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A-level Physics question help appreciated!

https://60abffc9b401b1c0936e01291c15112cab0778c4.googledrive.com/host/0B1ZiqBksUHNYVFI1YlJBOWhwRW8/June%202010%20QP%20-%20Unit%204B%20AQA%20Physics.pdf

markscheme says

On question 2a(i) I don't understand why the direction of the electric field is down/from the top plate to the bottom can anyone explain?

and 2bi why is there no electric field when the switch is moved to Y

(noob question) dc voltage supply does it have a positive and negative terminal in the diagram are you supposed to know which one is which?
Original post by blipson
https://60abffc9b401b1c0936e01291c15112cab0778c4.googledrive.com/host/0B1ZiqBksUHNYVFI1YlJBOWhwRW8/June%202010%20QP%20-%20Unit%204B%20AQA%20Physics.pdf

markscheme says

On question 2a(i) I don't understand why the direction of the electric field is down/from the top plate to the bottom can anyone explain?

and 2bi why is there no electric field when the switch is moved to Y

(noob question) dc voltage supply does it have a positive and negative terminal in the diagram are you supposed to know which one is which?


for ai, if the direction of the electric field is from the top to the bottom, the bottom part must be negative because the direction of electric field of a negative source is inwards, and the top part must be positive because the direction of the electric field of a positive source is outwards. Hence, the negatively charge particle will be pulled upwards by the coulomb force between the positive and negative which is true according to the statement provided that the particle move upwards.
Original post by blipson
https://60abffc9b401b1c0936e01291c15112cab0778c4.googledrive.com/host/0B1ZiqBksUHNYVFI1YlJBOWhwRW8/June%202010%20QP%20-%20Unit%204B%20AQA%20Physics.pdf

markscheme says

On question 2a(i) I don't understand why the direction of the electric field is down/from the top plate to the bottom can anyone explain?

and 2bi why is there no electric field when the switch is moved to Y

(noob question) dc voltage supply does it have a positive and negative terminal in the diagram are you supposed to know which one is which?


for bi, if u take a look to the circuit, u will notice that the plates will be both negatively charged from ai, and they will have the same amount of voltage because it is coming from the same source, thus the electric fields cancel each other.

u are supposed to know it from observing the movement of the negatively charged particle from ai
(edited 7 years ago)
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Thanks but I don't understand why the plates are both negatively charged
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Can anyone explain this?
Original post by blipson
https://60abffc9b401b1c0936e01291c15112cab0778c4.googledrive.com/host/0B1ZiqBksUHNYVFI1YlJBOWhwRW8/June%202010%20QP%20-%20Unit%204B%20AQA%20Physics.pdf

markscheme says

On question 2a(i) I don't understand why the direction of the electric field is down/from the top plate to the bottom can anyone explain?

and 2bi why is there no electric field when the switch is moved to Y

(noob question) dc voltage supply does it have a positive and negative terminal in the diagram are you supposed to know which one is which?


If they give you a battery, then you should know that the longer side is the positive terminal and the shorter side is the negative terminal.

In part (a) (i), you don't even need that. You know the particle is negatively charged and that the ball moves vertically upwards. That means the positive plate is above (because negative will be attracted to positive). The direction of an electric field is the path a small positive test charge will take, and hence in this case that will be from up (positive plate) to down (negative plate). Hence the electric field is downwards.

In part (b) (i), essentially, by switching the position of the switch to Y, you no longer have a power supply in your circuit - there is no emf provided to the plates. Instead, the negative and positive plate are just attached and you have something quite like a discharging capacitor. The plates will discharge and so there will be no electric field. And they couldn't do this before because the power supply was in the way.
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Original post by kingaaran
If they give you a battery, then you should know that the longer side is the positive terminal and the shorter side is the negative terminal.

In part (a) (i), you don't even need that. You know the particle is negatively charged and that the ball moves vertically upwards. That means the positive plate is above (because negative will be attracted to positive). The direction of an electric field is the path a small positive test charge will take, and hence in this case that will be from up (positive plate) to down (negative plate). Hence the electric field is downwards.

In part (b) (i), essentially, by switching the position of the switch to Y, you no longer have a power supply in your circuit - there is no emf provided to the plates. Instead, the negative and positive plate are just attached and you have something quite like a discharging capacitor. The plates will discharge and so there will be no electric field. And they couldn't do this before because the power supply was in the way.


Thanks so much man much appreciated I understand it now, do you know why the answer to q23 https://60abffc9b401b1c0936e01291c15112cab0778c4.googledrive.com/host/0B1ZiqBksUHNYVFI1YlJBOWhwRW8/January%202011%20QP%20-%20Unit%204A%20AQA%20Physics.pdf

is A?, my reasoning was that since all of the magnets have an electric field of north to south then due to flemmings left hand rule the force will be sideways? so it won't make a difference and all would fall at the same time but it's not right
Original post by blipson
Thanks so much man much appreciated I understand it now, do you know why the answer to q23 https://60abffc9b401b1c0936e01291c15112cab0778c4.googledrive.com/host/0B1ZiqBksUHNYVFI1YlJBOWhwRW8/January%202011%20QP%20-%20Unit%204A%20AQA%20Physics.pdf

is A?, my reasoning was that since all of the magnets have an electric field of north to south then due to flemmings left hand rule the force will be sideways? so it won't make a difference and all would fall at the same time but it's not right


Well, a change in magnetic flux will induce an emf into a wire. As Q falls through the loop, the magnetic flux will be increasing and an emf will be induced in the wire. However, by Lenz' law, the emf that it induces (and consequently the current) will create its own magnetic field such that it opposes the change that caused it. Hence, the magnetic field around the coil will oppose the magnetic field of Q and slow Q down. This means Q will travel slower than P (because there is no coil that P falls through and so it will experience no repulsion from it).

The tricky one here is R. And what we can see is that the loop here has a gap (in other words, it's an open circuit). Now you won't have current flowing round an open circuit and so when R falls through the loop, it won't induce a current into the loop. It will fall straight through just like P did.

Hence P and R will arrive at the same time, followed by Q.
(edited 7 years ago)
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Original post by kingaaran
Well, a change in magnetic flux will induce an emf into a wire. As Q falls through the loop, the magnetic flux will be increasing and an emf will be induced in the wire. However, by Lenz' law, the emf that it induces (and consequently the current) will create its own magnetic field such that it opposes the change that caused it. Hence, the magnetic field around the coil will oppose the magnetic field of Q and slow Q down. This means Q will travel slower than P (because there is no coil that P falls through and so it will experience no repulsion from it).

The tricky one here is R. And what we can see is that the loop here has a gap (in other words, it's an open circuit). Now you won't have current flowing round an open circuit and so when R falls through the loop, it won't induce a current into the loop. It will fall straight through just like P did.

Hence P and R will arrive at the same time, followed by Q.


Thanks, for this paper Q15 do you have any idea how to get the answer?
http://filestore.aqa.org.uk/subjects/AQA-PHYA41-W-QP-JAN12.PDF
Original post by blipson
Thanks, for this paper Q15 do you have any idea how to get the answer?
http://filestore.aqa.org.uk/subjects/AQA-PHYA41-W-QP-JAN12.PDF


Place the origin at the point charge of +8.0 +8.0 nC. Let the distance where the electric field strength is zero be d d from the origin. Use the principle of superposition for electric field strength for the two point charges and equate to zero and solve for d d.

Spoiler

You would arrive two numerical answers, so you need to reject one of them. Make sure that you know why you are rejecting it - not because it does not appear in one of the four choices.

Hope it helps.
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Original post by Eimmanuel
Place the origin at the point charge of +8.0 +8.0 nC. Let the distance where the electric field strength is zero be d d from the origin. Use the principle of superposition for electric field strength for the two point charges and equate to zero and solve for d d.

Spoiler

You would arrive two numerical answers, so you need to reject one of them. Make sure that you know why you are rejecting it - not because it does not appear in one of the four choices.

Hope it helps.


I am still confused tbh but understand it a little more,when you use the principle of superposition why have you not got 4pi in either of the equations and why is the electric constant on the top instead of the bottom usually the equation is Q/4pi.electric constant. r2
Original post by blipson
I am still confused tbh but understand it a little more,when you use the principle of superposition why have you not got 4pi in either of the equations and why is the electric constant on the top instead of the bottom usually the equation is Q/4pi.electric constant. r2


The electric constant is what other people would call Coloumb's constant.

ke=14πϵ0 k_e = \frac{1}{4 \pi \epsilon_0}

So the magnitude of electric field strength is ∣E⃗∣=keQr2 |\vec{E}| = \frac{k_e Q}{r^2}

To me, I would call ϵ0 \epsilon_0 as electric permittivity.
Original post by blipson
Thanks, for this paper Q15 do you have any idea how to get the answer?
http://filestore.aqa.org.uk/subjects/AQA-PHYA41-W-QP-JAN12.PDF


8-2 = 6

60/6 = 10

That means 1 nC of charge per 10mm, in essence.

The zero resultant electric field strength will be closer to the 2nC charge than the 8nC charge, which means it is 40mm (because it is the only one that is a multiple of 10 and closer to the 2nC charge). You could use the formula, but that's long - intuition is better

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