A paper filled capacitor is charged to a potential difference of 2.3 V and then

Question

A paper filled capacitor is charged to a potential difference of 2.3 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 Keeping the plates insulated, the paper filling is withdrawn, allowing air to fill the capacitor instead.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 off the capacitor.

(volts)

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 0.49 times the original potential difference (when paper filled the capacitor). What is the substances dielectric constant?

Explanation / Answer

V = Q/C

V is proportional to 1/k

where k is dielectric constant

as we remove dielectric

V 1= 2.3 * 3.7 = 8.51 volts

b)

V2-V1 = .49(8.51-2.3)

V2 = 11.5529 volts

k1 = 1 for air

V1k1 = V2k2

k2 = .74

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