A paper-filled capacitor is charged to a potential difference of 2.4 V and then

Question

A paper-filled capacitor is charged to a potential difference of 2.4 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 Keeping the plates insulated, the paper filling is withdrawn, allowing air to fill the capacitor instead.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. (V)

While continuing to keep the capacitor's plates insulated, an unknown substance is inserted between them. The plattes then attain a potential difference that is 0.47 times the original potential difference (when paper filled the capacitor). What is this substance's dielectric constant?

Explanation / Answer

V = 2.1 V, k = 3.7, capacitance = kC

Q = (kC)V = constant

Keeping the plates insulated, the paper filling is withdrawn, allowing air to fill the capacitor instead. The resulting potential difference off the capacitor = Q/C = kV = 7.77 V

new potential difference V' = 0.59V

Q = (k'C)V' = kCV

so k' = kV/V' = 3.7/0.59 = 6.27

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