Superposition and Electric Fields

11
PS 250: Lecture 3 Superposition and Electric Fields J. B. Snively August 28 th , 2015

Transcript of Superposition and Electric Fields

Page 1: Superposition and Electric Fields

PS 250: Lecture 3 Superposition and

Electric Fields

J. B. Snively August 28th, 2015

Page 2: Superposition and Electric Fields

Today’s Class

Electric Field Superposition of Electric Fields Example Problems Summary

Page 3: Superposition and Electric Fields

Electric FieldA force “Fo” is exerted on a test charge “qo” by an electric field “E”.

�Fo

= qo

�E

Given Coulomb’s law, the Electric Field due to a Point Charge can be found:

⇤Fo

= qo

⇤E = qo

1

4⇥�o

q

r2r̂ ⇤E

o

=1

4⇥�o

q

r2r̂

q qor �F

Page 4: Superposition and Electric Fields

Electric Field vs. Coulomb’s Law for Force Calculations

Electric field can be calculated for any charge distribution - However, Coulomb’s law (as was specified in 21.3) applies to point charges only.

... “F=qoE” only applies for test point charges “qo”, although “E” may be due to a charge distribution!

Electric fields exist without the presence of the test charges, such that:

�E = limq!0

�F

q

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Today’s Class

Electric Field Superposition of Electric Fields Example Problems Summary

Page 6: Superposition and Electric Fields

Superposition of Fields/Forces

+q +qo

Test Charger1

+q

r2

+q

r3

�F = �F1 + �F2 + �F3

�F = qo

�E = qo

�E1 + qo

�E2 + qo

�E3

�E =�F

qo

= �E1 + �E2 + �E3

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Superposition

Coulomb’s Law Superposition: Sum of vector forces upon a test charge in the presence of other point charges can be calculated.

Electric Field Superposition: The total vector electric field can be calculated from the sum of fields due to other charged bodies.

... Then, use “F=qoE” to obtain force!

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Charge ConfigurationsLine Charge: 1D Charge Distribution

Surface Charge: 2D Charge Distribution

Volume Charge: 3D Charge Distribution

Q =

ZdQ =

Z

l�dl

Q =

ZdQ =

Z

s�ds

Q =

ZdQ =

Z

v�dv

�(C/m)

�(C/m2)

dl

ds

dv

�(C/m3)

Page 9: Superposition and Electric Fields

Electric Field due to Charge Distribution

Consider infinitesimal E-field contributions (dE) due to each part of the distributed charge!

E =

ZdE =

1

4⇥�o

ZdQ

r2

This will require examples!

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Today’s Class

Electric Field Superposition of Electric Fields Example Problems Summary

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Summary / Next Class:

Read Textbook Sections: 21.6–21.7 (and begin Chapter 22 if feeling ambitious!)

Work on Homework for Wednesday.

Mastering Physics for Monday.

Prepare to discuss!