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Physics
Classical Physics
Mechanics
Derive energy density proportional to emitted power per unit area
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[QUOTE="gtoguy97, post: 6896828, member: 514117"] The energy density ##d\psi## at any point in the cylinder is proportional to the energy emitted per unit time and area from the surface ##dS##, i.e., ##d\psi \propto E/A \cdot t##.We can derive this relation by considering an element of surface area ##dA## of the surface ##dS## and the energy it emits in one second, ##E/A \cdot t##. This energy spreads out in a spherical wave with an intensity ##I##, where ##I = E/A \cdot t##. The total energy emitted by the surface is equal to the energy of the wave at any distance ##r## from the surface, i.e., ##E = 4 \pi r^2 I##.The energy density at any point in the cylinder is then given by the energy of the wave divided by the cylindrical volume, i.e., $$d\psi = \frac{4 \pi r^2 I}{\pi r^2 h} = \frac{4I}{h} $$where h is the height of the cylinder.Therefore, the energy density at any point in the cylinder is proportional to the energy emitted per unit time and area from the surface ##dS##, i.e., ##d\psi \propto E/A \cdot t##. [/QUOTE]
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Classical Physics
Mechanics
Derive energy density proportional to emitted power per unit area
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