Download Quantum ElectroDynamics

TOPIC 2
Electric Fields
www.cbooth.staff.shef.ac.uk/phy101E&M/
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Fields & Forces
Coulomb’s law
Q
r
q
F
qQ
40r
ˆ
r
2
How does q “feel” effect of Q?
Q modifies the surrounding space.
Sets up electrostatic field E
Force on charge q is
F=qE
E due to point charge Q is
E
Q
40r
ˆ
r
2
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Electrostatic Field Lines
Like charges
Unlike charges
Field lines:
• Start on positive charge, end on negative
• Number proportional to charge
• Strength of field = density of field lines
• Direction of force at point = tangent to field line.
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Conductors
Charges flow in response to a field
Equilibrium  no net field within a conductor
Free charges only exist on surface
Consider components of field at surface:
Eext
E
E//
+
+
+
+ +
Charge flows until E// = 0
 ETot is always perpendicular to surface of
conductor
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Continuous Charge Distributions
Divide into charge elements dq
dE 
Use superposition
E
dq
40r
ˆ
r
2
 4 r
dq
ˆ
r
2
0
In practice, express dq in terms of position r
 Use charge density
3D
2D surface
1D line
dq =  dV dV = element of volume
dq =  dA dA = element of area
dq =  d d = element of length
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Example 1
A rod of length L carries a charge Q distributed
uniformly along its length. If it is centred on the
origin and oriented along the y-axis, what is the
resulting electric field at points on the x-axis?
Solution available on web page
Example 2
A charge Q is uniformly distributed along the
circumference of a thin ring of radius R. What is the
electric field at points along the axis of the ring?
For next lecture: revise binomial theorem.
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Electric Dipoles
Pair of equal & opposite charges, Q & –Q, separated
by distance d
Dipole moment (vector)
p=Qd
(direction is from negative to positive charge)
Total charge is zero, but still produces and
experiences electric fields
In uniform electric field, dipole experiences a torque
(though no net force)
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Eext
+
d
F
F
θ
–
Pair of equal & opposite forces F = QE
Perpendicular separation between lines of forces = d sin 
Torque  = F d sin  = Q E d sin  = p E sin 
As vector,  = p  E
i.e. torque acting about centre of dipole, tending to
rotate it to align with electric field
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Eext
+
d
F
F
θ
–
Would have to do work to rotate dipole away from aligned
position – stored as potential energy.
Dipole does work (loses energy) rotating towards aligned
position.
Define zero of potential energy when dipole is perpendicular
to field –  = 90°.
Rotating to position shown, each charge does work:
work =forcedistance = F d/2 cos
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Energy of dipole U = – p E cos  = – p.E
Example 1
What is the electric field at points on the x-axis due
to a dipole formed by a charge Q at x = a/2 and a
charge –Q at x = –a/2 , for values of x >> a?
Example 2
Two dipoles, with the same charge and separation
as above, are placed parallel and a distance  apart:
(a) parallel to the line of the dipoles
(b) perpendicular to the line of the dipoles.
In which case is the force between the dipoles
greatest?
Part (b) is HARD!!
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