The use of a rectangular coordinate system (x,y,z) is again appropriate, as shown in the sketch above showing two test charges: +q and - q referred to an antenna of dimension S, with position vectors indicated. We can write:
q(+) = q 0 cos w t
q(-) = - q 0 cos w t
The dipole moment is then:
P = q 0 S cos w t z^ = q S
The electrical potential in the first instance is expressed:
VP =
1/ 4 p εo {q 0 cos w(t - R + /c) /R + - q 0 cos w(t - R - /c) /R -}
The law of cosines yields:
R+ 2 = r 2 + (S/2) 2 + (r S/2) 2 cos q
= r 2 + S2/4 + r S cos q
or:
R+ = [r 2 + S2/4 + r S cos q] 1/2
For: r >> S (Distance much larger than antenna dimension):
S2 < < rS (Then terms in S2 can be neglected)
=> R+ = [r 2 + r S cos q] 1/2
R+ = [r + r S cos q] 1/2
R+= r (1 + S/ r cos q] 1/2
= r (1 + S cos q/2 r)
Note the Trigonometric identity used
to simplify expressions:
cos (A + B) = cos A cos B + sin A sin B
Where we let A = w(t - r/c) , B = s w cos q/2c
Then: cos w(t - r/c) + cos (s w cos q/2c) =
cos w(t - r/c) cos (s w cos q/2c) + sin w(t - r/c) sin (s w cos q/2c)
We now substitute the above expression into expression for V (r, t):
V (r, q, t) =
1/ 4 p εo [q 0/ r ( 1 + S cos q/2 r) cos w(t - r/c) - S w cos q / 2c (sin w(t - r/c)) -
q 0/r ( 1 - S cos q/2 r) cos w(t - r/c) + S w cos q / 2c (sin w(t - r/c))]
Each term contains q 0 and 1 / r so these can be factored out to get:
Also cos w(t - r/c) cancels, as does s w cos q / 2c (sin w(t - r/c)) So:
V (r, q, t) =
q 0/4 p εo r [ s cos q / r (cos w(t - r/c)) - s w cos q / 2c (sin w(t - r/c))]
But: q 0S = P 0 and factor out cos q :
V (r, t) =
P 0 cos q /4 p εo r [ cos w(t - r/c)/ r - w / c sin w(t - r/c) ]
Note: As w -> 0
V (r, t) = P 0 cos q /4 p εo r 2
But in general given the previous approximations, conditions, the potential for the dipole radiation field will be expressed:
V (r, t) ~ -P 0 w /4 p c εo (cos q / r) sin w(t - r/c)
See Also:
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