Topic 8.4 Notes – Electric Fields of Charge Distributions

1. The Electric Field from a Continuous Charge Distribution

For a small piece of charge $dq$, Coulomb’s law gives

$$\vec{E} = \frac{1}{4\pi \varepsilon_0} \int \frac{dq}{r^2}\,\hat{r}$$

• $r$ is the distance from $dq$ to the field point
• $\hat{r}$ points from $dq$ toward the field point
• This is a vector integral

This is just superposition in calculus form:

$$\vec{E}_{\text{total}} = \sum \vec{E}_i \quad \longrightarrow \quad \int d\vec{E}$$

Charge densities

You rewrite $dq$ using the appropriate density:

• Linear: $\lambda = \frac{dq}{dl} \Rightarrow dq = \lambda\,dl$
• Surface: $\sigma = \frac{dq}{dA} \Rightarrow dq = \sigma\,dA$
• Volume: $\rho = \frac{dq}{dV} \Rightarrow dq = \rho\,dV$

In this unit, you’ll only integrate for specific line distributions and a ring/arc. No random 3D blobs.

2. The General Strategy for Field Integrals

Every problem follows the same pattern. If you internalize this flow, FRQs feel much cleaner.

1. Use symmetry first.
   Decide the direction of $\vec{E}$. Figure out which components cancel.

2. Choose coordinates that match the geometry.
   Cylindrical for lines, polar for rings.

3. Write $dq$ using the proper density.
   For example, along a wire: $dq = \lambda\,dz$.

4. Express $r$ and any trig relationships.

5. Keep only surviving components.
   Don’t integrate components you already know cancel.

6. Check limits.
   Far away, does it behave like a point charge? Does the direction make sense?

Students lose points when they grind through full vector integrals without using symmetry. If two components cancel by symmetry, say it clearly and move on.

3. Symmetry and What It Tells You Immediately

Symmetry predicts the direction before you ever integrate.

For example, an infinite uniformly charged plane produces a field that must be perpendicular to the surface and have the same magnitude on both sides.

Cylindrical symmetry (infinite line)

• Field points radially outward
• Depends only on distance $r$ from axis
• No preferred direction along the wire

Result:

$$E = \frac{\lambda}{2\pi \varepsilon_0 r}$$

Notice the $1/r$ dependence. The field spreads cylindrically, not spherically.

Planar symmetry (infinite plane)

• Field is perpendicular to the plane
• Same magnitude everywhere
• Independent of distance

$$E = \frac{\sigma}{2\varepsilon_0}$$

Constant field. No decay with distance.

Cancellation logic

Common patterns:

• Opposite sides cancel horizontal components.
• At the center of symmetric arcs or rings, only one axis survives.
• Some points must have zero field purely from symmetry.

On quizzes, sometimes the whole question is just “determine the direction.” Don’t overthink it.

4. Required Calculus-Based Charge Distributions

You’re expected to integrate only these.

Infinite uniform line of charge

Setup idea:

• Place wire on z-axis
• $dq = \lambda dz$
• Vertical components cancel
• Only radial component survives

Final result:

$$E = \frac{\lambda}{2\pi \varepsilon_0 r}$$

As wire length → ∞, finite-wire results reduce to this.

Thin ring of charge (field on axis)

Radius $R$, total charge $Q$, point a distance $x$ along axis.

• Radial components cancel.
• Axial components add.

$$E = \frac{1}{4\pi \varepsilon_0} \frac{Qx}{(R^2 + x^2)^{3/2}}$$

Important behaviors:

• At $x=0$, $E=0$
• For $x \gg R$, becomes $kQ/x^2$

That limiting check is a favorite justification step on FRQs.

Semicircular arc (field at center)

Uniform $\lambda$, radius $R$.

• Horizontal components cancel.
• Vertical components add.

$$E = \frac{2k\lambda}{R}$$

Direction points toward the open side for positive charge.

Finite line of charge

Two cases only:

1. Point collinear with the wire
2. Point on perpendicular bisector

Symmetry removes one component in each case. The algebra typically leads to expressions involving endpoint angles. As length → ∞, you recover the infinite-line result.

5. Comparing Field Behaviors