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Radiation from an accelerated charge
Erad

P
P’
• q is initially at rest
at P
• q is accelerated and
moves from P to P’
• After some time has
passed, the electric
field at point  is
changed by an
amount Erad
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29:130 S 2012 Notes for 4.12.13
y
M’
d
M

N’
P2
P1
O
• +q at rest at O; E lies along
line OL
L
• Q given uniform accel. a
along Oy for a short time
N
interval . At the end of this
short interval q will be at P1
and will be moving with
velocity v = a , where v << c.
• After an interval t >> , q will
reach P2, where P1P2 = v t.
• The “kink” in the electric field line OL, due to the accel. of q, moves
outward along OL with velocity c. At the end of the interval  + t, it will
have reached N.
• After passing P1, q moves with uniform velocity v and a point on the
electric field line which leaves q at P1 has moved uniformly outward and
the line is now in the position P2M, almost parallel to OL; also OP1 and
P1P2 are << compared to the radii of the two spheres, which are
separated by an amount d = c .
2
analysis
c
M’
M
Er
Et
N’
c
E
N
• The outer spherical shell moves
outward at c. In front of this shell,
the E field is stationary, while
behind it, there is a moving field.
• A disturbed field moves outward
to change the field due to the
acceleration. The disturbed field
has radial and tangential
components– Er and Et.
Et MN 

Er MM 
MM   c 
P2 , M

MN   OP2 sin  
 vt sin   , since t  
N’
O
3
Et vt sin  
q

, but Er 
,
2
Er
c
4 o r
vt sin  
q
 Et 

,
2
c
4 o r
but t  r c , and v   a,
qa sin  
 Et 
2
4 o c r
•
•
This is the radiated electric field due to the accelerated charge. Note that
if a = 0, Et = 0  an accelerated charge radiates electromagnetic waves.
 is the angle between the velocity of the charge and the observer; there is
no radiation in the forward direction ( = 0).
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The magnetic field can be obtained simply by realizing
that at large distances the radiation fields must conform
to plane waves, so that B = E c ,so with
Erad
qa sin  
1 qa sin  

, Brad 
.
2
2
4 o c r
c 4 o c r
The Poynting vector is then:
S=
Erad Brad
o
 qa sin    1

 3 2.
 4 o  c r
2
The angular distributionof the radiation is  sin 2   .
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Angular distribution of radiated power
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Total Radiated power
The total power radiated by the accelerated charge is
given by the integral of S over a sphere centered at
the instantaneous position of the charge.
2
 qa  1
P   S  da  
 3
 4 o  c
2


0
sin 2  
2

2

r
sin   d
2
r
2
 qa  2  3
 qa  2 4

 3 0 sin   d  
 3  .
 4 o  c
 4 o  c 3
1
Writing
 o , this can be expressed as
2
 oc
2 o q 2 a 2
P=
.
3 4 c
This is the famous Larmor power formula.
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