Consider the charging capacitor in Prob. 7.34. (a) Find the electric and magnetic fields in the gap, as functions of the distance s from the axis and the time t. (Assume the charge is zero at t = 0.) (b) Find the energy density uem and the Poynting vector S in the gap. Note especially the direct

Question

Consider the charging capacitor in Prob. 7.34.

(a) Find the electric and magnetic fields in the gap, as functions of the distance s from the axis and the time t. (Assume the charge is zero at t = 0.)

(b) Find the energy density [latex]u_{ em }[/latex] and the Poynting vector S in the gap. Note especially the direction of S. Check that Eq. 8.12 is satisfied.

[latex]\frac{\partial u}{\partial t}=- abla \cdot S[/latex] (8.12)

(c) Determine the total energy in the gap, as a function of time. Calculate the total power flowing into the gap, by integrating the Poynting vector over the appropriate surface. Check that the power input is equal to the rate of increase of energy in the gap (Eq. 8.9—in this case W = 0, because there is no charge in the gap). [If you’re worried about the fringing fields, do it for a volume of radius b < a well inside the gap.]

[latex]\frac{d W}{d t}=-\frac{d}{d t} \int_{ u } \frac{1}{2}\left(\epsilon_{0} E^{2}+\frac{1}{\mu_{0}} B^{2} ight) d au-\frac{1}{\mu_{0}} \oint_{ S }( E imes B ) \cdot d a[/latex] (8.9)

Accepted Answer

[latex] ext { (a) } E =\frac{\sigma}{\epsilon_{0}} \hat{ z } ; \sigma=\frac{Q}{\pi a^{2}} ; Q(t)=I t \Rightarrow E (t)=\frac{I t}{\pi \epsilon_{0} a^{2}} \hat{ z }[/latex] .

[latex]B 2 \pi s=\mu_{0} \epsilon_{0} \frac{\partial E}{\partial t} \pi s^{2}=\mu_{0} \epsilon_{0} \frac{I \pi s^{2}}{\pi \epsilon_{0} a^{2}} \Rightarrow B (s, t)=\frac{\mu_{0} I s}{2 \pi a^{2}} \hat{\phi}[/latex] .

[latex] ext { (b) } u_{ em }=\frac{1}{2}\left(\epsilon_{0} E^{2}+\frac{1}{\mu_{0}} B^{2} ight)=\frac{1}{2}\left[\epsilon_{0}\left(\frac{I t}{\pi \epsilon_{0} a^{2}} ight)^{2}+\frac{1}{\mu_{0}}\left(\frac{\mu_{0} I s}{2 \pi a^{2}} ight)^{2} ight]=\frac{\mu_{0} I^{2}}{2 \pi^{2} a^{4}}\left[(c t)^{2}+(s / 2)^{2} ight][/latex].

[latex]S =\frac{1}{\mu_{0}}( E imes B )=\frac{1}{\mu_{0}}\left(\frac{I t}{\pi \epsilon_{0} a^{2}} ight)\left(\frac{\mu_{0} I s}{2 \pi a^{2}} ight)(-\hat{ s })=-\frac{I^{2} t}{2 \pi^{2} \epsilon_{0} a^{4}} s \hat{ s }[/latex] .

[latex]\frac{\partial u_{ em }}{\partial t}=\frac{\mu_{0} I^{2}}{2 \pi^{2} a^{4}} 2 c^{2} t=\frac{I^{2} t}{\pi^{2} \epsilon_{0} a^{4}} ; \quad- abla \cdot S =\frac{I^{2} t}{2 \pi^{2} \epsilon_{0} a^{4}} abla \cdot(s \hat{ s })=\frac{I^{2} t}{\pi^{2} \epsilon_{0} a^{4}}=\frac{\partial u_{ em }}{\partial t}[/latex] .

[latex] ext { (c) } U_{ em }=\int u_{ em } w 2 \pi s d s=2 \pi w \frac{\mu_{0} I^{2}}{2 \pi^{2} a^{4}} \int_{0}^{b}\left[(c t)^{2}+(s / 2)^{2} ight] s d s=\left.\frac{\mu_{0} w I^{2}}{\pi a^{4}}\left[(c t)^{2} \frac{s^{2}}{2}+\frac{1}{4} \frac{s^{4}}{4} ight] ight|_{0} ^{b}[/latex] .

[latex]=\frac{\mu_{0} w I^{2} b^{2}}{2 \pi a^{4}}\left[(c t)^{2}+\frac{b^{2}}{8} ight] . ext { Over a surface at radius } b: P_{ ext {in }}=-\int S \cdot d a =\frac{I^{2} t}{2 \pi^{2} \epsilon_{0} a^{4}}[b \hat{ s } \cdot(2 \pi b w \hat{ s })]=\frac{I^{2} w t b^{2}}{\pi \epsilon_{0} a^{4}}[/latex] .

[latex]\frac{d U_{ em }}{d t}=\frac{\mu_{0} w I^{2} b^{2}}{2 \pi a^{4}} 2 c^{2} t=\frac{I^{2} w t b^{2}}{\pi \epsilon_{0} a^{4}}=P_{ in }[/latex] . (Set b = a for total.)

Question 8.2: Consider the charging capacitor in Prob. 7.34. (a) Find the ...

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