***I need help with the 2nd answer! A paper-filled capacitor is charged to a pot

Question

***I need help with the 2nd answer!

A paper-filled capacitor is charged to a potential difference of 2.5 V and then disconnected from the charging circuit. The dielectric constant of the paper is 3.7. Keeping the plates insulated, the paper filling is withdrawn, allowing air to fill the capacitor instead. Find the resulting potential difference of the capacitor.

= 9.25 V

While continuing to keep the capacitor's plates insulated, an unknown substance is inserted between them. The plates then attain a potential difference that is 0.67 times the original potential difference (when paper filled the capacitor). What is this substance's dielectric constant?

= _____________ ?

Explanation / Answer

The new voltage is supposed to be .67 times the original, which is (.67)(2.5) = 1.675

Since Q = CV and charge is conserved

C1V1 = C2V2

C2 = kC1 when you add the dielectric

C1V1 = kC1V2   (Note C1 cancels)

k = V1/V2

k = (9.25)/(1.675)

k = 5.52

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