Consider a capacitor comprising two parallel plates each...

Consider a capacitor comprising two parallel plates, each with an area of 2000 cm^2 and 1.00 cm apart. The capacitor is connected to a power supply and charged to a potential difference V0 = 3.00 kV. It is then disconnected from the power supply, and a sheet of insulating plastic material is inserted between the plates, completely filling the space between them. We find that the potential difference decreases to 1.00 kV while the charge on each capacitor plate remains constant. Compute (a) the original capacitance C0 (b) the magnitude of the charge Q on each plate (c) the capacitance C after the dielectric is inserted (d) the relative permittivity εr of the dielectric (e) the permittivity ε of the dielectric (f) the total energy stored in the electric field of the capacitor, both before and after the dielectric sheet is inserted. Note: ε0 = 8.85 × 10^-12 F/m

Transcript

00:01 Hi there, here's our problem.

00:04 We have a parallel plate capacitor of a given area of 2 ,000 square centimeter and separated by a distance of 1 centimeter.

00:12 It is connected to a power source and it's been charged to 3 kilovolts.

00:19 So charging stops the moment the potential difference across the plates is equivalent to the power source where it's connected to.

00:25 So when it's charged, its plates have a charge of positive q as of this time and the other one connected to the negative terminal will have a negative q charge.

00:37 Then it was disconnected from the voltage source and then an insulator or a dielectric is inserted completely filling the gap between the two plates.

00:50 After inserting the die electric, the potential difference or the voltage across the two plates decreased to one kilo volt but the charge in each plate remains constant.

01:02 So these are the five questions that we need to answer.

01:06 We are to determine the capacitance of the pilot the capacitor without the die electric.

01:13 Then we need to determine the charge on each plate, the capacitance of the capacitor with the die electric.

01:20 And then we are to determine the relative permittivity of the die electric in part d.

01:26 And then in part e, the permittivity of the die electric.

01:30 Okay, so let's do this.

01:32 For the first part, since all of the factors that we need are present, then we can just immediately use the definition of capacitans of a parallel plate capacitor without any di -electric, let's say vacuum or air, perhaps.

01:51 So this is just equivalent to the permittivity of free space or vacuum times the area of the plate.

01:59 And divided by the separation.

02:02 Since our constant here, the permittivity of free space is in international system, then we need to be consistent with the other units of measurement.

02:12 So for the area, we have to use the square meter, and for the separation, we need to use the meter.

02:18 So we have epsilon sub -not here as 8 .85 times 10 to the negative 12.

02:26 That's the permittivity of free space.

02:28 Area in square meter that's point two and then separation in meter that's 0 .01 so the capacitance without the die electric is given by 1 .77 times 10 to the minus 10 ferrad we can also write this in terms of nano nano is 10 to the minus 9 so we can move the decimal place here one place to the left and it becomes 0 .177 nano for 10 to the minus 10 fared.

03:03 Okay.

03:06 Let's box our final answer.

03:08 Let's go to letter b.

03:11 So for letter b, we are to determine the magnitude of the charge in each plate.

03:16 Since this charge did not change with and without the die electric, then we can just use the capacitance without the dielectric, since you already know it.

03:28 It's just the ratio of the charge to the initial voltage.

03:33 So that's delta v sub -not.

03:36 So that the q here will just be c -sab -0 times the potential difference or the voltage.

03:44 Let's move this up higher.

03:47 So we have here 1 .77 times 10 to the minus 10 ferrad.

03:53 And then this is, you have to get read of the kilo.

03:57 So that's in powers of 10.

03:59 That's 10 to positive 3.

04:00 So that's basically 3 ,000 volts.

04:04 Those are the things that you need to be careful with the prefixes.

04:09 So the charge, the magnitude of the charge in each plate is 5 .31 times 10 to the minus 7 columns.

04:19 Or we can also write this in micro, which is 10 to minus 6.

04:24 So this is just 0 .531 micropulums.

04:31 Okay, let's box our final answer.

04:35 Okay, so one plate that's connected to the positive terminal of the battery becomes positive.

04:42 0 .531 microculum charge and the other one's negative.

04:46 Same magnitude.

04:48 Letter c, we are to determine the capacitance this time with the dielectric.

04:55 So that would just be c without any subscript.

04:59 We just use the fundamental definition of capacitance, which is the ratio of the charge.

05:04 This time to the new voltage, which is one kilo -volt.

05:09 So this is still 5 .31.

05:14 Exponent negative 7, that's coulum, divided by 1 kilo, so that's 1 ,000 volts...

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