Ii two identical capacitors are connected in parallel and...

(II) Two identical capacitors are connected in parallel and each acquires a charge $Q_{0}$ when connected to a source of voltage $V_{0} .$ The voltage source is disconnected and then a dielectric $(K=3.2)$ is inserted to fill the space between the plates of one of the capacitors. Determine $(a)$ the charge now on each capacitor, and $(b)$ the voltage now across each capacitor.

Transcript

00:01 So here we have been given that there are two identical capacitors that's connected in parallel.

00:07 And the charge that's deposited is q0 when a potential difference of v0 is applied across these two capacitors in parallel.

00:18 And now a dielectric is introduced in the first capacitor.

00:22 And the dielectric constant is given out to be 3 .2.

00:27 And the battery is removed before the insertion of dielectric.

00:32 So we need to figure out the charge that's present on the plates of the capacitor in the second case after the battery is disconnected, as well as the potential difference in the second case.

00:48 So here, in the second case, we can see that the potential difference is same because they are connected in parallel.

00:55 So if we consider the capacitance of each capacitor as c in the first case, because they are identical.

01:01 In the second case, the capacitance of one capacitor will be same, and the other capacitors capacitance will become 3 .2 times its earlier capacitance because a dielectric of dielectric constant 3 .2 was inserted.

01:16 So now we can see that in the second case, when battery is removed, the potential difference has to be seen.

01:22 So basically we know the equation of capacitance, skew is equal to cv is followed.

01:27 So from here, if we get the expression, arranged to get the potential difference that will be q by c.

01:33 So as the capacitance here is altered and this capacitance is increased, so there will be some extra amount of charge deposited here and there will be some amount of charge which will be supplied by this capacitor.

01:48 So the voltage here for this capacitor will be its earlier charge that's q0 plus q divided by the capacitance that's 3 .2c.

02:00 And this voltage should be equal to the voltage across this capacitor, that's q0 minus q by c.

02:09 So from here we can solve and the c gets cut and we get q0 plus q equals to 3 .2 times q0 minus 3 .2 times q.

02:20 So upon rearranging, let's bring the q to the left hand side...

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