them — that gap is the visual signature of repulsion. For a charge and an equal opposite charge (a dipole) the lines stream directly from the positive to the negative charge, the picture of attraction.
Section 22-4

The Electric Field Due to a Point Charge

To find the field of a single point charge \(q\), imagine a test charge \(q_{0}\) a distance \(r\) away. Coulomb's law gives the force; dividing by \(q_{0}\) gives the field:

Field of a point charge
\[ E = \frac{1}{4\pi\varepsilon_{0}}\frac{|q|}{r^{2}} \qquad k = \frac{1}{4\pi\varepsilon_{0}} = 8.99\times10^{9}\,\mathrm{N\cdot m^{2}/C^{2}} \]
The field points away from a positive \(q\) and toward a negative \(q\). Notice the test charge has cancelled out: the result depends only on the source charge and the distance, confirming that the field belongs to \(q\) alone.
Superposition again. When several point charges are present, the net field at \(P\) is the vector sum of the individual fields, \(\vec{E}_{\text{net}} = \vec{E}_{1} + \vec{E}_{2} + \cdots\). Each source contributes as if the others were absent — exactly the same rule that governed forces in Chapter 21, now applied to fields.
Section 22-5

The Electric Field Due to an Electric Dipole

Two equal and opposite charges \(+q\) and \(-q\) separated by a small distance \(d\) form an electric dipole. On the dipole axis, a distance \(z\) from the centre, the two fields nearly cancel — and what survives, for \(z \gg d\), is:

Field of a dipole on its axis (far field)
\[ E = \frac{1}{2\pi\varepsilon_{0}}\frac{qd}{z^{3}} = \frac{1}{2\pi\varepsilon_{0}}\frac{p}{z^{3}}, \qquad \vec{p} = q\vec{d} \]
The product \(p = qd\) is the magnitude of the electric dipole moment \(\vec{p}\), a vector that points from the negative to the positive charge. The far field depends only on this product, not on \(q\) and \(d\) separately.
📉
Why a dipole fades faster
A dipole field falls off as 1/z³, not 1/r² like a single charge.

Double the distance from a point charge and the field drops to a quarter; double it from a dipole and the field drops to an eighth. From far away the two opposite charges almost — but not quite — overlap, so their fields almost cancel, leaving only the faint residue that the inverse-cube law describes.

Section 22-6

The Electric Field Due to a Line of Charge

Real charge is often spread continuously, so we describe it with a charge density: linear \(\lambda\) (C/m) along a line, surface \(\sigma\) (C/m²) over an area, or volume \(\rho\) (C/m³) through a solid. The recipe is always the same: slice the object into differential elements \(dq\), write the field \(d\vec{E}\) of each as a point charge, and integrate.

For a ring of total charge \(q\) and radius \(R\), every element's perpendicular field component is cancelled by the element on the opposite side; only the axial components survive. The field on the central axis, a distance \(z\) from the centre, is:

Table 22-2 · Measures of continuous charge
NameSymbolSI unit
ChargeqC
Linear charge densityλC/m
Surface charge densityσC/m²
Volume charge densityρC/m³
Field on the axis of a charged ring
\[ E = \frac{1}{4\pi\varepsilon_{0}}\frac{qz}{\left(z^{2}+R^{2}\right)^{3/2}} \]
At the centre (\(z = 0\)) the field is zero by symmetry. Far away (\(z \gg R\)) the bracket reduces to \(z^{3}\) and \(E \to kq/z^{2}\) — the ring looks like a point charge, exactly as it should.
Section 22-7

The Electric Field Due to a Charged Disk

Build a uniformly charged disk of radius \(R\) and surface charge density \(\sigma\) out of nested rings, integrate the ring result over all radii, and the central-axis field becomes:

Field on the axis of a charged disk
\[ E = \frac{\sigma}{2\varepsilon_{0}}\left(1 - \frac{z}{\sqrt{z^{2}+R^{2}}}\right) \]
As the disk grows toward an infinite sheet (\(R \to \infty\), or simply \(z \ll R\)) the fraction vanishes and the field becomes uniform, \(E = \sigma/2\varepsilon_{0}\) — independent of distance. This uniform-sheet field is the workhorse behind capacitors in later chapters.
The limit is the lesson. Near a large flat charged surface the field hardly changes as you step toward or away from it — the lines are evenly spaced and parallel. This is why the gap inside a parallel-plate capacitor holds a clean, uniform field.
Section 22-8

A Point Charge in an Electric Field

The two halves of field theory are now joined. Sections 22-2 through 22-7 found the field produced by charge; we now find the force a field exerts on a charge placed in it. Rearranging the definition of the field gives the simplest and most-used equation of the chapter:

🎯
Force on a charge in a field
F = qE

The force on a positive charge is along the field; on a negative charge it is opposite to the field. The force depends only on the charge and on the field at its location, regardless of what produced that field.

Two classic applications live here. In the Millikan oil-drop experiment, Robert Millikan suspended tiny charged oil drops between charged plates, balancing gravity against the electric force \(qE\); by measuring the field needed, he showed the charge always came in whole multiples of \(e = 1.602\times10^{-19}\,\mathrm{C}\) — direct proof that charge is quantized. In an ink-jet printer, drops are given a controlled charge and then steered onto the page by the field between two deflecting plates, the deflection set by \(F = qE\).

Section 22-9

A Dipole in an Electric Field

Place a dipole in a uniform external field. The two equal and opposite charges feel equal and opposite forces, so there is no net force — but the forces do not share a line, so they twist the dipole. The result is a torque that tries to swing \(\vec{p}\) into alignment with \(\vec{E}\):

Torque and energy of a dipole
\[ \tau = pE\sin\theta \;\Rightarrow\; \vec{\tau} = \vec{p}\times\vec{E}, \qquad U = -pE\cos\theta = -\,\vec{p}\cdot\vec{E} \]
The energy is lowest when \(\vec{p}\) is aligned with \(\vec{E}\) (\(\theta = 0\), \(U = -pE\)) and highest when anti-aligned (\(\theta = 180^{\circ}\), \(U = +pE\)). Like a compass needle in a magnetic field, the dipole oscillates about the aligned orientation.
Why microwaves heat food. A water molecule is a permanent dipole. An oscillating field flips its orientation back and forth; the molecules jostle their neighbours as they turn, and that agitation is heat. The dipole torque is doing the cooking.
Worked Examples

Putting It to Work

1 Field of a single point charge

Problem. Find the magnitude of the electric field produced by a charge \(q = +2.0\times10^{-8}\,\mathrm{C}\) at a point \(r = 0.30\,\mathrm{m}\) away.

Solution. Apply the point-charge field directly.

E = kq/r²
\[ E = (8.99\times10^{9})\frac{2.0\times10^{-8}}{(0.30)^{2}} \approx 2.0\times10^{3}\,\mathrm{N/C} \]

The field points radially away from the positive charge. A second positive charge placed there would be pushed outward; a negative one would be pulled in.

2 Net field from three charges

Problem. Three particles with charges \(q_{1} = +2Q\), \(q_{2} = -2Q\), and \(q_{3} = -4Q\) each lie a distance \(d\) from the origin in different directions. Outline how to find the net field at the origin.

Solution. Each charge produces a field of magnitude \(E_{i} = k|q_{i}|/d^{2}\). Point each vector correctly (toward a negative source, away from a positive one), resolve into \(x\) and \(y\) components, and add.

Superpose as vectors
\[ \vec{E}_{\text{net}} = \vec{E}_{1} + \vec{E}_{2} + \vec{E}_{3}, \qquad E_{x} = \sum_i E_{i}\cos\theta_{i}, \quad E_{y} = \sum_i E_{i}\sin\theta_{i} \]

The factor \(kQ/d^{2}\) is common to all three, so it can be pulled out and the geometry handled with the dimensionless coefficients 2, 2, and 4. As always, the net direction must fall among the contributing vectors.

3 Where the ring field peaks

Problem. A ring of radius \(R\) carries uniform charge \(q\). At what axial distance \(z\) is the field strongest?

Solution. Set \(dE/dz = 0\) for \(E = kqz(z^{2}+R^{2})^{-3/2}\).

Maximize E(z)
\[ \frac{dE}{dz} = 0 \;\Rightarrow\; z^{2}+R^{2} - 3z^{2} = 0 \;\Rightarrow\; z = \frac{R}{\sqrt{2}} \]

The field is zero at the centre, rises to a maximum at \(z = R/\sqrt{2}\), then falls off as \(1/z^{2}\) far away — a useful check that the limiting behaviours match physical intuition.

4 Deflecting an ink-jet drop

Problem. An ink drop of mass \(m\) and charge magnitude \(Q\) enters a field \(E\) between deflecting plates of length \(L\) at speed \(v\). How far is it deflected as it crosses the plates?

Solution. The field gives a constant transverse acceleration \(a = QE/m\); the time to cross is \(t = L/v\). This is just projectile motion.

y = ½at² with a = QE/m, t = L/v
\[ y = \tfrac{1}{2}\,a t^{2} = \frac{QE L^{2}}{2 m v^{2}} \]

Gravity is negligible next to \(QE\) here, so it drops out. The deflection scales with the charge placed on the drop — which is precisely the quantity the printer controls to aim each drop.

5 Torque and energy of a dipole

Problem. A dipole of moment \(p = 3.0\times10^{-29}\,\mathrm{C\cdot m}\) sits in a uniform field \(E = 1.5\times10^{4}\,\mathrm{N/C}\) at \(\theta = 90^{\circ}\) to the field. Find the torque on it, and the work needed to rotate it from aligned (\(0^{\circ}\)) to anti-aligned (\(180^{\circ}\)).

Solution. Torque is \(pE\sin\theta\); the work equals the change in \(U = -pE\cos\theta\).

τ = pE sinθ; W = U(180°) − U(0°) = 2pE
\[\begin{aligned} \tau &= (3.0\times10^{-29})(1.5\times10^{4})\sin 90^{\circ} \approx 4.5\times10^{-25}\,\mathrm{N\cdot m} \\ W &= 2pE = 2(3.0\times10^{-29})(1.5\times10^{4}) \approx 9.0\times10^{-25}\,\mathrm{J} \end{aligned}\]

The torque is maximum at \(90^{\circ}\) and zero at alignment. Flipping the dipole all the way over costs \(2pE\) — the gap between the lowest and highest energy orientations.

Review

Chapter Summary

✓ The field concept

A charge fills space with a field; a second charge responds to the field where it sits, not to the source directly.

✓ Definition of E

\(\vec{E} = \vec{F}/q_{0}\), in N/C, pointing along the force on a positive test charge.

✓ Field lines

Start on +, end on −; \(\vec{E}\) is tangent; crowding shows strength; lines never cross.

✓ Point charge

\(E = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{|q|}{r^{2}}\); add several by vector superposition.

✓ Dipole

\(E = \dfrac{1}{2\pi\varepsilon_{0}}\dfrac{p}{z^{3}}\), \(p = qd\); falls off as \(1/z^{3}\).

✓ Ring & disk

Ring: \(E = \dfrac{kqz}{(z^{2}+R^{2})^{3/2}}\). Disk: \(E = \dfrac{\sigma}{2\varepsilon_{0}}\!\left(1-\dfrac{z}{\sqrt{z^{2}+R^{2}}}\right)\).

✓ Force on a charge

\(\vec{F} = q\vec{E}\) — Millikan's oil drop and the ink-jet printer both run on it.

✓ Dipole in a field

\(\vec{\tau} = \vec{p}\times\vec{E}\), \(U = -\vec{p}\cdot\vec{E}\); lowest energy when aligned.

✓ Continuous charge

Use densities \(\lambda, \sigma, \rho\); slice into \(dq\), write \(d\vec{E}\), integrate.

Practice

Problems

Take \(k = 8.99\times10^{9}\,\mathrm{N\cdot m^{2}/C^{2}}\), \(\varepsilon_{0} = 8.85\times10^{-12}\,\mathrm{C^{2}/N\cdot m^{2}}\), and \(e = 1.602\times10^{-19}\,\mathrm{C}\). For several charges, add fields as vectors; for continuous charge, integrate over elements; for a charge in a field, use \(\vec{F} = q\vec{E}\).

  1. What is the magnitude of a point charge whose electric field 50 cm away has magnitude 2.0 N/C?
  2. Two point charges \(+q\) and \(+q\) are a distance \(d\) apart. Find the electric field at the midpoint between them, and explain the result with symmetry.
  3. At what distance from a charge \(q = 1.0\,\mu\mathrm{C}\) does the field equal the air-breakdown value \(3.0\times10^{6}\,\mathrm{N/C}\)?
  4. An electric dipole consists of charges \(\pm 1.0\times10^{-9}\,\mathrm{C}\) separated by \(2.0\,\mathrm{mm}\). Find its dipole moment and the axial field 25 cm away.
  5. A ring of radius \(R = 5.0\,\mathrm{cm}\) carries \(q = 8.0\,\mathrm{nC}\). Find the field on the axis at \(z = 5.0\,\mathrm{cm}\).
  6. For the ring of Problem 5, find the field very far away and confirm it matches the point-charge formula.
  7. A disk of radius \(R = 2.5\,\mathrm{cm}\) has surface charge density \(\sigma = 5.3\,\mu\mathrm{C/m^{2}}\). Find the axial field at \(z = 12\,\mathrm{cm}\).
  8. Show that very close to the disk of Problem 7 (\(z \ll R\)) the field approaches \(\sigma/2\varepsilon_{0}\), and evaluate it.
  9. An electron is released from rest in a uniform field \(E = 1.0\times10^{4}\,\mathrm{N/C}\). Find its acceleration and its speed after travelling 2.0 cm.
  10. In a Millikan apparatus, an oil drop of mass \(1.2\times10^{-14}\,\mathrm{kg}\) hangs motionless in a field \(E = 1.8\times10^{5}\,\mathrm{N/C}\). How many excess electrons does it carry?
  11. An ink drop (\(Q = 1.5\times10^{-13}\,\mathrm{C}\), \(m = 1.3\times10^{-10}\,\mathrm{kg}\)) enters deflecting plates of length 1.6 cm at 18 m/s in a field of \(1.4\times10^{6}\,\mathrm{N/C}\). Find its deflection as it leaves the plates.
  12. A dipole of moment \(p = 1.5\times10^{-29}\,\mathrm{C\cdot m}\) sits at \(30^{\circ}\) to a field \(E = 2.0\times10^{4}\,\mathrm{N/C}\). Find the torque on it.
  13. For the dipole of Problem 12, find the work needed to rotate it from \(\theta = 0^{\circ}\) to \(\theta = 90^{\circ}\).
  14. Two charges \(+q\) and \(-q\) sit at opposite corners of a square of side \(a\). Find the magnitude and direction of the field at a third corner.
Tip: three habits carry most of this chapter. First, decide the direction of each field from the sign of its source (away from +, toward −) before computing magnitudes — the formula uses only magnitudes. Second, for continuous charge, always exploit symmetry to kill the components that cancel before you integrate; you usually integrate only the surviving axial component. Third, keep the two roles of a field separate in your head: one set of formulas tells you the field a charge produces, and \(\vec{F} = q\vec{E}\) tells you the force a field exerts.