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Electric Field from Charged Ring Lecture
Sep 10, 2024
Electromagnetics Theory Lecture: Electric Field on the Axis of a Uniformly Charged Ring
Introduction
Professor Itesh Dolakia presents the topic.
Focus: Electric field on the axis of a uniformly charged ring.
Overview of the Ring
Charged Ring
: A ring with total charge
Q
.
Radius
: Denoted as
a
.
Line Charge Density
:
Formula: ( , \lambda = \frac{Q}{2\pi a} ) (total charge divided by the length of the ring).
Electric Field on the Axis
Axis Definition
: Defined as the x-axis extending from the center of the ring.
Point of Interest
: Distance
x
from the center of the ring.
Differential Charge Element
:
Consider a small charge element
dq
, contributing to the electric field at a point on the axis.
Components of Electric Field
Distance Calculation
:
Distance from dq to the point on the axis: ( r = \sqrt{x^2 + a^2} ).
Electric Field Contribution
:
Electric field due to
dq
: ( dE = \frac{K , dq}{r^2} ).
Angle Analysis
:
Angles formed by the electric field components are considered (angles
theta
).
Component Breakdown
:
Cosine Component
: ( dE \cos(\theta) ) contributes to the net electric field.
Sine Component
: Opposite directions cancel out.
Calculation of Electric Field
Cosine of Angle
:
From triangle: ( \cos(\theta) = \frac{x}{r} ).
Net Electric Field Calculation
:
( dE_{net} = \frac{K , x , dq}{(x^2 + a^2)^{3/2}} ) directed along
ax
._
Total Electric Field
Integration
: To find the total electric field:
Total electric field:
[ E = \int dE_{net} = \frac{K , x , Q}{(x^2 + a^2)^{3/2}} \text{ in the direction of } ax. ]
Alternative Expression
:
Using line charge density:
[ E = \frac{K , x , \lambda 2\pi a}{(x^2 + a^2)^{3/2}} \text{ with } \lambda = \frac{Q}{2\pi a}. ]_
Key Takeaways
Remember both formulas for calculating the electric field due to a uniformly charged ring.
The formulas will be useful for future calculations, including those related to a uniformly charged disc.
Next Lecture
Topic: Calculation of electric field due to a uniformly charged disc.
Encouragement for questions in the comment box.
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Full transcript