Physics 241: Lecture Two Slides

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Learning Objectives for Today

What is the electric field?
What are electric field lines?
How to calculate the electric field from a point charge
How to calculate the electric field from a continuous charge distribution

Electric Field

A far more useful concept than electric force is the electric field E. We think of the electric field as a condition set up in all space by a charge or a distribution of charges. The electric field is defined via the test charge, which is usually taken to be positive. The field at a point is the force exerted on a small test charge per unit charge in the limit as that test charge goes to zero:

\mathbf{E} = \lim{q0 \to 0} \frac{\mathbf{F}}{q_0}

Thus, for a test charge $q0$, the force is $\mathbf{F} = q0 \mathbf{E}$ at the location of the test charge. The field viewpoint is often preferred because the field itself is a more fundamental quantity than the force acting on a particular test charge.

Maxwell’s equations describe the laws for electric and magnetic fields, and they underpin electromagnetism in general. The units of the electric field are commonly written as N/C, and later we will also use V/m.

Notes emphasize that the test charge is usually treated as positive, and that the field at a point is produced by the source charges rather than by the test charge itself. The field perspective is a central organizing concept in electromagnetism.

Scalar Field vs Vector Field

A scalar field assigns a single value to every point in space; an example is temperature, which is a scalar field.
A vector field assigns a vector to every point in space; the wind speed and direction constitute a vector field, and the electric field is also a vector field. In general, a scalar field is described by a function $\phi(\mathbf{r})$, while a vector field is described by $\mathbf{F}(\mathbf{r})$.

Electric Field Lines

Electric field lines (or field lines) are a visualization tool: lines are drawn tangent to the electric field at each point. There are infinitely many lines because the field is defined everywhere, but we draw a few to visualize the configuration of charges.
Rules for field lines:
- Lines begin on positive charges and end on negative charges.
- The number of lines entering or leaving a charge is proportional to the magnitude of that charge.
- The density of field lines is proportional to the magnitude of the field (field strength).
- Field lines cannot cross; if two lines crossed, that would imply two different field directions at the same point, which is impossible.
Point-charge configurations and field line patterns illustrate how the field distributes in space. For example, two positive charges produce lines that originate on each positive charge and repel away from both.

Electric Field from a Point Charge

The electric field created by a point charge $Q$ at a point located at position $\mathbf{r}$ (with respect to the charge) is:

\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} \hat{\mathbf{r}}

where $r = |\mathbf{r}|$ is the distance from the charge to the field point, and $\hat{\mathbf{r}}$ is the unit vector pointing from the charge to the field point. The constant $k$ is often introduced as $k = \dfrac{1}{4\pi\varepsilon_0}$, so the expression can be written as

\mathbf{E}(\mathbf{r}) = k \frac{Q}{r^2} \hat{\mathbf{r}}.

Dipole Field and Far-Field Behavior

A common configuration is a dipole, consisting of two charges separated by a small distance with total charge zero. The dipole moment is\ $\mathbf{p} = q \mathbf{d}$, where $\mathbf{d}$ points from the negative to the positive charge with magnitude equal to the separation.

The electric field of a dipole at a position $\mathbf{r}$ (in the far-field limit $r \gg d$) is

\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \frac{1}{r^3} \left[ 3(\mathbf{p}\cdot \hat{\mathbf{r}}) \hat{\mathbf{r}} - \mathbf{p} \right],

where $\hat{\mathbf{r}}$ is the unit vector in the direction from the dipole center to the field point. This expression shows the characteristic $1/r^3$ falloff of the dipole field in the far field.

Along a specific axis (e.g., the $x$-axis) one can simplify the expression depending on the orientation of $\mathbf{p}$ relative to the observation direction.

For small separations, a binomial or multipole expansion can be used to approximate the potential and the field; the leading term in the far field varies as $1/r^3$ for a dipole,

V(\mathbf{r}) \approx \frac{1}{4\pi\varepsilon_0} \frac{\mathbf{p}\cdot \hat{\mathbf{r}}}{r^2},

and $\mathbf{E} = -\nabla V$ follows accordingly.

Electric Field from a Continuous Charge Distribution

When charge is distributed over a region of space, the field can be computed by summing the contributions from each infinitesimal charge element $dq$. The elementary expression is

\mathbf{E} = \frac{1}{4\pi\varepsilon_0} \int \frac{dq\, (\mathbf{r} - \mathbf{r'})}{|\mathbf{r} - \mathbf{r'}|^3},

where $\mathbf{r}$ is the field point and $\mathbf{r'}$ is the source-point. The differential charge $dq$ is related to a density times a differential element, depending on the geometry:

Line charge: $dq = \lambda \, dl$, hence

\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \int \frac{\lambda \, dl \, (\mathbf{r} - \mathbf{r'})}{|\mathbf{r} - \mathbf{r'}|^3}.
Surface charge: $dq = \sigma \, dA$, hence

\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \int \frac{\sigma \, dA \, (\mathbf{r} - \mathbf{r'})}{|\mathbf{r} - \mathbf{r'}|^3}.
Volume charge: $dq = \rho \, dV$, hence

\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \int \frac{\rho \, dV \, (\mathbf{r} - \mathbf{r'})}{|\mathbf{r} - \mathbf{r'}|^3}.

Densities are related to the total quantities as

\lambda = \frac{Q}{L}, \quad \sigma = \frac{Q}{A}, \quad \rho = \frac{Q}{V}.

The technique is to exploit symmetry to perform the integrals, since in many problems some components of the electric field vanish by symmetry, leaving only a single component (often along a symmetry axis).

Continuous Charge Distributions: Examples and Methods

Uniform line charge: use $dq = \lambda dl$ and integrate along the line.
Uniform surface charge: use $dq = \sigma dA$ and integrate over the surface.
Uniform volume charge: use $dq = \rho dV$ and integrate over the volume.

A classic set of problems involves calculating the field on the axis of a uniformly charged disk or ring, and the field due to a uniformly charged disk on its axis is a standard result obtained by symmetry and integration. For a ring of radius $R$ and total charge $Q$, the field on the axis a distance $z$ away is

\mathbf{E}(z) = \frac{1}{4\pi\varepsilon_0} \frac{Q z}{\left(z^2 + R^2\right)^{3/2}} \hat{\mathbf{z}}.

For a uniformly charged disk of radius $R$ with surface charge density $\sigma$, the field on axis is given by

\mathbf{E}(z) = \frac{\sigma}{2\varepsilon_0} \left(1 - \frac{z}{\sqrt{z^2 + R^2}}\right) \hat{\mathbf{z}}.

Equivalently, in terms of $k = 1/(4\pi\varepsilon_0)$, this can be written as

\mathbf{E}(z) = 2 \pi k \sigma \left(1 - \frac{z}{\sqrt{z^2 + R^2}}\right) \hat{\mathbf{z}}.

These results illustrate how the geometry of the charge distribution (line, ring, disk) shapes the field via symmetry and the integral.

The dipole field discussed above shows that a distribution with zero net charge but finite dipole moment produces a field that scales as $1/r^3$ in the far field, a key feature distinguishing near-field from far-field behavior.

How to Calculate E from a Continuous Charge Distribution: Step by Step

1) Identify the symmetry of the problem to determine which components of $\mathbf{E}$ will be nonzero.

2) Write the differential contribution $d\mathbf{E}$ from an infinitesimal charge element $dq$:

d\mathbf{E} = \frac{1}{4\pi\varepsilon_0} \frac{dq\, (\mathbf{r} - \mathbf{r'})}{|\mathbf{r} - \mathbf{r'}|^3}.

3) Replace $dq$ with the appropriate density times a differential element:

For a line: $dq = \lambda dl$.
For a surface: $dq = \sigma dA$.
For a volume: $dq = \rho dV$.

4) Set up the integral for $\mathbf{E}$ with the appropriate limits, exploiting symmetry to simplify.

5) Perform the integration (analytically if possible, numerically if necessary).

6) Verify units and check limiting cases (e.g., far away from the distribution, the field should behave as expected).

Differential Elements and Coordinate Systems

To perform these integrals efficiently, we use appropriate coordinate systems and differential elements. The possible differential elements are:

Cartesian coordinates: $dx$, $dy$, $dz$
Cylindrical coordinates: $(\rho, \phi, z)$ with differential elements
- $dV = \rho \, d\rho \, d\phi \, dz$,
- $dA$ on a curved surface: $dA = \rho \, d\phi \, dz$,
- and other combinations for different surfaces.
Spherical coordinates: $(r, \theta, \phi)$ with differential elements
- $dV = r^2 \sin\theta \, dr \, d\theta \, d\phi$,
- $dA$ on a spherical surface: $dA = r^2 \sin\theta \, d\theta \, d\phi$.

These forms are summarized in the standard table of differential areas and volumes:

Cartesian: $dV = dx\, dy\, dz$; $dA$ components on coordinate planes follow from $dx$, $dy$, $dz$.
Cylindrical: $dV = \rho \, d\rho \,, d\phi \,, dz$; $dA$ for surfaces include $dA = \rho \, d\phi \,, dz$ on curved surfaces.
Spherical: $dV = r^2 \sin\theta \; dr \; d\theta \; d\phi$; $dA$ on a spherical surface is $dA = r^2 \sin\theta \; d\theta \; d\phi$.

In problems, the choice of coordinates is guided by symmetry to simplify the integral and reduce the number of nonzero components of $\mathbf{E}$.

Volume of a Sphere in Spherical Coordinates (Example)

Using spherical coordinates, the volume of a sphere of radius $R$ is computed as:

V = \iiint dV = \int{0}^{R} \int{0}^{\pi} \int_{0}^{2\pi} r^2 \sin\theta \, d\phi \, d\theta \, dr = \frac{4}{3} \pi R^3.

Examples from the Lecture Slides and Practice Problems

Quick vector review concepts: If $\mathbf{E} = (Ex, Ey)$, the magnitude is $|\mathbf{E}| = \sqrt{Ex^2 + Ey^2}$ and the direction angle $\theta$ satisfies $\theta = \tan^{-1}\left( \frac{Ey}{Ex} \right)$. Given $Ex = 10 x$ and $Ey = 5 y$, the magnitude and direction will depend on the resultant vector.
In-class problems frequently ask you to:
- Determine the direction of the electric field at a point.
- Draw electric field lines for a given charge configuration (e.g., dipole, two like charges, etc.).
- Compute on-axis fields for rings and disks.

Connections to Foundational Concepts and Real-World Relevance

The electric field formalism underpins much of electrostatics, electrodynamics, and applications such as shielding, capacitors, and sensors.
Field lines provide intuition for how charges influence space and how the magnitude relates to line density.
Understanding continuous charge distributions is essential for modeling real-world objects (wires, plates, dielectric materials) and for applying Gauss’s law in symmetric situations, even though Gauss’s law is not explicitly covered in these slides.

Notable Equations and Key Concepts (Summary)

Electric field definition (point-like test charge):

\mathbf{E} = \lim{q0 \to 0} \frac{\mathbf{F}}{q_0}.

Electric field due to a point charge:

\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} \hat{\mathbf{r}}.

Dipole field (far-field):

\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \frac{1}{r^3} \left[ 3(\mathbf{p}\cdot \hat{\mathbf{r}}) \hat{\mathbf{r}} - \mathbf{p} \right],\quad \mathbf{p} = q\mathbf{d}.

Electric field from a continuous distribution (vector form):

\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \int \frac{\rho(\mathbf{r'}) (\mathbf{r} - \mathbf{r'})}{|\mathbf{r} - \mathbf{r'}|^3} \, dV'.

Line, surface, and volume elements for dq:

dq = \lambda \, dl, \quad dq = \sigma \, dA, \quad dq = \rho \, dV.

On-axis field results (quick references):
Ring of radius $R$, charge $Q$ on axis at distance $z$:

\mathbf{E}(z) = \frac{1}{4\pi\varepsilon_0} \frac{Q z}{(z^2 + R^2)^{3/2}} \hat{\mathbf{z}}.

Disk of radius $R$ with surface density $\sigma$ on axis at distance $z$: