- #1
DaTario
- 1,051
- 37
Hi All,
Going directly to a concrete numerical problem: Consider two capacitors in series in a one-loop circuit with a switch S. S is initially open. At t=0 S is closed.
Let [itex]C_1 = C_2 = 2F[/itex], the capacitances.
Let [itex]Q_1(t=0) = 4 C[/itex] and [itex]Q_2(t=0) = 0 C[/itex] (second capacitor initially discharged)
At equilibrium, after S is closed we have
[itex]\frac{Q_1}{C_1 }= \frac{Q_2}{C_2}[/itex] which by symmetry implies that
[itex]Q_{1f} = Q_{2f} = 2C [/itex]
Now comes the problem. Initially the energy content is located at capacitor 1.
[itex]E_i = \frac{{Q_1}^2}{2\, C_1 } = 4 J [/itex]
At the end of the process (flow of charges until the equilibrium) there is energy in both capacitors:
[itex] E_f = \frac{{Q_{1f}}^2}{2\, C_1} + \frac{{Q_{2f}}^2}{2\, C_2 } = 1+1 = 2 J[/itex]
Is the missing 2 Joules distributed in space in the form of a propagating EM wave since there were accelerated charges in the process?
Does this system provides us a way to infer (derive, calculate) the law of emission of radiation of an accelerated charge?
If we add a resistor we will have an extra energy cost with the term:
[itex] \Delta E = \int R (i(t))^2 \; dt [/itex] due to Joule effect. Is it correct?Best wishes
DaTario
Going directly to a concrete numerical problem: Consider two capacitors in series in a one-loop circuit with a switch S. S is initially open. At t=0 S is closed.
Let [itex]C_1 = C_2 = 2F[/itex], the capacitances.
Let [itex]Q_1(t=0) = 4 C[/itex] and [itex]Q_2(t=0) = 0 C[/itex] (second capacitor initially discharged)
At equilibrium, after S is closed we have
[itex]\frac{Q_1}{C_1 }= \frac{Q_2}{C_2}[/itex] which by symmetry implies that
[itex]Q_{1f} = Q_{2f} = 2C [/itex]
Now comes the problem. Initially the energy content is located at capacitor 1.
[itex]E_i = \frac{{Q_1}^2}{2\, C_1 } = 4 J [/itex]
At the end of the process (flow of charges until the equilibrium) there is energy in both capacitors:
[itex] E_f = \frac{{Q_{1f}}^2}{2\, C_1} + \frac{{Q_{2f}}^2}{2\, C_2 } = 1+1 = 2 J[/itex]
Is the missing 2 Joules distributed in space in the form of a propagating EM wave since there were accelerated charges in the process?
Does this system provides us a way to infer (derive, calculate) the law of emission of radiation of an accelerated charge?
If we add a resistor we will have an extra energy cost with the term:
[itex] \Delta E = \int R (i(t))^2 \; dt [/itex] due to Joule effect. Is it correct?Best wishes
DaTario