(a) In the case of an oscillating electric dipole,
$p(t) = p_0 \cos(ωt), \ddot{p}(t) = −ω^2 p_0 \cos(ωt),$
and we recover the results of Sect. 11.1.2.

(b) For a single point charge q, the dipole moment is
$p(t) = qd(t),$

Question

(a) In the case of an oscillating electric dipole,

[latex]p(t) = p_0 \cos(ωt), \ddot{p}(t) = −ω^2 p_0 \cos(ωt),[/latex]

and we recover the results of Sect. 11.1.2.

(b) For a single point charge q, the dipole moment is

[latex]p(t) = qd(t),[/latex]

Accepted Answer

where d is the position of q with respect to the origin. Accordingly,

[latex]\ddot{p}(t)=qa(t),[/latex]

where a is the acceleration of the charge. In this case the power radiated (Eq. 11.60) is

[latex]P_{rad}(t_0)\cong \frac{\mu_0}{6\pi c}[\ddot{p}(t_0)]^2. [/latex] (11.60)

[latex]P=\frac{\mu_0q^2a^2}{6\pi c}. [/latex] (11.61)

This is the famous Larmor formula; I’ll derive it again, by rather different means, in the next section. Notice that the power radiated by a point charge is proportional to the square of its acceleration.

Question 11.2: (a) In the case of an oscillating electric dipole,p(t) = p0 ...

Related Answered Questions