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Physics 2102
Jonathan Dowling
Physics 2113
Lecture 08: FRI 12 SEP
Electric Fields III
Charles-Augustin
de Coulomb
(1736-1806)
Electric Charges and Fields
First: Given Electric Charges, We
Calculate the Electric Field Using
E=kqr/r3.
Charge
Produces EField
Example: the Electric Field Produced
By a Single Charge, or by a Dipole:
Second: Given an Electric
Field, We Calculate the Forces
on Other Charges Using F=qE
Examples: Forces on a Single
Charge When Immersed in the
Field of a Dipole, Torque on a
Dipole When Immersed in an
Uniform Electric Field.
E-Field Then
Produces Force
on Another
Charge
Continuous Charge
Distribution
• Thus Far, We Have Only Dealt With
Discrete, Point Charges.
q
• Imagine Instead That a Charge q Is
Smeared Out Over A:
q
– LINE
q
– AREA
– VOLUME
• How to Compute the Electric Field E?
Calculus!!!
q
Charge Density
• Useful idea: charge density
• Line of charge:
charge
per unit length = 
• Sheet of charge:
 = q/L
 = q/A
charge
per unit area = 
• Volume of charge:
per unit volume = 
charge
 = q/V
Computing Electric Field
of Continuous Charge
Distribution
• Approach: Divide the Continuous Charge
Distribution Into Infinitesimally Small
Differential Elements
dq
• Treat Each Element As a POINT Charge
& Compute Its Electric Field
• Sum (Integrate) Over All Elements
• Always Look for Symmetry to Simplify
Calculation!
dq =  dL
dq =  dS
dq = ρ dV
Differential Form of Coulomb’s
Law
E-Field
at Point
q2
P1
Differential
dE-Field at
Point
P1
P2
dq2
ICPP: Arc of Charge
• Figure shows a uniformly charged
rod of charge –Q bent into a
circular arc of radius R, centered at
(0,0).
• What is the direction of the electric
field at the origin?
(a) Field is 0.
(b) Along +y
(c) Along -y
y
x
• Choose symmetric elements
• x components cancel
Arc of Charge: Quantitative
• Figure shows a uniformly charged rod
of charge –Q bent into a circular arc of
radius R, centered at (0,0).
• ICPP: Which way does net E-field
point?
• Compute the direction & magnitude of
E at the origin.
dEx = dE cosq =
Ex =
p /2
ò
0
kdq
cosq
2
R
kl
Ey =
R
–Q
450
x
dq = l Rdq
y
dq
k (lRdq ) cosq kl
=
2
R
R
kl
Ex =
R
y
p /2
ò cosqdq
q
0
kl
E= 2
R
Q
Q
2Q
l= =
=
L 2p R / 4 p R
E = Ex2 + Ey2
x
ICPP: Field on Axis of Charged Disk
• A uniformly charged circular disk
(with positive charge)
• What is the direction of E at point P
on the axis?
z
+
(a) Field is 0
(b) Along +z
(c) Somewhere in the x-y plane
P
y
x
Force on a Charge in Electric
Field
Definition of
Electric Field:
Force on
Charge Due to
Electric Field:
Force on a Charge in Electric Field
+++++++++
E
––––––––––
Positive Charge
Force in Same
Direction as EField (Follows)
+++++++++
E
––––––––––
Negative Charge
Force in Opposite
Direction as EField (Opposes)
(a) left
(b) left
-
(c) decrease
Electric Dipole in a Uniform Field
• Net force on dipole = 0;
center of mass stays where it
is.
• Net TORQUE  : INTO page.
Dipole rotates to line up in
direction of E.
• |  | = 2(qE)(d/2)(sin q)
= (qd)(E)sinq
= |p|
E sinq
= |p x E|
• The dipole tends to “align”
itself with the field lines.
• ICPP: What happens if the
field is NOT UNIFORM??
Distance Between
Charges = d
-
p = qd
-
+
+
-
-
-
Electric Dipole in a Uniform Field
• Net force on dipole = 0;
center of mass stays where it
is.
• Potential Energy U is smallest
when p is aligned with E and
largest when p anti-aligned
with E.
• The dipole tends to “align”
itself with the field lines.
U = - pE cos0° = - pE
U = - pE cos180° = + pE
Distance Between
Charges = d
-
1 and 3 are “uphill”.
2 and 4 are “downhill”.
U1 = U3 > U2 = U4
-
= 45°
-
t 1 = pE sin ( 45° + 45° + 45°) = pE sin (135°) = 0.71pE
t 2 = pE sin ( 45°) = 0.71pE
t 3 = pE sin ( -135°) = 0.71pE
t 4 = pE sin ( -45°) = 0.71pE
t1 = t 2 = t 3 = t 4
U1 = - pE cos (135°) == +0.71pE
U2 = - pE cos ( +45°) == -0.71pE
U3 = - pE cos ( -135°) == +0.71pE
U4 = - pE cos ( -45°) == -0.71pE
Summary
• The electric field produced by a system of charges
at any point in space is the force per unit charge
they produce at that point.
• We can draw field lines to visualize the electric
field produced by electric charges.
• Electric field of a point charge: E=kq/r2
• Electric field of a dipole:
E~kp/r3
• An electric dipole in an electric field rotates to
align itself with the field.
• Use CALCULUS to find E-field from a continuous
charge distribution.