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 q=2.95 × 10-6 C on each plate. The space between the plates is then filled with a dielectric material, with dielectric constant k = 5.49. What must the magnitude of the charge in micro coulombs on each capacitor plate now be, to produce the same potential difference between the plates as before? Express the answer with two decimal places.

I am completely lost.. if you could please show me steps/formulas you used to get to the answer that would be great! thank you!

Explanation / Answer

q=cv=2.95 × 10-6 C now second time capacitance is increased 5 times now q'=5.49*c*v c=k*area*eo/d q'=5.49*2.95 × 10-6 C =16.1955 × 10-6 C

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