A parallel plate capacitor has square plates of side L,...

A parallel plate capacitor has square plates of side L, separated by a distance d, as shown in the figure. Capacitor is charged with a battery with a potential difference $V_0$, the battery is then disconnected. a. Determine the capacitance, $C_0$, and the electric potential energy, $U_0$, and energy density $u_0$ stored in the capacitor at this point. b. A slab made of a dielectric, with relative permittivity of $\kappa$ ($\kappa$ > 1) is then inserted so that it fills a ratio $r$ of the volume between the plates ($r = 2/3$), as shown in the figure. Determine the new capacitance $C_1$, the new potential difference between the plates $V_1$ and the new electric potential energy $U_1$, stored in the capacitor. c. Neglecting gravity, did the inserter of the dielectric slab have to do work or not?

Transcript

00:01 In this question, we have a parallel plate capacitor.

00:06 He has an area of 10 cm by 10 cm.

00:13 The d is 2 .5 m m.

00:19 It is charged to 70 volts.

00:23 So it has plus q and minus q here.

00:26 And then it is disconnected from the battery.

00:29 After that, we insert that.

00:33 Dielectric in between the plates, okay, so that it fuels two -third of the space inside the capacitor.

00:53 Okay, so this dielectric is plexiglas, it has a dielectric constant of 3 .4.

01:01 So we want to calculate the capacitance without the dielectric and the electric potential energy before, before we insert the dielectric and then calculate the c prime b prime and u prime after we insert the dielectric then so let's start to solve this problem okay so part a i'll be using c0 equals to f dot a over d okay and then you also be using u0 is half c0 b square so um just substitute the numbers the area is 10 cm by 10 cm, 0 .1 times 0 .1.

02:04 Ad is 2 .5 m m.

02:10 The capacitance is 3 .54 times 10 to the 0 of negative 11 ferret.

02:21 Then we calculate u .0, which is half, c0, b square, half times 3 .54 times 10 to the negative 11, times, times 70 square.

02:36 The answer is 9 .96 times 10 to the power negative 8, juice.

02:47 Okay, then in part b, okay, now we insert dielectric and so that only two -third of the volume is field.

02:58 So the setup, the setup can be thought of as a parallel plate two parallel plate capacitors in parallel.

03:19 So with the dielectric occupying two -third of the space, this setup can be treated as two capacitors in parallel.

03:58 So the setup will look like this.

04:02 So they have the same d.

04:09 So they have the same d and then on the top we have 2 3 a bottom we have a over 3 and then the top one has a kappa of 3 .4 below is just 1 okay so the new capacitance c prime okay it's just kappa, epsilon a over a d times 2 3rd plus epsilon a over d times 1 3rd.

04:53 So you can pull out the epsilon 0a over d...

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