A parallel plate capacitor has a charge magnitude Q = 1.5 x 10-8 C on each plate

Question

A parallel plate capacitor has a charge magnitude Q = 1.5 x 10-8 C on each plate. The material between the plates is free space. (You can assume that Q remains constant throughout this problem.) (a) If the voltage between the plates is 12 V, what is the capacitance? (b) If we double the distance between the plates, what is the resulting voltage and capacitance? (c) If we take the configuration from part (b) and add a material with a dielectric constant %u03BA = 2.1 between the plates, what is the resulting voltage and capacitance?

Explanation / Answer

q=cv

C=Q/V=1.5 x 10-8/12=1.25nF

b)

capacitance half when distance doubled

v=12

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