A parallel-plate capacitor has plates with an area of 054m^2 and a separation of

Question

A parallel-plate capacitor has plates with an area of 054m^2 and a separation of 58mm, the space between the plates is filled with a dielectric whose dielectric constant is 1.50. (a) What is the potential difference between the plates when the charge on the capacitor plates is 3.5 mu c? (b) Will your answer to the part (a) increase, decrease, or stay the same if the dielectric constant is decreased? Explain. (c) Calculate the potential difference for the case where the dielectric constant is 1.2.

Explanation / Answer

A=0.054m^2

d=0.58mm

K=1.5

Q=3.5×10^-6C

C=permitivity×A/d=8.854×10^-12×0.054/0.58×10^-3=0.824×10^-9

V=Q/kC=3.5×10^-6/1.5×0.824×10^-9=2.831×10^3..........eq 1

If dielectric constant is decreased V wil increase as both are inversely proportional

If battery remains connected to the capacitor then V wil remain same

If k=1.2 then using eq 1 we get

V=3.538×10^3V

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