A parallel-plate capacitor, with air between the plates, is connected to a batte

Question

A parallel-plate capacitor, with air between the plates, is connected to a battery. The battery establishes a potential difference between the plates by placing charge of magnitude 5.56 x 10-6 C on each plate. The space between the plates is then filled with a dielectric material, with dielectric constant K = 18.1. What must the magnitude of the charge in micro-coulomb on each capacitor plate now be, to produce the same potential difference between the plates as before? Express the answer with one decimal place.

Explanation / Answer

Charge q = 5.56 x 10 -6 C Dielectric constant k = 18.1 Potential difference is same in two situations. C q ( C ' / C ) = ( q ' / q) ( k C / C ) = ( q ' / q ) Since after insertion of dielectric the capacitance is increases to k times. So, q ' / q = k From this charge on each plate after insertion of dielectric q ' = kq                                                                                               = 100.636 x 10 -6 C                                                                                               ~ 100.6 x 10 -6 C

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