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January 14, 09
Physics 2102Physics 2102Lecture: 03 FRI 16 JANLecture: 03 FRI 16 JAN
Electric Fields IElectric Fields ICharles-Augustin
de Coulomb (1736-1806)
Physics 2102
Jonathan Dowling
Version: 1/14/09
What Are We Going to Learn?What Are We Going to Learn?A Road MapA Road Map
• Electric charge- Electric force on other electric charges- Electric field, and electric potential
• Moving electric charges : current• Electronic circuit components: batteries, resistors,
capacitors• Electric currents - Magnetic field
- Magnetic force on moving charges• Time-varying magnetic field è Electric Field• More circuit components: inductors.• Electromagnetic waves - light waves• Geometrical Optics (light rays).• Physical optics (light waves)
2q!
12F
1q+
21F
12r
CoulombCoulomb’’s Laws Law
2
12
2112
||||||
r
qqkF =
2
212
00
1085.8with 4
1
mN
Ck
!"== #
$#
2
2
91099.8
C
mN!k =
For Charges in aVacuum
Often, we write k as:
E-Field is E-Force Divided by E-ChargeE-Field is E-Force Divided by E-Charge
Definition ofElectric Field:
!
! E =
! F
q
!
|! F 12|=
k |q1| |q
2|
r12
2
+q1 –q2
!
! F 12
P1 P2
!
|! E 12|=
k |q2|
r12
2
–q2
!
! E 12
P1 P2
Units: F = [N] = [Newton]; E = [N/C] = [Newton/Coulomb]
E-Forceon
Charge
E-Fieldat
Point
Electric FieldsElectric Fields
• Electric field E at some point in
space is defined as the force
experienced by an imaginary point
charge of +1 C, divided by 1 C.
• Note that E is a VECTOR.
• Since E is the force per unit charge,
it is measured in units of N/C.
• We measure the electric field using
very small “test charges”, and
dividing the measured force by the
magnitude of the charge.2
||||R
qkE =
–q
R
E+1C
Electric Field of a Point Charge
Superposition of F and ESuperposition of F and E• Question: How do we
figure out the force orfield due to severalpoint charges?
• Answer: consider onecharge at a time,calculate the field (avector!) produced byeach charge, and thenadd all the vectors!(“superposition”)
• Useful to look out forSYMMETRY to simplifycalculations!
Example
• 4 charges are placed at thecorners of a square as shown.
• What is the direction of theelectric field at the center ofthe square?
(a) Field is ZERO!(b) Along +y(c) Along +x
-q
-2q
+2q
+q
y
x
Total electric field
Electric Field Lines• Field lines: useful way to
visualize electric field E
• Field lines start at apositive charge, end atnegative charge
• E at any point in spaceis tangential to field line
• Field lines are closerwhere E is stronger
Example: a negative pointcharge — note sphericalsymmetry
Direction of Electric Field Lines
E-Field VectorsPoint Away fromPositive Charge —Field Source!
E-Field VectorsPoint TowardsNegative Charge— Field Sink!
Electric Field of a DipoleElectric Field of a Dipole
• Electric dipole: two pointcharges +q and –qseparated by a distance d
• Common arrangement inNature: molecules,antennae, …
• Note axial or cylindricalsymmetry
• Define “dipole moment”vector p: from –q to +q,with magnitude qd
Cancer, Cisplatin and electric dipoles:http://chemcases.com/cisplat/cisplat01.htm
Electric Field On Axis of DipoleElectric Field On Axis of Dipole
!+ += EEE :ionSuperposit
2
2!"
#$%
& '
=+
ax
kqE
2
2!"
#$%
&+
'='a
x
kqE
!!!!
"
#
$$$$
%
&
'(
)*+
,+
-
'(
)*+
, -
=22
2
1
2
1
ax
ax
kqE 22
2
4
2
!!"
#$$%
&'
=
ax
xakq
Pa x
-q +qa x
-q +q
Electric Field Electric Field OnOn Axis of Dipole Axis of Dipole
22
2
22
2
4
2
4
2
!!"
#$$%
&'
=
!!"
#$$%
&'
=
ax
kpx
ax
xakqE
What if x>> a? (i.e. very far away)
p = qa“dipole moment”
a VECTOR
- +
34
22
x
kp
x
kpxE =!
E = p/r3 is actually true for ANY point far from adipole (not just on axis)
3r
pE
!!!
Force on a Charge in Electric FieldForce on a Charge in Electric Field
Definition ofElectric Field:
!
! E =
! F
q
Force onCharge Due toElectric Field:
!
! F = q
! E
Force on a Charge in Electric FieldForce on a Charge in Electric Field
E
E
Positive ChargeForce in SameDirection as E-Field
Negative ChargeForce in OppositeDirection as E-Field
+ + + + + + + + +
+ + + + + + + + +
– – – – – – – – – –
– – – – – – – – – –
Electric Dipole in a Uniform FieldElectric Dipole in a Uniform Field• Net force on dipole = 0;
center of mass stays whereit is.
• Net TORQUE τ : INTO page.Dipole rotates to line up indirection of E.
• | τ | = 2(qE)(d/2)(sin θ)= (qd)(E)sinθ= |p| E sinθ= |p x E|
• The dipole tends to “align”itself with the field lines.
• What happens if the field isNOT UNIFORM??
Distance BetweenCharges = d