Vacuum Variable Capacitor
Vacuum Variable Capacitor
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Energy of a capacitor?
Hello everyone thanks for taking the time of reading my questions which is a hw question that I can't get through because I am really bad solving stuff with variables in terms of other variables if you have any tips or suggestions of what should I do I would appreciateit a lot thanks.
A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is U. The battery is removed, and then a dielectric material with dielectric constant K is inserted into the capacitor, filling the space between the plates. Finally, the capacitor is fully discharged through a resistor (which is connected across the capacitor terminals).
Find U_r, the the energy dissipated in the resistor.
Express your answer in terms of U and other given quantities.
It is widely known that if a dielectric is present between the plates of a capacitor, capacitor increases by a factor K, where K is the dielectric constant. Less known are other effects of this, namely, that the field inside (between plates), and the voltage across, decreases by the same factor. This is true if the capacitor is disconnected from battery prior to dielectric insertion. If the battery remains connected as the dielectric is placed, none of this happens; instead, additional charge will flow into capacitor's plates.
Another rather obscure effect is that the dielectric slab is pulled inside the armature (plates) by the attractive forces between plate and induced electrostatic charges in the dielectric material. Thus, the system is capable of doing work against any resisting force as it pulls the slab into position. This should make clear that, as dielectric is introduced, electric energy U in the capacitor decreases precisely by the amount of work done. This seems simple, but actually is far from being so. For some fraction of this work can easily be converted to kinetic energy , giving rise to a mechanical oscillatory motion(of slab). In an even greater degree, this would happen also if there were no resisting force to do work against.
Anyhow, it is possible to determine the electric energy stored in the capacitor entirely by electrical means, after the dielectric has been inserted. If we call U this electric energy, one expression giving energy in terms of voltage is
U = ½ qV
If battery voltage is denoted by V₀, and since voltage decreases by K, then the energy remaining in the capacitor is just
U = ½ qV = ½ q V₀ / K.
This is also the energy dissipated in the resistor as it discharges the capacitor.
Anticipating that you could object this solution, because battery voltage is unknown, just recall that V = q / CK. Substituting this in the equation above,
U = ½ q² / CK²
Nonetheless, be aware that neither capacitance C, nor charge q are specified in problem statement, all the same.
Hope this helps.