Back to the on-screen lesson ·
Coulomb's law, fields of point charges added as vectors, charge densities, and the fields of rings and lines by integration and symmetry.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to find electric fields of point charges and of continuous distributions, using symmetry and integration.
From Physics 2 you know Coulomb's law, the idea of an electric field, and how to add vectors by components. From calculus you can set up and evaluate integrals. This lesson turns the sum over point charges into an integral over continuous distributions, which is the central skill of the electricity half of Physics C.
| Term | What it means |
|---|---|
| Coulomb's law | $F = kq_1q_2/r^2$, with $k = 1/4\pi\varepsilon_0 = 8.99 \times 10^9$ N·m²/C². |
| Electric field | $\vec{E} = \vec{F}/q_0$, the force per unit positive test charge, in N/C. |
| Superposition | The field of several charges is the vector sum of their separate fields. |
| Linear charge density | $\lambda$, charge per unit length, in C/m. |
| Surface charge density | $\sigma$, charge per unit area, in C/m². |
| Charge element | $dq$, a piece small enough to treat as a point charge. |
| Permittivity of free space | $\varepsilon_0 = 8.85 \times 10^{-12}$ C²/N·m². |
A point charge $q$ makes a field
$$\vec{E} = \frac{kq}{r^2}\,\hat{r},$$
pointing away from a positive charge and toward a negative one. The field of several charges is the vector sum of their fields. For charge spread continuously, cut it into elements $dq$, each a point charge, and integrate:
$$\vec{E} = \int \frac{k\,dq}{r^2}\,\hat{r}, \qquad dq = \lambda\,dx \text{ or } \sigma\,dA.$$
Because $\hat{r}$ changes direction from element to element, integrate components, and use symmetry first: pairs of elements placed symmetrically often cancel one component exactly, leaving a single integral. On the axis of a ring, for example, the sideways components cancel and $E = kQz/(z^2 + R^2)^{3/2}$.
Another way: picture
Picture standing on the axis of a hula hoop covered in charge. Each bit of the hoop pushes you away along a line from itself to you. Those lines all tilt outward by the same angle, in every direction around the hoop. The outward tilts cancel between opposite bits, and only the push along the axis adds up.
Another way: steps
Two point charges exert equal and opposite forces along the line joining them, of size $kq_1q_2/r^2$: like charges repel, unlike attract. The constant $k$ is written $1/4\pi\varepsilon_0$ in most of electromagnetism, because Gauss's law and the capacitor formulas come out simpler that way.
The field separates the source from the test charge. Charge $q$ creates $\vec{E}$ everywhere around it; a charge $q_0$ placed at a point feels $\vec{F} = q_0\vec{E}$. A field of $10^4$ N/C pushes a proton, $1.6 \times 10^{-19}$ C, with $1.6 \times 10^{-15}$ N, which gives it an acceleration of about $10^{12}$ m/s². Electric forces on small particles dwarf gravity.
Fields from separate charges simply add, as vectors. For two equal positive charges at $x = \pm a$, a point on the $y$-axis is the same distance $r = \sqrt{a^2 + y^2}$ from each. Their fields have equal size, $kq/r^2$, and mirror-image directions: the $x$-components cancel and the $y$-components add, each $kq/r^2 \cdot y/r$. So $E = 2kqy/r^3$, straight along the axis.
With unlike charges the $y$-components cancel instead and the $x$-components add, which is the dipole field on the bisector: $E = 2kqa/r^3$, pointing from the positive charge toward the negative side, parallel to the line of the charges. Always draw the two field vectors before deciding which components cancel.
Real charged objects have too many charges to add one by one, so treat the charge as continuous. A thin rod has a linear density $\lambda$, and a piece $dx$ long carries $dq = \lambda\,dx$. A surface has $\sigma$, and a patch $dA$ carries $\sigma\,dA$. If the density varies, the total is an integral: a rod with $\lambda = cx$ from $0$ to $L$ carries $\int_0^L cx\,dx = cL^2/2$.
The field is set up the same way: each $dq$ contributes $k\,dq/r^2$ in its own direction. The distance $r$ and the direction usually vary along the distribution, so they must be written in terms of the integration variable before integrating.
For a ring of radius $R$ and charge $Q$, a point on the axis at distance $z$ is $r = \sqrt{z^2 + R^2}$ from every element. Each element's field has an axial fraction $z/r$; the rest cancels around the ring. So $E = \int k\,dq\,z/r^3 = kQz/(z^2 + R^2)^{3/2}$, with no integral left to do, because $z$ and $r$ are the same for every element.
At the center, $z = 0$, the field is zero. Far away, $z \gg R$, it becomes $kQ/z^2$, a point charge. A disk is a set of rings: integrating the ring result over radius gives the field on a disk's axis, and letting the disk grow without limit gives $E = \sigma/2\varepsilon_0$, the uniform field of an infinite sheet.
For a long straight line of density $\lambda$, take a point a distance $r$ away and an element at position $x$ along the line. Its distance is $\sqrt{x^2 + r^2}$, and by symmetry only the perpendicular component survives. The integral $\int_{-\infty}^{\infty} k\lambda r\,dx/(x^2 + r^2)^{3/2}$ evaluates, with the substitution $x = r\tan\theta$, to $2k\lambda/r$.
The field falls as $1/r$, more slowly than a point charge's $1/r^2$, because moving away brings more of the line into view. A sheet goes further: its field does not fall at all. Point, line, sheet: $1/r^2$, $1/r$, constant. The next lesson finds all three far more quickly with Gauss's law.
Checking an answer. Units must be N/C: $kQ/\text{m}^2$. The field must point away from positive charge. Far away, any finite distribution must look like a point charge, $kQ/r^2$. And at points of symmetry, such as the center of a ring, the field must vanish if the symmetry demands it.
Treating $dq$ as a point charge is justified because the element is small compared with its distance to the field point; the error shrinks to zero as the elements do, which is what the integral means. Superposition is an experimental fact: the force between two charges is not changed by the presence of a third.
Cancelling components by symmetry requires the distribution to be symmetric about the field point, in charge as well as shape. A ring with more charge on one side has a sideways field on its axis, and the shortcut fails. Checking that the charge, not just the object, is symmetric keeps the method honest.
In a metal, charges move freely, so in equilibrium the field inside must be zero, or the charges would still be moving. Any extra charge therefore sits on the surface, and the field just outside points straight out from it, with strength $\sigma/\varepsilon_0$.
Charge crowds onto sharp points, where the surface curves most, so the field is strongest there. That is why lightning rods are pointed, why sparks jump from fingertips and door keys, and why Van de Graaff generators use large, smooth spheres: a smooth surface holds the most charge before the field anywhere reaches the breakdown strength of air.
The same reasoning explains shielding. A closed metal box has zero field inside no matter what charges sit outside, because its surface charges rearrange to cancel any outside field. Cars protect their passengers from lightning this way, and sensitive electronics, from radios to hospital monitors, are wrapped in metal cans for the same reason.
Coulomb's constant is enormous. Two charges of one coulomb each, a meter apart, would push on each other with $9 \times 10^9$ N, the weight of a large ship. Ordinary objects never hold anything like a coulomb of net charge, because such charges would tear themselves apart; the charge on a rubbed balloon is a few hundred nanocoulombs.
Compared with gravity, the electric force is stronger by an almost unimaginable factor. Between two protons the ratio of electric repulsion to gravitational attraction is about $10^{36}$, independent of their distance, since both forces fall as $1/r^2$. Gravity dominates the motions of planets only because large bodies are almost perfectly neutral: their positive and negative charges cancel to better than one part in $10^{18}$. At the scale of atoms, molecules and everyday contact forces, electricity governs everything: friction, the normal force, tension in a rope and the stiffness of a spring are all electric forces between atoms, summed over enormous numbers of them.
A Van de Graaff generator carries charge up on a moving rubber belt and deposits it on a large metal sphere. The charge spreads over the outside, and the sphere charges up until the field at its surface, $kQ/R^2$, reaches the breakdown strength of air, about $3 \times 10^6$ N/C, when sparks jump.
The generator at the Museum of Science in Boston, built by Robert Van de Graaff at MIT in the 1930s, has spheres about $4.5$ m across and makes sparks several meters long. Its large radius lets it hold far more charge, at a far higher voltage, than a small sphere could. The same machines, enclosed in pressurized gas, accelerated particles for early nuclear physics.
Xerography, invented by Chester Carlson and developed by the Haloid Company in Rochester, New York, which became Xerox, is electrostatics at work. A drum coated with a photoconductor is charged uniformly; light from the image discharges it where the page is white, leaving a pattern of charge where the text will be.
Charged toner powder is attracted to the charged regions, then transferred to paper by a stronger field behind the sheet, and fused with heat. The fields involved are a few million newtons per coulomb over millimeters, and the forces on each toner grain are thousands of times its weight. Every laser printer repeats the process for each page.
With two charges each giving a field of $100$ N/C at a point, it is tempting to say the field there is $200$ N/C. That is true only if the two fields point the same way. If they point opposite ways, the field is zero; at right angles, $141$ N/C. Always resolve into components before adding.
The same error appears in integrals: writing $\int k\,dq/r^2$ without a direction cosine adds up magnitudes. For a ring, that would give $kQ/r^2$ on the axis instead of $kQz/r^3$, too large because it ignores the cancelling sideways parts.
A $+4.0$ nC charge is at $x = 0$ and a $-1.0$ nC charge at $x = 0.30$ m. Find the field of the first at $x = 0.60$ m.
$E_1 = \dfrac{9.0 \times 10^9 \times 4.0 \times 10^{-9}}{0.60^2} = 100\ \text{N/C}, +x$
Away from the positive charge.
Find the field of the second charge there.
$E_2 = \dfrac{9.0 \times 1.0}{0.30^2} = 100\ \text{N/C}, -x$
Toward the negative charge.
Add them as vectors.
$E = 100 - 100 = 0$
They cancel at this point.
Find the field at $x = 0.90$ m.
$E = \dfrac{36}{0.81} - \dfrac{9.0}{0.36} = 44.4 - 25.0 = 19.4\ \text{N/C}, +x$
Beyond the null point the positive charge wins.
Find the force on a $2.0$ nC charge placed there.
$F = 2.0 \times 10^{-9} \times 19.4 = 3.9 \times 10^{-8}\ \text{N}$
$F = qE$, along $+x$.
A ring of radius $0.30$ m carries $5.0$ nC. Find the distance from the ring to the point $0.40$ m along its axis.
$r = \sqrt{0.40^2 + 0.30^2} = 0.50\ \text{m}$
Every element is this far away.
Find the field of the whole charge at that distance.
$\dfrac{kQ}{r^2} = \dfrac{9.0 \times 5.0}{0.25} = 180\ \text{N/C}$
Before taking components.
Find the axial fraction.
$\dfrac{z}{r} = \dfrac{0.40}{0.50} = 0.80$
The rest cancels around the ring.
Multiply for the field.
$E = 180 \times 0.80 = 144\ \text{N/C}$
Along the axis, away from the ring.
Compare with a point charge at the center.
$\dfrac{kQ}{z^2} = \dfrac{45}{0.16} = 281\ \text{N/C}$
The ring's charge is farther away and tilted.
Find the field at the center of the ring.
$z = 0: E = 0$
Every element's push is cancelled by the opposite one.
A rod from $x = 0$ to $x = 1.0$ m has $\lambda = 6.0x$ nC/m. Find its total charge.
$Q = \displaystyle\int_0^{1.0} 6.0x\,dx = 3.0\ \text{nC}$
Integrate the density.
Set up its field at the origin.
$dE = \dfrac{k\lambda\,dx}{x^2} = \dfrac{k(6.0x)\,dx}{x^2}$
Each element is a distance $x$ away.
Simplify the integrand.
$dE = \dfrac{6.0k\,dx}{x}$
One power of $x$ cancels.
Notice the problem at the end.
$\displaystyle\int_0^{1.0}\dfrac{dx}{x} \text{ diverges}$
The rod touches the origin, so the field there is infinite; move the field point off the rod.
Take the field at $x = -0.50$ m instead.
$E = 6.0k\displaystyle\int_0^{1.0}\dfrac{x\,dx}{(x + 0.50)^2}$
Now every element is at least $0.50$ m away.
Evaluate the integral.
$\left[\ln(x + 0.5) + \dfrac{0.5}{x + 0.5}\right]_0^{1} = \ln 3 - \tfrac{2}{3} = 0.432$
Write $x = (x + 0.5) - 0.5$ and split.
Multiply by the constants.
$E = 6.0 \times 9.0 \times 0.432 = 23\ \text{N/C}, -x$
Pointing away from the rod.
Find the distance to each charge.
$r = \sqrt{0.40^2 + 0.30^2} = 0.50\ \text{m}$
The same for both.
Find the field of one charge.
$E_1 = \dfrac{9.0 \times 3.0}{0.25} = 108\ \text{N/C}$
Coulomb's law per unit charge.
Add the $y$-components.
Two small charged spheres repel each other with a force of $72$ mN. They are moved to three times their original separation. What is the force now?
Complete the worked solution: a ring of radius $0.9$ m carries $8$ nC spread evenly around it. With $k = 9.0 \times 10^9$ N·m²/C², find the distance from the ring to a point on its axis $1.2$ m from the center, the fraction of each element's field that points along the axis, and the field there in N/C.
Find the distance from any element to the point.
$r = \sqrt{z^2 + R^2} =$ r
Every element is equally far away.
Find the axial fraction of each element's field.
$\cos\alpha = \dfrac{z}{r} =$ c
The sideways parts cancel between opposite elements.
Add the axial parts of all the elements.
$E = \dfrac{kQ}{r^2}\cdot\dfrac{z}{r} = \dfrac{kQz}{r^3} =$ e
The integral of $dq$ around the ring is just $Q$.
Check the far limit.
$z \gg R: E \to \dfrac{kQ}{z^2}$
From far away the ring looks like a point charge.
Match each charge distribution to the magnitude of its electric field.
| $kq/r^2$ | $kQz/(z^2 + R^2)^{3/2}$ | $2k\lambda/r$ | $\sigma/2\varepsilon_0$ | |
|---|---|---|---|---|
| point charge | ||||
| ring, on its axis | ||||
| long line | ||||
| large sheet |
A $6$ μC point charge sits in empty space. With $k = 9.0 \times 10^9$ N·m²/C², fill in the field $3$ m away in N/C, the force on a $2.0$ μC charge placed there in N, and the field twice as far away in N/C.
| value | |
|---|---|
| field (N/C) | |
| force on 2.0 μC (N) | |
| field at twice the distance (N/C) |
A thin rod lies along the $x$-axis from $x = 0$ to $x = L$ (m). Its charge per unit length grows along it, $\lambda = 8x$ nC/m. Write the rod's total charge, in nC, as a formula in $L$.
Answer:
Two $+8$ nC charges sit at $x = \pm0.6$ m on the $x$-axis. With $k = 9.0 \times 10^9$ N·m²/C², how strong is the field at the point $y = 0.8$ m on the $y$-axis, in N/C?
Answer: N/C on the bisector
A science museum's Van de Graaff generator has a metal sphere of radius $0.6$ m. Air breaks down and sparks when the field at the surface reaches $3.0 \times 10^6$ N/C. With $k = 9.0 \times 10^9$ N·m²/C², how much charge can the sphere hold, in μC?
Answer: μC before sparking
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A thin rod lies along the $x$-axis from $x = 0$ to $x = L$ (m). Its charge per unit length grows along it, $\lambda = 8x$ nC/m. Write the rod's total charge, in nC, as a formula in $L$.
Answer:
You can find electric fields by superposition. Explain to someone why the field at the center of a uniformly charged ring is zero.
20. Your turn: two $+3.0$ nC charges at $x = \pm 0.40$ m. Find the field at $y = 0.30$ m on the $y$-axis., step 3
$E = 2 \times 108 \times \dfrac{0.30}{0.50} = 130\ \text{N/C}$
The $x$-components cancel.