Question 11.26: An ideal electric dipole is situated at the origin; its dipo...

An ideal electric dipole is situated at the origin; its dipole moment points in the \hat{ z } direction, and is quadratic in time:

p (t)=\frac{1}{2} \ddot{p}_{0} t^{2} \hat{ z }, \quad(-\infty<t<\infty)

where \ddot{p}_{0} is a constant.

(a) Use the method of Section 11.1.2 to determine the (exact) electric and magnetic
fields, for all
r > 0 (there’s also a delta-function term at the origin, but we’re not concerned with that).

\left[\text { Partial Answer }: V=\frac{\mu_{0} \ddot{p}_{0}}{8 \pi} \cos \theta\left[(c t / r)^{2}-1\right], A =\frac{\mu_{0} \ddot{p}}{4 \pi c}[(c t / r)-1] \hat{ z } .\right]

(b) Calculate the power, P(r, t), passing through a sphere of radius r.

\left[\text { Answer: } \frac{\ddot{p}_{0}^{2}}{12 \pi \epsilon_{0} r^{3}} t\left[t^{2}+(r / c)^{2}\right] .\right]

(c) Find the total power radiated (Eq. 11.2), and check that your answer is consistent with Eq. 11.60 .^{21}

P_{ rad }\left(t_{0}\right)=\lim _{r \rightarrow \infty} P\left(r, t_{0}+\frac{r}{c}\right)                                 (11.2)

P_{ rad }\left(t_{0}\right) \cong \frac{\mu_{0}}{6 \pi c}\left[\ddot{p}\left(t_{0}\right)\right]^{2}                                (11.60)

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\text { (a) Write } p (t)=q(t) d \hat{ z }, \text { with } q(t)=k t^{2} \text {, where } k d=(1 / 2) \ddot{p}_{0} . \text { As in Eq. } 11.5,

V( r , t)=\frac{1}{4 \pi \epsilon_{0}}\left\{\frac{q_{0} \cos \left[\omega\left(t-ᴫ_{+} / c\right)\right]}{ᴫ_{+}}-\frac{q_{0} \cos \left[\omega\left(t-ᴫ_{-} / c\right)\right]}{ᴫ_{-}}\right\}                         (11.5)

V( r , t)=\frac{1}{4 \pi \epsilon_{0}}\left[\frac{k(t-ᴫ+/ c)^{2}}{ᴫ+}-\frac{k(t-ᴫ-/ c)^{2}}{ᴫ-}\right]

 

=\frac{k}{4 \pi \epsilon_{0}}\left[\frac{\left(t^{2}-(2 t / c) ᴫ++ᴫ_{+}^{2} / c^{2}\right)}{ᴫ+}-\frac{\left(t^{2}-(2 t / c) ᴫ-+ᴫ_{+}^{2} / c^{2}\right)}{ᴫ-}\right]

 

=\frac{k}{4 \pi \epsilon_{0}}\left[t^{2}\left(\frac{1}{ᴫ+}-\frac{1}{ᴫ-}\right)+\frac{1}{c^{2}}(ᴫ+-ᴫ-)\right].

From Eqs. 11.8 and 11.9,

ᴫ_{\pm} \cong r\left(1 \mp \frac{d}{2 r} \cos \theta\right)                                             (11.8)

\frac{1}{ᴫ_{\pm}} \cong \frac{1}{r}\left(1 \pm \frac{d}{2 r} \cos \theta\right)                                     (11.9)

ᴫ_{\pm}=r\left(1 \mp \frac{d}{2 r} \cos \theta\right), \quad \frac{1}{ᴫ_{\pm}}=\frac{1}{r}\left(1 \pm \frac{d}{2 r} \cos \theta\right)

so

V( r , t)=\frac{k}{4 \pi \epsilon_{0}}\left[\frac{t^{2}}{r}\left(\frac{d}{r} \cos \theta\right)-\frac{r}{c^{2}}\left(\frac{d}{r} \cos \theta\right)\right]=\frac{k}{4 \pi \epsilon_{0} c^{2}} d \cos \theta\left[\left(\frac{c t}{r}\right)^{2}-1\right]=\frac{\mu_{0} \ddot{p}_{0}}{8 \pi} \cos \theta\left[\left(\frac{c t}{r}\right)^{2}-1\right].

As in Eq. 11.15, I (t)=(d q / d t) \hat{ z }=2 k t \hat{ z } , so (following Eqs. 11.16, and 11.17),

A ( r , t)=\frac{\mu_{0}}{4 \pi} \int_{-d / 2}^{d / 2} \frac{-q_{0} \omega \sin [\omega(t-ᴫ / c)] \hat{ z }}{ᴫ} d z                                         (11.16)

A (r, \theta, t)=-\frac{\mu_{0} p_{0} \omega}{4 \pi r} \sin [\omega(t-r / c)] \hat{ z }                              (11.17)

A ( r , t)=\frac{\mu_{0}}{4 \pi} \hat{ z } \int_{-d / 2}^{d / 2} \frac{2 k(t-ᴫ / c)}{ᴫ} d z=\frac{\mu_{0}}{4 \pi} 2 k \frac{(t-r / c)}{r} d \hat{ z }=\frac{\mu_{0} \ddot{p}_{0}}{4 \pi c}\left[\left(\frac{c t}{r}\right)-1\right] \hat{ z }.

E =-\nabla V-\frac{\partial A }{\partial t}=-\frac{\mu_{0} \ddot{p}_{0}}{8 \pi}\left\{\cos \theta\left[-2 \frac{(c t)^{2}}{r^{3}}\right] \hat{ r }-\frac{1}{r} \sin \theta\left[\left(\frac{c t}{r}\right)^{2}-1\right] \hat{ \theta }\right\}-\frac{\mu_{0} \ddot{p}_{0}}{4 \pi c}\left(\frac{c}{r}\right) \hat{ z }

 

=\frac{\mu_{0} \ddot{p}_{0}}{4 \pi r}\left[\cos \theta\left(\frac{c t}{r}\right)^{2} \hat{ r }+\frac{1}{2} \sin \theta\left(\frac{c t}{r}\right)^{2} \hat{ \theta }-\frac{1}{2} \sin \theta \hat{ \theta }-(\cos \theta \hat{ r }-\sin \theta \hat{ \theta })\right]

 

=\frac{\mu_{0} \ddot{p}_{0}}{4 \pi r}\left\{\left[\left(\frac{c t}{r}\right)^{2}-1\right] \cos \theta \hat{ r }+\frac{1}{2}\left[\left(\frac{c t}{r}\right)^{2}+1\right] \sin \theta \hat{ \theta }\right\}.

B =- \nabla \times A =\frac{\mu_{0} \ddot{p}_{0}}{4 \pi c} \nabla \times\left\{\left[\left(\frac{c t}{r}\right)-1\right](\cos \theta \hat{ r }-\sin \theta \hat{ \theta })\right\}

 

=\frac{\mu_{0} \ddot{p}_{0}}{4 \pi c r}\left\{-\frac{\partial}{\partial r}\left(r\left[\left(\frac{c t}{r}\right)-1\right] \sin \theta\right)-\frac{\partial}{\partial \theta}\left(\left[\left(\frac{c t}{r}\right)-1\right] \cos \theta\right)\right\} \hat{\phi}

 

=\frac{\mu_{0} \ddot{p}_{0}}{4 \pi c r}\left\{\sin \theta+\left[\left(\frac{c t}{r}\right)-1\right] \sin \theta\right\} \hat{\phi}=\frac{\mu_{0} \ddot{p}_{0} t}{4 \pi r^{2}} \sin \theta \hat{\phi}.

(b) The Poynting vector is

S =\frac{1}{\mu_{0}}( E \times B )=\frac{\mu_{0} \ddot{p}_{0}^{2} t}{16 \pi^{2} r^{3}} \sin \theta\left\{\left[\left(\frac{c t}{r}\right)^{2}-1\right] \cos \theta(-\hat{ \theta })+\frac{1}{2}\left[\left(\frac{c t}{r}\right)^{2}+1\right] \sin \theta \hat{ r }\right\}.

S \cdot d a =\frac{\mu_{0} \ddot{p}_{0}^{2} t}{32 \pi^{2}}\left[\left(\frac{c t}{r}\right)^{2}+1\right] \frac{\sin ^{2} \theta}{r^{3}}\left(r^{2} \sin \theta d \theta d \phi\right).

P(r, t)=\frac{\mu_{0} \ddot{p}_{0}^{2} t}{32 \pi^{2} r}\left[\left(\frac{c t}{r}\right)^{2}+1\right] 2 \pi \int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{\mu_{0} \ddot{p}_{0}^{2} t}{12 \pi r}\left[\left(\frac{c t}{r}\right)^{2}+1\right].

\text { (c) } P\left(r, t_{0}+r / c\right)=\frac{\mu_{0} \ddot{p}_{0}^{2}}{12 \pi r}\left(t_{0}+\frac{r}{c}\right)\left[\frac{c^{2}}{r^{2}}\left(t_{0}^{2}+2 t_{0} \frac{r}{c}+\frac{r^{2}}{c^{2}}\right)+1\right]=\frac{\mu_{0} \ddot{p}_{0}^{2}}{12 \pi c}\left(1+\frac{c t_{0}}{r}\right)\left[2+2 \frac{c t_{0}}{r}+\left(\frac{c t_{0}}{r}\right)^{2}\right].

P_{ rad }\left(t_{0}\right)=\lim _{r \rightarrow \infty} P\left(r, t_{0}+r / c\right)=\frac{\mu_{0} \ddot{p}_{0}^{2}}{6 \pi c}.

in agreement with Eq. 11.60.

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