Electric Fields: Concepts, Calculations, and Applications

Electric Fields

The Electric Field

The electric field is a vector field that describes the force per unit charge exerted on a small test charge placed at a point in space. It is a fundamental concept in electromagnetism, allowing us to analyze the effects of charges without direct reference to the forces between them.

• Definition: The electric field E at a point r is defined as the force F experienced by a small test charge q_0 at that point, divided by the magnitude of the test charge:

$E = \frac{F}{q_0}$

• Force on a Charge: If the electric field at a point is E, a charge q_0 placed at that point experiences a force:

$F = q_0 E$

• Units: The SI unit of electric field is newtons per coulomb (N/C).

• Electric Field of a Point Charge: For a point charge q at the origin, the electric field at a distance r is:

$E = k \frac{q}{r^2} \hat{r}$

• Where k = 1/(4\pi\varepsilon_0) and \hat{r} is the unit vector from the charge to the field point.

Electric Fields from Particular Charge Distributions

Different charge distributions produce different electric field patterns and magnitudes. Here are some common cases:

• Electric Dipole: A pair of equal and opposite charges (±q) separated by a small distance d. The electric dipole moment is \vec{p} = q\vec{d}.

$E_z = \frac{1}{2\pi\varepsilon_0} \frac{p}{z^3} = k \frac{2qd}{z^3}$

• Valid for points on the axis of the dipole, far from the dipole (z ≫ d).

• Line of Charge: For a ring of radius R and total charge q, the electric field at a point on the axis a distance z from the center is:

$E = \frac{qz}{4\pi\varepsilon_0 (z^2 + R^2)^{3/2}}$

• Charged Disk & Infinite Sheet: For a disk of radius R and surface charge density \sigma, the field at a point on the axis a distance z from the center is:

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

• For an infinite sheet (R \to \infty):

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

Forces on Charges in Electric Fields

Charges in electric fields experience forces and, in some cases, torques:

• Force on a Point Charge: $F = qE$

• Direction: For positive q, force is in the direction of E; for negative q, force is opposite to E.

• Torque on a Dipole: A dipole in a uniform field experiences a torque:

$\tau = \vec{p} \times \vec{E}$

• Potential Energy of a Dipole:

$U = -\vec{p} \cdot \vec{E}$

Electric Field Lines

Electric field lines provide a visual representation of the direction and relative strength of the electric field:

• Field lines originate on positive charges and terminate on negative charges.

• Field lines never cross.

• The number of lines is proportional to the magnitude of the charge.

Worked Examples

Example 1: Floating Charged Object in an Electric Field

• Problem: An object with charge $q = 24\ \mu C$ floats in a uniform electric field $E = 610\ N/C$. What is its mass?

• Solution: The electric force balances gravity: $|q|E = mg$

$m = \frac{|q|E}{g} = \frac{(24 \times 10^{-6}\ C)(610\ N/C)}{9.80\ m/s^2} = 1.5 \times 10^{-3}\ kg$

• Answer: $1.5\ g$

Example 2: Acceleration of an Electron in an Electric Field

• Problem: An electron is released from rest in a uniform field $E = 2.00 \times 10^4\ N/C$. Find its acceleration.

• Solution: $F = |q|E$, $a = F/m$

$a = \frac{(1.602 \times 10^{-19}\ C)(2.00 \times 10^4\ N/C)}{9.11 \times 10^{-31}\ kg} = 3.51 \times 10^{15}\ m/s^2$

• Direction: Opposite to the field (since electron is negative).

Example 3: Point Charge Producing a Given Electric Field

• Problem: What charge produces $E = 1.00\ N/C$ at $r = 1.00\ m$?

• Solution: $E = k|q|/r^2 \implies |q| = Er^2/k$

$|q| = \frac{(1.00\ N/C)(1.00\ m)^2}{8.99 \times 10^9\ N\cdot m^2/C^2} = 1.11 \times 10^{-10}\ C$

Example 4: Electric Field at a Point Due to Multiple Charges

• Problem: Three positive charges at the corners of a square; find $E$ at point $P$.

• Solution: By symmetry, two contributions cancel; the third gives:

$E_{net} = \frac{q}{\pi \varepsilon_0 a^2}$ (points at $45^\circ$)

Example 5: Electric Field at the Center of a Square

• Problem: Four charges at the corners of a square (some positive, some negative). Find $E$ at the center.

• Solution: Calculate the vector sum of the fields from each charge, using symmetry and trigonometry.

$E_{Total} = 0.0i + (1.02 \times 10^5\ N/C)j$ (points in $+y$ direction)

Example 7: Electron Oscillating Through a Charged Ring

• Problem: An electron oscillates along the axis of a ring of charge. Find the angular frequency of oscillation.

• Solution: For small displacements $|z| \ll R$, the restoring force is linear:

$F_z \approx -\frac{eq}{4\pi\varepsilon_0 R^3} z$

$\omega = \sqrt{\frac{eq}{4\pi\varepsilon_0 m R^3}}$

Example 9: Electric Field of a Charged Disk

• Problem: Disk of radius $2.5\ cm$, surface charge density $5.3\ \mu C/m^2$, find $E$ at $z = 12\ cm$ on axis.

• Solution: Use disk formula:

$E = \frac{\sigma}{2\varepsilon_0} \left(1 - \frac{z}{\sqrt{z^2 + r^2}} \right) = 629\ N/C$

Example 10: Motion of Electron and Proton in an Electric Field

• Problem: Electron and proton in $E = 520\ N/C$, find speed after $48\ ns$.

• Solution: $F = |q|E$, $a = F/m$, $v = at$

Electron: $v = 4.4 \times 10^6\ m/s$ Proton: $v = 2.4 \times 10^3\ m/s$

Example 11: Acceleration and Speed of a Proton in a Field

• Problem: Proton in $E = 2.00 \times 10^4\ N/C$, (a) find acceleration, (b) speed after $1.00\ cm$.

• Solution:

(a) $a = 1.92 \times 10^{12}\ m/s^2$ (b) $v = 1.96 \times 10^5\ m/s$

Example 12: Water Drop Suspended in an Electric Field

• Problem: Water drop of diameter $1.20\ \mu m$ suspended in $E = 462\ N/C$. (a) Find weight, (b) number of excess electrons.

• Solution:

(a) $W = 8.87 \times 10^{-15}\ N$ (b) $q = -1.92 \times 10^{-17}\ C$, $N = 120$ electrons

Example 13: Electron and Proton Released Between Plates

• Problem: Electron and proton released from opposite plates, $D = 5.0\ cm$ apart. Where do they meet?

• Solution: Set up equations of motion, solve for $x_{meet}$:

$x_{meet} = \frac{D m_e}{m_e + m_p} = 2.73 \times 10^{-5}\ m = 27\ \mu m$

• Note: The answer does not depend on the electric field strength.

Example 14: Stopping Electrons with an Electric Field

• Problem: Electrons with $K = 1.60 \times 10^{-17}\ J$ must be stopped in $d = 10.0\ cm$. Find $E$.

• Solution: Use work-energy theorem:

$W = \Delta K = -1.60 \times 10^{-17}\ J$ $W = -F d \implies F = 1.60 \times 10^{-16}\ N$ $E = F/e = 1.00 \times 10^3\ N/C$

Summary Table: Key Electric Field Formulas

Configuration	Electric Field (E)	Direction
Point charge (q)	$E = k \frac{q}{r^2}$	Radially outward (q>0), inward (q<0)
Dipole (on axis, far field)	$E = \frac{1}{2\pi\varepsilon_0} \frac{p}{z^3}$	Along axis
Ring (on axis)	$E = \frac{qz}{4\pi\varepsilon_0 (z^2 + R^2)^{3/2}}$	Along axis
Disk (on axis)	$E = \frac{\sigma}{2\varepsilon_0} \left(1 - \frac{z}{\sqrt{z^2 + R^2}} \right)$	Along axis
Infinite sheet	$E = \frac{\sigma}{2\varepsilon_0}$	Perpendicular to sheet

Additional info: The notes above expand on the original content by providing definitions, context, and step-by-step explanations for each example, as well as a summary table for quick reference. All equations are provided in LaTeX format as required.