Map Sapling Learning A paper-filled capacitor is charged to a potential differen

Question

Map Sapling Learning 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 Keeping the plates insulated, the paper filling is withdrawn, allowing air to fill the capacitor instead.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 of the capacitor Number 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 0.61 times the original potential difference (when paper filled the capacitor). What is this substance's dielectric constant? Number Previous Check Answer Next Hint

Explanation / Answer

Part A -

Suppose C is the initial capacitance of the capacitor.

So, the new capacitance of the capacitor after removing the paper, C1 = C / 3.7

Suppose new potential difference across the plate = V1

Now, charge across the plate will be the same in both the cases.

So, Q = C*V = C1*V1

=> C*2.3 = (C / 3.7)*V1

=> V1 = 2.3 * 3.7 = 8.51 V

So, the resulting potential difference = 8.51 V

Part B -

In this case, suppose the new capacitance is C2.

So, C1*8.51 = C2*0.61*2.3

=> C2 = (C1*8.51) / (0.61*2.3) = 6.06 * C1

So, the dielectric constant of the substance = 6.06

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