AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread]

Original post by PotterPhysics

Or by the definition of speed -- the length of water coming out per second -- we know that we just have to find the length. This is found by dividing the volume by the cross sectional area, because volume=length*area so volume/area=length.

Yeah, that would do too :smile:

Reply 1963

OP

Original post by PotterPhysics

Or by the definition of speed -- the length of water coming out per second -- we know that we just have to find the length. This is found by dividing the volume by the cross sectional area, because volume=length*area so volume/area=length.

That, too aha! That's how I thought of it in my class mock.

Original post by huniibehi

Your explanations are always on point! And yeah you should deffo get an A for unit 4, I'm terrified for both papers tbh

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Ah I'm terrified for PHYA5 more than PHYA4 if I'm honest! Glad I've got a week in between them :smile:

Reply 1964

knighty437

1

Hey guys, I was going through the JUNE 2014 paper and the last question is about charging and discharging a capacitor through a resistor. I've noticed on the mark scheme for the discharging part it says
"#VC decreases from -E to zero and VR decreases from E to zero
# at any time VC = - VR
# both VC and VR decrease exponentially with time"
Can someone please explain this to me. How can VC decrease from a negative number to zero? Shouldn't it increase? I know the voltage across a capacitor decreases when discharging, I'm just confused by how they've said it. If anyone is curious the CGP A2 physics revision guide has a diagram on pg24 showing the graphs at the bottom and one is negative but increases to zero.

Reply 1965

OP

Original post by knighty437

Hey guys, I was going through the JUNE 2014 paper and the last question is about charging and discharging a capacitor through a resistor. I've noticed on the mark scheme for the discharging part it says
"#VC decreases from -E to zero and VR decreases from E to zero
# at any time VC = - VR
# both VC and VR decrease exponentially with time"
Can someone please explain this to me. How can VC decrease from a negative number to zero? Shouldn't it increase? I know the voltage across a capacitor decreases when discharging, I'm just confused by how they've said it. If anyone is curious the CGP A2 physics revision guide has a diagram on pg24 showing the graphs at the bottom and one is negative but increases to zero.

The mark scheme is stating (albeit ambiguously) that the magnitude of the PD is decreasing to zero, which is correct for a discharging capacitor.

The negative sign indicates the PD is in the opposite direction to the changing current. Numerically, you are correct in that the number does nominally increase from a negative number to zero, but this is not true for the "actual" magnitude potential across the capacitor. The capacitor discharges until the PD across its plates equals zero.

In short, the minus sign is only there for convention (due to the direction of the initial current) rather than to be taken literally.

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Reply 1966

knighty437

1

Thank-you! That made a lot of sense. I've also noticed that is says the voltage across the resistor decreases from the emf to zero when the capacitor is discharging. But when the capacitor is charging it also says it that the voltage across the resistor decreases from the emf to zero. My question is, when does it change from zero after charging to the emf before discharging? Sorry if that's worded badly, the mark scheme confuses me so much!

Reply 1967

OP

Original post by knighty437

Thank-you! That made a lot of sense. I've also noticed that is says the voltage across the resistor decreases from the emf to zero when the capacitor is discharging. But when the capacitor is charging it also says it that the voltage across the resistor decreases from the emf to zero. My question is, when does it change from zero after charging to the emf before discharging? Sorry if that's worded badly, the mark scheme confuses me so much!

When charging:
• The initial current is at a maximum.
• The initial PD across the resistor is a maximum (E)
• The PD across the resistor decreases to zero as the PD across the capacitor increases to E.
• At any time, the PD across the resistor and the PD across the capacitor sum to E.

When discharging:
• The current reverses direction.
• The initial current is again at a maximum.
• The initial PD across the resistor is equal to -E (or E, if you call the capacitor PD -E).
• The PD across the resistor decreases from -E (or E) to zero as the PD across the capacitor decreases from E (or -E) to zero.
• At any time, the PD across the resistor and the PD across the capacitor are the negative of one another (due to the current convention).

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Reply 1968

zaybun

Original post by Mehrdad jafari

Well, I think that's what makes it harder or even medical physics one basically. Astro is 4 chapters and medical 5 chapters whereas turning points is only 3 chapters. Even though turning points may be harder, the content is logical and reasonable and there is not so much in it that seem being beyond the scope of the A2 specification.

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Fair enough. But I guess it just depends on the person then haha

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Reply 1969

Original post by zaybun

Fair enough. But I guess it just depends on the person then haha

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That's true. I haven't looked at the astro chapter to see what is taught otherwise I may in fact have enjoyed studying astro more than turning points :smile:

Reply 1970

JJBinn

Hi guys please can somebody explain 11 to me? And for 14, I don't understand why the answer is negative, I thought energy was a scalar?

http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-W-QP-JUN11.PDF

Reply 1971

OP

Original post by JJBinn

Hi guys please can somebody explain 11 to me? And for 14, I don't understand why the answer is negative, I thought energy was a scalar?

http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-W-QP-JUN11.PDF

Question 11, the answer is C:
ImageUploadedByStudent Room1432720472.751052.jpg

Question 14, the answer is B:

[br]{E_{EP}} = \dfrac{-1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{4 \pi {\epsilon_0} \times 1.0 \times 10^{-10}}[br]

This gives:

[br]{E_{EP}} = -2.3 \times 10^{-18} J[br]

As the energy is negative, it means negative work has to be done to attract the proton. It seems counter intuitive, but no work is done in moving two opposite charges towards one another, but energy still has to be transferred due to the conservation of energy.

If it was two like charges, then the energy would be positive as work must be done to move the two charges together.

In this case, the electron and proton will attract one another without any work being done on either particle. As such, work done is negative in this case.

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Reply 1972

JJBinn

Original post by CD223

Question 11, the answer is C:
ImageUploadedByStudent Room1432720472.751052.jpg

Question 14, the answer is B:

[br]{E_{EP}} = \dfrac{-1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{4 \pi {\epsilon_0} \times 1.0 \times 10^{-10}}[br]

This gives:

[br]{E_{EP}} = -2.3 \times 10^{-18} J[br]

As the energy is negative, it means negative work has to be done to attract the proton. It seems counter intuitive, but no work is done in moving two opposite charges towards one another, but energy still has to be transferred due to the conservation of energy.

If it was two like charges, then the energy would be positive as work must be done to move the two charges together.

In this case, the electron and proton will attract one another without any work being done on either particle. As such, work done is negative in this case.

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Oops, I meant 8, not 11, sorry. Thanks for the explanation on 14, makes more sense now :smile:

Reply 1973

OP

Original post by JJBinn

Oops, I meant 8, not 11, sorry. Thanks for the explanation on 14, makes more sense now :smile:

Lol, no worries :tongue:

Question 8, the answer is B:
ImageUploadedByStudent Room1432722367.980613.jpg

Reply 1974

Original post by CD223

As the energy is negative, it means negative work has to be done to attract the proton. It seems counter intuitive, but no work is done in moving two opposite charges towards one another, but energy still has to be transferred due to the conservation of energy.

If it was two like charges, then the energy would be positive as work must be done to move the two charges together.

In this case, the electron and proton will attract one another without any work being done on either particle. As such, work done is negative in this case.

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Oh man, your explanation made me realise a very silly mistake i was making when we had discussion about the gravitational potential.

That's true, the potential energy between two oppositely charged particles is negative because energy is released if the oppositely charged particles attract each other and the negative sign is an indication of the energy released because we consider the energy of the system rather ourselves to say that the negative sign indicates the energy required. As a result, the oppositely charged particles after attracting each other will have a less energy content.

The same is with gravitational potential. No work (from us) needs to be done to move a mass from infinity to to a gravitational potential point in space, and therefore energy is released when a mass is moved from infinity to that point because gravity is attractive. Hence, the negative sign is an indication of the energy released.

I don't know why i got that the other way round? :frown:

Reply 1975

OP

Original post by Mehrdad jafari

Oh man, your explanation made me realise a very silly mistake i was making when we had discussion about the gravitational potential.

That's true, the potential energy between two oppositely charged particles is negative because energy is released if the oppositely charged particles attract each other and the negative sign is an indication of the energy released because we consider the energy of the system rather ourselves to say that the negative sign indicates the energy required. As a result, the oppositely charged particles after attracting each other will have a less energy content.

The same is with gravitational potential. No work (from us) needs to be done to move a mass from infinity to to a gravitational potential point in space, and therefore energy is released when a mass is moved from infinity to that point because gravity is attractive. Hence, the negative sign is an indication of the energy released.

I don't know why i got that the other way round? :frown:

Ah yeah that makes sense. All very in depth and counter-intuitive stuff! :tongue:

Reply 1976

Original post by CD223

Ah yeah that makes sense. All very in depth and counter-intuitive stuff! :tongue:

Yeah, true :smile:

Now i understand why the gravitational potential at infinity is the greatest. This is because the work done by the mass creating the potential on a test mass is zero at that point, or in other words the test mass will be in a higher energy state at infinity that, if released, a huge amount of energy is released when reaching the mass creating the potential.

Hope this has dispelled the confusions we encountered before :smile:

I always prefer the understanding of things as the mathematics of them don't always convince me :wink:

Reply 1977

OP

Original post by Mehrdad jafari

Yeah, true :smile:

Now i understand why the gravitational potential at infinity is the greatest. This is because the work done by the mass creating the potential on a test mass is zero at that point, or in other words the test mass will be in a higher energy state at infinity that, if released, a huge amount of energy is released when reaching the mass creating the potential.

Hope this has dispelled the confusions we encountered before :smile:

I always prefer the understanding of things as the mathematics of them don't always convince me :wink:

Yeah the maths conventions don't always make logical sense. Like the capacitor PD being the negative of the resistor PD during a capacitor discharge :tongue:

Reply 1978

Original post by CD223

Yeah the maths conventions don't always make logical sense. Like the capacitor PD being the negative of the resistor PD during a capacitor discharge :tongue:

Lol, yeah

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Reply 1979

JJBinn