Year 1 Sem 2/LC Electromagnetism I/Lecture 02

Lecture 2: Electrostatics II: Field Lines and Continuous Charges
Thu 22 Jan 2026 11:00

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Field Lines

To visualise a field, we draw field lines. These are drawn so that the tangent of these lines at any point in spaces gives the direction of $\underline{E}$ at that point.

The lines point radially outwards from a positive charge, and radially towards a negative charge. The density of these lines is proportional to the magnitude of $\underline{E}$, but the absolute number of lines doesn’t convey much.

Field lines must start at a positive charge and end on a negative charge. They are continuous and can never cross each other, they only start/stop at charges or proceed to infinity.

Continuous Charge Distributions

What do we do if we have a charge which isn’t a point particle?

We break the object up into a distribution of infinitesimal point charges.
We treat each piece as a point charge and add up the separate $\underline{E}$ fields across an integral.
This gets messy for anything non-trivial, resulting in a complex 3D integral often requiring computational techniques.

Charges can be distributed across an object in a number of ways:

Charge per unit length across a line, given by $\lambda$:
$$q = \int \lambda dl$$
Charge per unit area across a surface area, given by $\sigma$:
$$q = \int \sigma dA$$
Charge per unit volume, given by $\rho$:
$$q = \int \rho dV$$

Example: E-Field from a Line of Charge

Consider an E-field from a line of length $2L$, measured at a point along the midpoint’s perpendicular. This point sits $x$ length units along the perpendicular.

We consider some element of charge $\delta Q = \lambda \delta y$ along the line at y. There are two components to the E-field, one pointing in the positive x direction and one in the negative y direction.

When we consider all of the $\delta Q$s along the length of the charged rod, each point on the positive y-axis along the length will have a corresponding point on the negative y-axis. These two points will have equal and opposite magnitude, so $E_y = 0$ by symmetry. There is therefore no resultant $y$ component and the final field will point along $\hat{x}$.

For some infinitesimal $\delta q$ generically along the line of charge:

$$\delta E_x = \frac{\delta q}{4 \pi \epsilon_0 r^2} \cos \alpha \tag{1}$$

And for each $\delta Q$:

$$\delta Q = \lambda \delta y \tag{2}$$

And from trig:

$$r^2 = y^2 + x^2 \tag{3}$$

$$\cos \alpha = \frac{x}{r} \tag{4}$$

Substituting 2, 3, 4 into 1:

$$\delta E_x = \frac{\lambda \delta y}{4 \pi \epsilon_0} \frac{x}{\left[y^2 + x^2\right]^{3/2}}$$

Hence:

$$E_x = \frac{\lambda x}{4 \pi \epsilon 0} \int_{-L}^{L} \frac{1}{\left[y^2 + x^2\right]^{3/2}} \, dy$$

To solve this, we would use the standard integral given in the question. Alternatively, if a standard integral isn’t given in the question we could express in the integral in terms of $\alpha$ and integrate from the minimum to maximum alpha.

Example Part B: The Easier Route

As before:

$$\delta E_x = \frac{\lambda \delta y}{4 \pi \epsilon_0 r^2} \cos \alpha \tag{1}$$

$$\cos \alpha = \frac{x}{r} \implies r = \frac{x}{\cos \alpha} \tag{2}$$

And:

$$\tan \alpha = \frac{y}{x} \implies y = x \tan \alpha \tag{3}$$

As:

$$\frac{d}{d \alpha} \tan \alpha = \frac{1}{\cos^2 \alpha}$$

We have:

$$\delta y = \frac{x \delta \alpha}{\cos^2 \alpha} \tag{4}$$

Hence:

$$\delta E_x = \frac{\lambda}{4 \pi \epsilon_0} \underbrace{\frac{x \delta x}{\cos^2 \alpha}}_\text{$\delta y$} \underbrace{\frac{\cos^2 \alpha}{x^2}}_\text{$1/r^2$} \cos \alpha$$

$$= \frac{\lambda}{4 \pi \epsilon_0 x} \cos \alpha$$

Which is much easier to integrate from the minimum to maximum values of $\alpha$.

$$E_x = \frac{\lambda}{4 \pi \epsilon_0 x} \int_{\alpha_\text{min}}^{\alpha_\text{max}} \cos \alpha \, d \alpha$$

And from the geometry of the problem:

$$\alpha_\text{max} = \arctan \left(\frac{L}{x}\right) \qquad \alpha_\text{min} = - \arctan \left(\frac{L}{x}\right)$$

Which is easy to integrate.

If the problem was setup such that the rod was infinitely long, then:

$$E_x = \frac{\lambda}{4 \pi \epsilon_0 x} \int_{-\pi / 2}^{\pi / 2} \cos \alpha \, d \alpha$$

$$E_x = \frac{\lambda}{2 \pi \epsilon_0 x}$$

This is approximately true for the case where $x \ll L$.

Example: Ring of Charge

Consider a ring of charge with radius $a$. We step out some perpendicular constant distance $x$ from the centre of the ring to point $P$. What is the resultant field at $P$?

The ring has total charge $Q$, distributed with charge per unit length $\lambda$.

Again, we can use symmetry. For every point on the ring, there will be a point one diameter away creating an equal and opposite $\delta E$. Hence, the $y$ and $z$ components cancel leaving only $E_x$, pointing in the $\hat{x}$ direction.

$$E_y = E_z = 0 \text{ by symmetry}$$

We have:

$$\delta E_x = \frac{\delta q}{4 \pi \epsilon_0 r^2} \cos \theta$$

$$\cos \theta = \frac{x}{y}$$

$$r = \sqrt{a^2 + x^2}$$

Hence:

$$\delta E_x = \frac{\delta q}{4 \pi \epsilon_0} \frac{x}{\left[a^2 + x^2\right]^{3/2}}$$

Here, a and x are constants, so the only variable we can integrate across is $\delta q$:

$$E_x = \frac{x}{4 \pi \epsilon_0} \frac{1}{\left[a^2 + x^2\right]^{3/2}} \int dq$$

$$= \frac{x Q}{4 \pi \epsilon_0} \frac{1}{\left[a^2 + x^2\right]^{3/2}}$$

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