A paper-filled capacitor is charged to a potential difference of 2.2 V and then

Question

A paper-filled capacitor is charged to a potential difference of 2.2 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 Keeping the plates insulated, the paper filling is withdrawn, allowing air to fill the capacitor instead.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

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 0.41 times the original potential difference (when paper filled the capacitor). What is this substance's dielectric constant?

Explanation / Answer

We know that the capacitance of the capacitor increases by k times when we interduce a dielectric constant

Let us now consider the potential difference between the plates of the capacitor is Vo

When the battery is disconnected the charge stored remains the same on the capacitors

Now the potential difference between the plates of the capacitor is

V =Vo/k

The dielectric constant of paper is k= 3.7

Then Vo =3.7V

Given that V =2.2V

Then Vo =3.7*2.2 =8.14V

Now unknow material of dielectric constant k is placed between the plates

Therefore the potenial diference between the paltes is V1 =0.41(2.2V) =0.902V

Then V1 =Vo/k

Then the substances dielectric constant is given by

k =V0/V1 =8.14V/0.902V =9.024

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