Continuous charge distributions, field lines and motion in a uniform field

Last week we treated charges as points. In reality, charge is spread over a rod, a ring or a plate. This week you learn two things: finding the field of spread-out charge by integration, and computing how a particle moves in a uniform field. The second is really the projectile motion from Physics I, with $qE/m$ in place of $g$. On the exam the ring and disk formulas are usually either given or you are asked to derive them; in both cases you need to know which component cancels and why.

Golden rules:

1. Treat a small piece as a point charge. Write $dE = k\,dq/r^2$, then add up (integrate).
2. Symmetry first. Before integrating, decide which component will be zero.
3. From far away, everything is a point charge. Check whether your result goes to $kQ/x^2$ in the limit $x \gg$ size.
4. Uniform field = constant acceleration. $a = qE/m$, then the kinematics you already know.

Charge densities

If charge is spread uniformly over an object, we describe it with a density:

The general recipe:

$\vec E = k_e \int \frac{dq}{r^2}\,\hat r$

This is a vector integral. You can’t add the magnitudes directly; first you split into components.

Field on the axis of a rod

Consider a rod of length $L$ and total charge $Q$. The point $P$ is on the extension of the rod, a distance $d$ from the near end. Every piece of the rod gives a field at $P$ in the same direction, so there is no need to split into components.

For the piece at distance $x$ from $P$: $dq = \lambda\,dx$, $dE = k\lambda\,dx/x^2$. Here $x$ runs from $d$ to $d + L$:

$E = k\lambda \int_d^{d+L} \frac{dx}{x^2} = k\lambda\left(\frac{1}{d} - \frac{1}{d + L}\right) = \frac{k_e Q}{d(d + L)}$

If $d \gg L$, then $E \to kQ/d^2$: a point charge from far away. The check works.

Example 1. $L = 0.40$ m, $Q = 6.0$ nC, $P$ is 0.10 m from the near end.

$E = \frac{(8.99 \times 10^{9})(6.0 \times 10^{-9})}{(0.10)(0.50)} = 1.08 \times 10^{3}\ \text{N/C}$

If you had put all the charge at the middle of the rod (0.30 m away) you would get 599 N/C. Very different. The near pieces dominate because of $1/r^2$.

Field on the axis of a ring

A ring of radius $a$ and charge $Q$. The point $P$ is on the axis, a distance $x$ from the center. Every piece $dq$ is at distance $r = \sqrt{x^2 + a^2}$ from $P$.

Symmetry: the components perpendicular to the axis from two opposite pieces of the ring cancel. Only the component along the axis remains:

$dE_x = k\frac{dq}{r^2}\cos\theta = k\frac{dq}{r^2}\cdot\frac{x}{r}$

$x$ and $r$ are the same for all pieces, so they come out of the integral:

$E_x = \frac{k_e x}{(x^2 + a^2)^{3/2}}\int dq = \frac{k_e Q x}{(x^2 + a^2)^{3/2}}$

Checks: at $x = 0$, $E = 0$ (symmetry at the center). For $x \gg a$, $E \to kQ/x^2$. There is a maximum in between: solving $dE/dx = 0$ gives $x = a/\sqrt{2}$.

Example 2. $a = 0.10$ m, $Q = 5.0$ nC, $x = 0.20$ m.

x² + a² = 0.040 + 0.010 = 0.050 m²
(0.050)^(3/2) = 0.01118 m³
E = (8.99e9)(5.0e-9)(0.20) / 0.01118 = 804 N/C   (away from the ring)

PITFALL · Forgetting the cos θ factor. If you treat the ring as a point charge at distance $r$ and write $kQ/r^2$, you get 899 N/C in Example 2. The perpendicular components cancel; only the fraction $x/r$ remains.

Field on the axis of a disk

Divide the disk into nested rings. A ring of radius $r'$ and thickness $dr'$ has charge $dq = \sigma (2\pi r')\,dr'$. The field of each ring is the formula above. They all point the same way and add directly:

$E = k\sigma \pi x\int_0^R \frac{2r'\,dr'}{(x^2 + r'^2)^{3/2}} = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right)$

In the limit $R \to \infty$ (or very close to the disk, $x \ll R$) the bracket becomes 1:

$E_{\text{infinite sheet}} = \frac{\sigma}{2\varepsilon_0}$

It does not depend on distance. This result is the basis for why capacitors produce a uniform field.

Example 3. $R = 0.15$ m, $\sigma = 2.0\ \mu\text{C/m}^2$, $x = 0.05$ m.

σ/(2ε0) = 2.0e-6 / (1.77e-11) = 1.13e5 N/C
bracket: 1 − 0.05/√(0.0025 + 0.0225) = 1 − 0.316 = 0.684
E = 1.13e5 × 0.684 = 7.7e4 N/C

TIP · Which formula for which case? On the exam, first identify your geometry: is the point on the extension of the rod or on its perpendicular bisector? For the ring and disk, the formulas are valid only on the axis. If a point off the axis is asked, no closed formula is expected; either symmetry cancels a component or the question stays qualitative.

Electric field lines

A way to show the field with a picture. The rules:

• The line is tangent to the field vector at every point.
• It starts on a positive charge and ends on a negative charge (or at infinity).
• The number of lines is proportional to the magnitude of the charge: $2q$ has twice as many lines leaving as $q$.
• Where lines are dense the field is strong, where they are sparse it is weak.
• Two lines never cross. If they did, the field would have two directions at that point.

For a dipole (two equal and opposite charges), the lines leave the positive charge, curve, and enter the negative one. For two equal positive charges, the lines repel each other and there is a point in the middle where the field is zero. Between two parallel oppositely charged plates the lines are straight, parallel and equally spaced: a uniform field.

TIP · Field lines are not trajectories. When a particle is released, its acceleration is tangent to the line, not its velocity. If the lines are curved, the particle does not follow the line. Only for straight lines and a start from rest does the path lie along the line.

Charged particle in a uniform field

In a uniform field the force $\vec F = q\vec E$ is constant. Newton’s second law:

$\vec a = \frac{q\vec E}{m}$

Since the acceleration is constant, the kinematic equations from Physics I work as they are. For electrons and protons the gravitational force is neglected: $eE$ is billions of times larger than $mg$.

Example 4 (starting from rest). A proton is released from rest in a field $E = 5.0 \times 10^{3}$ N/C. What is its speed after travelling 2.0 cm?

a = eE/mp = (1.60e-19)(5.0e3)/(1.67e-27) = 4.79e11 m/s²
v² = 2 a Δx = 2(4.79e11)(0.020)  →  v = 1.38e5 m/s

Example 5 (deflection between plates). An electron enters between two plates 10 cm long with a horizontal speed $v_0 = 3.0 \times 10^{6}$ m/s. The field points downward, $E = 200$ N/C. What is the vertical deflection as it leaves the plates?

The electron is negative and the field points down: the force is upward. There is no horizontal force, so the horizontal speed is constant. This is a projectile that “falls” upward.

Horizontal:  t = L / v0 = 0.10 / 3.0e6 = 3.33e-8 s
Vertical:    a = eE/me = (1.60e-19)(200)/(9.11e-31) = 3.51e13 m/s²  (upward)
             y = ½ a t² = ½ (3.51e13)(3.33e-8)² = 0.0195 m ≈ 1.95 cm  (upward)
             vy = a t = 1.17e6 m/s
Exit angle: tan θ = vy / v0 = 0.390  →  θ = 21.3°

The path is a parabola. After leaving the plates there is no field, and the electron goes straight at this angle. Old tube televisions and oscilloscopes used exactly this.

PITFALL · Mixing up the mass. An electron and a proton have charges of the same magnitude, but their masses differ by a factor of 1836. In the same field the electron’s acceleration is 1836 times larger. If you get a result above the speed of light, you most likely used the electron mass for a proton.

Common mistakes