A parallel plate capacitor is filled with air and has a capacitance of 25 Mf. A non-conductive liquid with dielectric constant 6.50 is deposited in the space between the plates and fills a fraction f of said space a) Find the new capacitance as a function of f b) What would you expect the new capacitance to be if f = 0? Does it match the value I calculate in the first part? c) What would you expect the new capacitance to be if f = 1? Does it match the value I calculate in the first part?

A parallel plate capacitor is filled with air and has a capacitance of 25 Mf. A non-conductive liquid with dielectric constant 6.50 is deposited in the space between the plates and fills a fraction f of said space a) Find the new capacitance as a function of f b) What would you expect the new capacitance to be if f = 0? Does it match the value I calculate in the first part? c) What would you expect the new capacitance to be if f = 1? Does it match the value I calculate in the first part?

Similar questions

Item 8 Learning Goal: To review the meaning of capacitance and ways of changing the capacitance of a parallel-plate capacitor. Capacitance is one of the central concepts in electrostatics. Understanding its meaning and the difference between its definition and the ways of calculating capacitance can be challenging at first. This tutorial is meant to help you become more comfortable with capacitance. Recall the fundamental formula for capacitance: C = Q/AVC, where C is the capacitance in farads, Q is the charge stored on the plates in coulombs, and AVC is the potential difference (or voltage) between the plates. In the following problems it may help to keep in mind that the voltage is related to the strength of the electric field E and the distance between the plates, d, by AVc = Ed. Part A What property of objects is best measured by their capacitance? the ability to conduct electric current the ability to distort an external electrostatic field the ability to store charge Submit Part…
с A slab of copper of thickness b is thrust into a parallel-plate capacitor of plate area A and plate separation d, as shown in the figure above; the slab is exactly halfway between the plates. NOTE: Express your answers in terms of the given variables using eo and q when necessary. (a) What is the capacitance after the slab is introduced? = W b (b) If a charge q is maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted? d U₁ Uf (c) How much work is done on the slab as it is inserted? = = (d) Is the slab sucked in or must it be pushed in? [Choose one ▾
n Increasing the separation of the two charged parallel plates of a capacitor, which are disconnected from a battery, will produce what effect on the capacitor? O increase charge decrease charge increase capacitance O decrease capacitance
A 12.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 26.0 V across the plates. A piece of material having a dielectric constant of 3.90 is placed between the plates, completely filling the space between them. a) How much energy is stored in the capacitor before the dielectric is inserted? b) How much energy is stored in the capacitor after the dielectric is inserted? c) By how much did the energy change during the insertion?
A 3.9-μF capacitor is fully charged by a 9.0-V battery. The battery is then disconnected. The capacitor is not ideal and the charge slowly leaks out from the plates. The next day, the capacitor has lost half its stored energy. A. Calculate the amount of charge lost.
B