PHY 2 Electric Fields

Introduction

Welcome to the 4th and 5th lectures of the course. This lecture series aims to enhance your understanding of electrostatics and its applications.

Goals

• Catch up on content from previous lectures

• Complete discussions on chapters 2 and 3, focusing on fundamental concepts and calculations of electric fields resulting from various charge distributions.

Next Week

We will begin with chapter 24, which features McGowan's law and its implications in electromagnetic theory.

Announcement

The second homework assignment has been made available on Quest since Tuesday, and it is due on Thursday. Hints for the upcoming homework 3 will also be released soon to aid your preparations.

Charge Distributions

There are three prominent types of charge distributions that we will study in detail:

1. Line Charge Distribution: Charge is spread uniformly along a straight line, characterized by a linear charge density denoted as (λ). It is essential in analyzing electric fields generated by wires and filamentary structures.

2. Surface Charge Distribution: Charge is present on a two-dimensional surface, described by surface charge density (σ). This type is crucial for understanding fields around charged plates and surfaces.

3. Volume Charge Distribution: Charge is distributed throughout a three-dimensional volume, represented by volume charge density (ρ). This concept is important when dealing with charged objects that occupy a defined space.

Calculating Electric Fields

To find electric fields generated by these charge distributions, we break down complex shapes into small charge pieces, termed dq. The relationships for calculating dq are as follows:

• Line Charge: dq = λ * dl (where dl is an infinitesimal length element along the charge distribution).

• Surface Charge: dq = σ * dA (where dA represents an infinitesimal area on the charged surface).

• Volume Charge: dq = ρ * dV (where dV is an infinitesimal volume element).

Example: Uniformly Charged Rod

Consider a rod with a total charge (Q) and length (L) having a uniform charge distribution. To compute the electric field at a point on the axis of the rod:

• The electric field produced by the rod points to the right for a positive charge value.

• The result derived from integration is given by:

  [ E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{d^2 + L^2} , \text{ at a distance } d \text{ from the rod.} ]

Note: The integral calculations for off-axis electric fields are not covered in this lecture; refer to your textbook for comprehensive examples and additional context.

Example: Long Straight Charged Wire

When considering an infinitely long uniformly charged wire:

• The electric field at any point is determined by integrating the contributions from infinitely many charge segments.

• Direction: The electric field vectors point radially outward from the wire for positive charge distributions.

• The uniform linear charge density (λ) leads to:

  [ E = \frac{\lambda}{2 \pi \epsilon_0 r} \hat{r} ]

• Notice the magnitude of the electric field diminishes as 1/r, which is different from point charges where the dependence is on 1/r².

Units Check

• The units of the electric field are expressed in Newtons per Coulomb (N/C). Consistent cross-checking across all equations confirms that the units remain coherent throughout the expressions used.

Example: Charged Plate

Consider an infinite charged plate with a uniform surface charge density (σ):

• The direction of the electric field:

  • Above the plate: the electric field points upward.

  • Below the plate: the electric field points downward.

• The magnitude of the electric field is given by:

  [ E = \frac{\sigma}{2 \epsilon_0} ]

• This electric field is remarkable because its magnitude is constant regardless of distance from the plate, which is a key characteristic of infinite charged plates.

Multiple Charge Plates

When considering two infinite plates, one positively charged (+σ) and the other negatively charged (-σ):

• In between the plates: The electric fields produced by both plates combine linearly, yielding a resultant field magnitude of:

  [ E = \frac{\sigma}{\epsilon_0} ]

• Outside the plates: The fields from opposite plates cancel each other out, resulting in no electric field outside the pair of plates.

Understanding Forces on Charged Particles

A charged particle (q) placed in a uniform electric field (E) experiences a constant force:

• The relationship is expressed as: [ F = qE ].

• This consistent force leads to constant acceleration, allowing us to apply familiar kinematic equations based on the initial velocity and elapsed time.

Electric Dipoles

An electric dipole consists of a positive charge and a negative charge separated by a distance d (defined as the dipole moment). When exposed to electric fields, dipoles experience a torque that tends to orient the dipole in line with the field direction.

• The potential energy associated with dipole alignment is given by:

  [ U = - \mathbf{p} \cdot \mathbf{E} ]

• Here, p is the dipole moment vector, and E is the electric field vector.

Electric Field Lines

The method to visualize electric fields involves drawing field lines that start from positive charges and end on negative charges:

• Key characteristics:

  • Electric field lines cannot intersect.

  • The density of lines indicates field strength: regions with tightly clustered lines represent stronger electric fields.

• An example illustrating field configurations between equal and opposite charges demonstrates how lines depict electric field directions effectively.

Conclusion

In summary, this lecture recapped the various charge distributions and their implications for electric fields.

• Emphasis was placed on enhancing hands-on familiarity with calculations and relationships relating to electric fields and the forces acting on charged particles.

• Engaging in the assigned homework and further readings is encouraged to deepen your understanding and improve your skills in this subject area.