yungman
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This is not homework even thought this is an example in the book of Griffiths "Introduction of Electrodynamics" 3rd edition. This is a mix retarded field and Poynting theorem question:
\vec E \cdot \vec J \;=\;-\frac 1 2 \frac {\partial}{\partial t} ( \epislon_0 E^2 +\frac 1 {\mu_0} B^2) \;-\;\frac 1 {\mu_0} \nabla \cdot (\vec E X \vec B) (8.8)
\frac {dW}{dt} \;=\; \int _{v'} (\vec E \cdot \vec J) d \vec {v'} \;=\; -\frac 1 2 \frac {\partial}{\partial t} \int _{v'} ( \epsilon_0 E^2 +\frac 1 {\mu_0} B^2) d \vec {v'}\;-\;\frac 1 {\mu_0} \int _{s'} (\vec E X \vec B) d \vec {s'} (8.9)
In this example, the book given:
V=0,\;\;\hbox{ and } \;\;\vec A=\frac{\mu_0 k}{4c}(ct-|x|)^2 \hat z \;\hbox { for x = +ve and }\;\; \vec A=0 \;\;\hbox { for x = -ve.}
c=\frac 1 {\sqrt{\mu_0 \epsilon_0}}
From this, you get:
\vec E= -\frac {\partial \vec A}{\partial t} \;=\; -\frac {\mu_0 k}{2} (ct-|x|)\hat z \;\;\;\hbox { and }\;\;\; \vec B = \nabla X \vec A = ^+_- \hat y \frac{\mu_0 k}{2c}(ct-|x|)
Then the book want to determine the energy per unit time flowing into a box between t1 and t2 with given:
1) Dimensions of the box are 0<y<w, 0<z<l and d<x<(d+h) where w,l, d and h are all +ve number.
2) t1= d/c and t2 =(d+h)/c where c=\frac 1 {\sqrt{\epsilon_0 \mu_0}}.
Since x is +ve and t1= d/c and t2 =(d+h)/c therefore:
\vec E= -\frac {\mu_0 k}{2} (d+h-x)\hat z \;\;\;\hbox { and }\;\;\; \vec B = \hat y \frac{\mu_0 k}{2c}(d+h-x)
Since the question is energy flow into the box, I use second integration term of (8.9)
W \;=\; [\frac 1 {\mu_0} \int _{s'} (\vec E X \vec B) d \vec {s'}]
But the book do this instead:
W\;=\; \frac 1 2 \int _{v'} ( \epsilon_0 E^2 + \frac 1 {\mu_0} B^2) d v'
My understanding is \frac 1 {\mu_0} \int _{s'} (\vec E X \vec B) d \vec {s'} is the EM power flowing through the surface s'. Why is the book use the stored energy or the energy to assemble the charge and current to do the calculation? This is retarded field problem where the EM field just reach the box at t1 and nothing exit the box at t2.
\vec E \cdot \vec J \;=\;-\frac 1 2 \frac {\partial}{\partial t} ( \epislon_0 E^2 +\frac 1 {\mu_0} B^2) \;-\;\frac 1 {\mu_0} \nabla \cdot (\vec E X \vec B) (8.8)
\frac {dW}{dt} \;=\; \int _{v'} (\vec E \cdot \vec J) d \vec {v'} \;=\; -\frac 1 2 \frac {\partial}{\partial t} \int _{v'} ( \epsilon_0 E^2 +\frac 1 {\mu_0} B^2) d \vec {v'}\;-\;\frac 1 {\mu_0} \int _{s'} (\vec E X \vec B) d \vec {s'} (8.9)
In this example, the book given:
V=0,\;\;\hbox{ and } \;\;\vec A=\frac{\mu_0 k}{4c}(ct-|x|)^2 \hat z \;\hbox { for x = +ve and }\;\; \vec A=0 \;\;\hbox { for x = -ve.}
c=\frac 1 {\sqrt{\mu_0 \epsilon_0}}
From this, you get:
\vec E= -\frac {\partial \vec A}{\partial t} \;=\; -\frac {\mu_0 k}{2} (ct-|x|)\hat z \;\;\;\hbox { and }\;\;\; \vec B = \nabla X \vec A = ^+_- \hat y \frac{\mu_0 k}{2c}(ct-|x|)
Then the book want to determine the energy per unit time flowing into a box between t1 and t2 with given:
1) Dimensions of the box are 0<y<w, 0<z<l and d<x<(d+h) where w,l, d and h are all +ve number.
2) t1= d/c and t2 =(d+h)/c where c=\frac 1 {\sqrt{\epsilon_0 \mu_0}}.
Since x is +ve and t1= d/c and t2 =(d+h)/c therefore:
\vec E= -\frac {\mu_0 k}{2} (d+h-x)\hat z \;\;\;\hbox { and }\;\;\; \vec B = \hat y \frac{\mu_0 k}{2c}(d+h-x)
Since the question is energy flow into the box, I use second integration term of (8.9)
W \;=\; [\frac 1 {\mu_0} \int _{s'} (\vec E X \vec B) d \vec {s'}]
But the book do this instead:
W\;=\; \frac 1 2 \int _{v'} ( \epsilon_0 E^2 + \frac 1 {\mu_0} B^2) d v'
My understanding is \frac 1 {\mu_0} \int _{s'} (\vec E X \vec B) d \vec {s'} is the EM power flowing through the surface s'. Why is the book use the stored energy or the energy to assemble the charge and current to do the calculation? This is retarded field problem where the EM field just reach the box at t1 and nothing exit the box at t2.
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