OCR Physics A G485 - Frontiers of Physics - 18th June 2015

Can someone please help me with question 1dii in paper June 2010 G485. In the question it mentions that they are connected in parallel so that means both of the capacitors have different charges but why is the voltage calculated by dividing the charge of the capacitor we calculated in the above question by the total capacitance?

Reply 442

randlemcmurphy

11

Original post by sagar448

Can someone please help me with question 1dii in paper June 2010 G485. In the question it mentions that they are connected in parallel so that means both of the capacitors have different charges but why is the voltage calculated by dividing the charge of the capacitor we calculated in the above question by the total capacitance?

You are right, but total charge is conserved. So the charge you calculated in b)i) is the total charge, but each capacitor gets different amounts of it, when S2 is closed. We are asked to calculate the final p.d across the capacitors so we need total charge/sum of capacitors.

Reply 443

Original post by sagar448

Can someone please help me with question 1dii in paper June 2010 G485. In the question it mentions that they are connected in parallel so that means both of the capacitors have different charges but why is the voltage calculated by dividing the charge of the capacitor we calculated in the above question by the total capacitance?

I think it's because as soon as switch 2 is closed the pd across the 1.5 microF capacitor is zero. Therefore it's charge is equal to zero using Q=CV. The question states to find the pd across both capacitors, so you do total charge, which is 28.35+0, divided by the total capacitance, which is 1.5+4.5 microF.

Reply 444

Original post by randlemcmurphy

You are right, but total charge is conserved. So the charge you calculated in b)i) is the total charge, but each capacitor gets different amounts of it, when S2 is closed. We are asked to calculate the final p.d across the capacitors so we need total charge/sum of capacitors.

Oh that makes sense.

Reply 445

Original post by randlemcmurphy

You are right, but total charge is conserved. So the charge you calculated in b)i) is the total charge, but each capacitor gets different amounts of it, when S2 is closed. We are asked to calculate the final p.d across the capacitors so we need total charge/sum of capacitors.

OHHH how did I miss that? yes!! The charge that I calculated gets distributed to the capacitor so it is the total charge I see, yes. And to think I've done this paper before XD. Thanks bruh. :3

Reply 446

Original post by sagar448

If you don't mind I would like to know what question that is and from what past paper? Thanks :smile:

Q1 June 2011
http://www.ocr.org.uk/Images/62267-question-paper-unit-g485-fields-particles-and-frontiers-of-physics.pdf

Reply 447

Original post by chem@uni

How much and what sort of revision are people doing for this exam?

i.e. how many PP or how many notes, how much time a day?

For this exam only? Maybe an average of 1.5 hours a day. Sometimes 0, sometimes about 3.

Reply 448

Original post by randlemcmurphy

You are right, but total charge is conserved. So the charge you calculated in b)i) is the total charge, but each capacitor gets different amounts of it, when S2 is closed. We are asked to calculate the final p.d across the capacitors so we need total charge/sum of capacitors.

Original post by a123a

I think it's because as soon as switch 2 is closed the pd across the 1.5 microF capacitor is zero. Therefore it's charge is equal to zero using Q=CV. The question states to find the pd across both capacitors, so you do total charge, which is 28.35+0, divided by the total capacitance, which is 1.5+4.5 microF.

For d.i) on that same question, why is the total capacitance just the sum of the capacitors? I'd have thought that because the p.d.s across them are different at t=0 that means they're in series and therefore you have to use the inverse addition formula.

Reply 449

Original post by Elcor

For d.i) on that same question, why is the total capacitance just the sum of the capacitors? I'd have thought that because the p.d.s across them are different at t=0 that means they're in series and therefore you have to use the inverse addition formula.

Because when both switches are closed the capacitors are in parallel and so the capacitance just add normally. It tells you in the question, calculate the total capacitance is switch S2 is closed.

:biggrin:

Reply 450

Original post by sagar448

Because when both switches are closed the capacitors are in parallel and so the capacitance just add normally. It tells you in the question, calculate the total capacitance is switch S2 is closed.

:biggrin:

I agree that they're in parallel when both switches are open as they both have the same p.d. (0V). When S2 is closed though one has 6.3V and one has 0V. Different p.d.s so they're in series so use inverse addition formula. Where am I wrong?

Reply 451

Original post by Elcor

I agree that they're in parallel when both switches are open as they both have the same p.d. (0V). When S2 is closed though one has 6.3V and one has 0V. Different p.d.s so they're in series so use inverse addition formula. Where am I wrong?

Ok so before anything is closed both are uncharged. Then switch S1 closes and so the capacitor 4.5 has a 6.3 pd across it. It specifically states that the capacitors are in parallel and in parallel the capacitance just adds so we get a total of 6uF. The Switch S2 is still open. The charge on the 4.5uF capacitor is 28.35. Now the S2 is closed and S1 is open. We know that the charge on the 1.5uf capacitor was 0 before the switch was closed that means before the 4.5uf capacitor starts to discharge there is only 28.35 of charge available for BOTH capacitors. Therefore that is the total charge. And so to find the total pd across both the capacitors you add the capacitance (they are still in parallel) and you divide the charge by the capacitance to give you the total PD.

Hope that helps. :biggrin:

Ask again if you still don't understand. :3

Reply 452

Original post by Elcor

For d.i) on that same question, why is the total capacitance just the sum of the capacitors? I'd have thought that because the p.d.s across them are different at t=0 that means they're in series and therefore you have to use the inverse addition formula.

Hmm. I get what you mean. I think you use the sum of the capacitors because in the question it states that another capacitor is added in parallel.

Reply 453

chem@uni

9

Would someone explain when you use and what each variable means in

Z=Pc

Ir/Io=(z2-z1)^2/(z2+z1)^2

Dlamba/lamva=v/c

All on the very last page of formula book

Reply 454

Original post by chem@uni

Would someone explain when you use and what each variable means in

Z=Pc

Ir/Io=(z2-z1)^2/(z2+z1)^2

Dlamba/lamva=v/c

All on the very last page of formula book

Z is the acoustic impedance of a substance, rho is its density and c is the speed of ultrasound in the substance.

Ir/Io is the ratio of the intensity of reflected ultrasound waves to the intensity of incident ultrasound waves.

The last equation is for Doppler shift, and I believe it only works for galaxies (someone please correct me if that's wrong), where v is the recessional speed of the galaxy, delta lambda is the change in wavelength of light coming off it and lambda is the original wavelength of the light.

Reply 455

ForgottenApple

Original post by chem@uni

Would someone explain when you use and what each variable means in

Z=Pc

Ir/Io=(z2-z1)^2/(z2+z1)^2

Dlamba/lamva=v/c

All on the very last page of formula book

Someone call me out if I'm wrong!

Z is the acoustic impedance; P is the density of the medium, C is the speed of ultrasound in the medium

Delta Lamda is the Change in wavelength of a source that has had its wavelength changed due to motion (increased due to red shift for most)
Lamda is the original/non shifted wavelength, V is the recessional speed of a galaxy and c is the speed of light

Ir is in intensity reflected, Io is the incident intensity, Z are the acoustic impedances of different medium

Reply 456

chem@uni

9

Can anyone expalin how you can use 3/2kt=1/2mv^2 to do the very last question on Jan 2012

Reply 457

randlemcmurphy

11

Original post by chem@uni

Can anyone expalin how you can use 3/2kt=1/2mv^2 to do the very last question on Jan 2012

From that equation we can see that T is directly proportional to v², so we can say that T/v² is a constant, we can therefore say that T^1/2/v is a constant, you should be able to figure it out from there, by rearranging to form the ratio in terms of the temperature.

Reply 458

chem@uni

9

Original post by randlemcmurphy

From that equation we can see that T is directly proportional to v², so we can say that T/v² is a constant, we can therefore say that T^1/2/v is a constant, you should be able to figure it out from there, by rearranging to form the ratio in terms of the temperature.

Alright thanks that really helped

Reply 459