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The physics a2 megathread
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Original post by BobbJo
sorry very much
CIE does not require you to know variation of pd with time so no need to fret over it
they will give you the graph
CIE does not require you to know variation of pd with time so no need to fret over it
they will give you the graph
No problem.
By the way, where does the charge lost from the capacitor exactly go if even the resistor doesnt store it?
(edited 5 years ago)
Original post by Clark20
Alright. The graph didnt show the negative sign with the resistors voltage.
Both get to 0, but one decreases and the other increases.
Also by the way, if resistor doesnt store charge then where exactly does the charge lost by the capacitor go to?
Both get to 0, but one decreases and the other increases.
Also by the way, if resistor doesnt store charge then where exactly does the charge lost by the capacitor go to?
electrons flow from negative to positive
positive charges "neutralised"
Original post by old_teach
Just stumbled onto this.
I feel there is some confusion in the above posts.
This may help:
When discharging, the pd across capacitor = pd across resistor (it's a parallel circuit)
The + charge from one plate on the capacitor is moving through the resistor to cancel out the - charge on the other capacitor plate, until there is no pd across the capacitor. (conventional current flow)
I feel there is some confusion in the above posts.
This may help:
When discharging, the pd across capacitor = pd across resistor (it's a parallel circuit)
The + charge from one plate on the capacitor is moving through the resistor to cancel out the - charge on the other capacitor plate, until there is no pd across the capacitor. (conventional current flow)
why not pd across capacitor = - pd across resistor.
emf is 0.
unless you are talking about magnitudes
thank you
Original post by old_teach
Just stumbled onto this.
I feel there is some confusion in the above posts.
This may help:
When discharging, the pd across capacitor = pd across resistor (it's a parallel circuit)
The + charge from one plate on the capacitor is moving through the resistor to cancel out the - charge on the other capacitor plate, until there is no pd across the capacitor. (conventional current flow)
I feel there is some confusion in the above posts.
This may help:
When discharging, the pd across capacitor = pd across resistor (it's a parallel circuit)
The + charge from one plate on the capacitor is moving through the resistor to cancel out the - charge on the other capacitor plate, until there is no pd across the capacitor. (conventional current flow)
Alright, makes sense.
When the charge of the capacitor gets to 0/is cancelled out, the pd is zero too. Why again does the pd of the resistor become 0 too then?
(edited 5 years ago)
Original post by Clark20
Alright, makes sense.
When the charge of the capacitor gets to 0, the pd is zero too. Why again does the pd of the resistor become 0 too then?
When the charge of the capacitor gets to 0, the pd is zero too. Why again does the pd of the resistor become 0 too then?
KVL
Emf = sum of pds in a loop
0 = pd across C + pd across R
when pd across C is 0, so is pd across R
i hope this helps
Original post by BobbJo
KVL
Emf = sum of pds in a loop
0 = pd across C + pd across R
when pd across C is 0, so is pd across R
i hope this helps
Emf = sum of pds in a loop
0 = pd across C + pd across R
when pd across C is 0, so is pd across R
i hope this helps
Oh, I get it now. Thank you.
"why not pd across capacitor = - pd across resistor.
emf is 0."
I see your point. However, I'd connect one end of my voltmeter to the bottom wire, and one to the top wire, so I'd get the same pd across both.
"Why again does the pd of the resistor become 0 too then?"
pd across resistor = I x R, so when the capacitor has discharged, current is zero and so Vresistor = 0
Hope I'm being helpful!
emf is 0."
I see your point. However, I'd connect one end of my voltmeter to the bottom wire, and one to the top wire, so I'd get the same pd across both.
"Why again does the pd of the resistor become 0 too then?"
pd across resistor = I x R, so when the capacitor has discharged, current is zero and so Vresistor = 0
Hope I'm being helpful!
Original post by old_teach
"why not pd across capacitor = - pd across resistor.
emf is 0."
I see your point. However, I'd connect one end of my voltmeter to the bottom wire, and one to the top wire, so I'd get the same pd across both.
emf is 0."
I see your point. However, I'd connect one end of my voltmeter to the bottom wire, and one to the top wire, so I'd get the same pd across both.
Thanks
Original post by old_teach
"why not pd across capacitor = - pd across resistor.
emf is 0."
I see your point. However, I'd connect one end of my voltmeter to the bottom wire, and one to the top wire, so I'd get the same pd across both.
"Why again does the pd of the resistor become 0 too then?"
pd across resistor = I x R, so when the capacitor has discharged, current is zero and so Vresistor = 0
Hope I'm being helpful!
emf is 0."
I see your point. However, I'd connect one end of my voltmeter to the bottom wire, and one to the top wire, so I'd get the same pd across both.
"Why again does the pd of the resistor become 0 too then?"
pd across resistor = I x R, so when the capacitor has discharged, current is zero and so Vresistor = 0
Hope I'm being helpful!
Thank you.
Need help with both parts of part (b) of the question.
I know we use the equation (change in V)×(q)=1/2mv^2
My main question is, between which 2 points/distances do we calculate the changes in electric potential for use in both the parts to calculate the asked speeds? And why do we choose those 2 specific points for change in potential for the required speed?
I know we use the equation (change in V)×(q)=1/2mv^2
My main question is, between which 2 points/distances do we calculate the changes in electric potential for use in both the parts to calculate the asked speeds? And why do we choose those 2 specific points for change in potential for the required speed?
(edited 5 years ago)
For max speed, use the biggest change in potential, and for speed at B, use the potential at B, I guess.
(comparing those potentials to the potential at the start to get change potential)
(comparing those potentials to the potential at the start to get change potential)
(edited 5 years ago)
Original post by old_teach
For max speed, use the biggest change in potential, and for speed at B, use the potential at B, I guess.
(comparing those potentials to the potential at the start to get change potential)
(comparing those potentials to the potential at the start to get change potential)
Ok. Thats also how its solved in the mark scheme.
So, the point where the particle initially starts moving from is taken as our initial point, and we take final potential at the point at which speed is required, doing final-initial giving change in potential?
(edited 5 years ago)
Yes - that's how potential works, all you need is start and end. Potential is much easier to use than electric field (being a scalar helps).
Original post by BobbJo
…
why not pd across capacitor = - pd across resistor.
emf is 0.
unless you are talking about magnitudes
thank you
why not pd across capacitor = - pd across resistor.
emf is 0.
unless you are talking about magnitudes
thank you
Original post by BobbJo
KVL
Emf = sum of pds in a loop
0 = pd across C + pd across R
….
Emf = sum of pds in a loop
0 = pd across C + pd across R
….
Original post by old_teach
"why not pd across capacitor = - pd across resistor.
emf is 0."
I see your point. However, I'd connect one end of my voltmeter to the bottom wire, and one to the top wire, so I'd get the same pd across both.
…
emf is 0."
I see your point. However, I'd connect one end of my voltmeter to the bottom wire, and one to the top wire, so I'd get the same pd across both.
…
I suspect there is a bit misunderstanding on something in terms of
Is pd across capacitor = - pd across resistor or pd across capacitor = pd across resistor?
In fact, I would say both are correct as both are describing two “different” scenarios.
If a charged capacitor is connected in series with a resistor to undergo discharging, we can write KVL by starting from the negative side of the capacitor to the positive side of the capacitor and across the resistor:
p.d. across capacitor + p.d. across resistor = 0 ----Eqn(1)
If we rewrite the Eqn (1) using the direction of conventional current, we would get
p.d. across capacitor - IR = 0 ----Eqn(2)
This is because the current is in the direction of high potential to low potential. Next, we arrange it to become
p.d. across capacitor = IR ----Eqn(3) (this is what old_teach implies of pd across capacitor = pd across resistor)
And if we rearrange eqn(1) straightaway without considering the direction of conventional current, we would get
p.d. across capacitor = -p.d. across resistor ----Eqn(4) (this is what BobbJo implies)
Is eqn (4) wrong? Absolutely not, one needs to interpret the equation in context.
Original post by old_teach
Yes - that's how potential works, all you need is start and end. Potential is much easier to use than electric field (being a scalar helps).
Start potential remains same. In the eq Vq=1/2mv^2 we get speed at the point that we use as the end potential, right?
For e.g we needed the speed reaching at sphere B, we used the end potential as the potential at sphere B and the initial potential as the potential where the particle starts moving from.
That's it!
An additional note to Clark20, make sure you understand why the change in potential which is usually the final potential minus initial potential where it ends up having a negative sign but it is somehow disappeared when change in potential times charge is equated to the change in KE. This is where students tend to lose the mark without giving a clear working or explanation.
Original post by Eimmanuel
An additional note to Clark20, make sure you understand why the change in potential which is usually the final potential minus initial potential where it ends up having a negative sign but it is somehow disappeared when change in potential times charge is equated to the change in KE. This is where students tend to lose the mark without giving a clear working or explanation.
I see.
Also, reading my above replies, did I understand the question correctly?
I understood that the point where we need our speed at, we take that point as final potential. For e.g in (bii) of the question a particle travels from the surface of one sphere to anothers (Sphere B). We're required to calculate the speed of the particle when it reaches the surface of Sphere B. So, we used the potential at the surface of sphere B as final potential and from where the particle starts moving as initial.
Also, does a particle move fastest where the potential is lowest/magnitude of strength of electric field is highest? If yes, then why?
(edited 5 years ago)
"Also, does a particle move fastest where the potential is lowest/magnitude of strength of electric field is highest? If yes, then why?"
It moves fastest when it has the most ke, when there is the greatest potential energy decrease, so, in this question, when the potential is lowest.
At that point the gradient of the potential is zero, so the electric field is zero.
Remember electric field strength tells you force, when force is zero that means acceleration is zero. In this case it's been getting faster, so a=0 means it's now at max speed, as it moves further the force changes direction and the acceleration becomes negative, the particle slows down till it reaches sphere B.
I hope that helps your understanding.
It moves fastest when it has the most ke, when there is the greatest potential energy decrease, so, in this question, when the potential is lowest.
At that point the gradient of the potential is zero, so the electric field is zero.
Remember electric field strength tells you force, when force is zero that means acceleration is zero. In this case it's been getting faster, so a=0 means it's now at max speed, as it moves further the force changes direction and the acceleration becomes negative, the particle slows down till it reaches sphere B.
I hope that helps your understanding.
Original post by Clark20
I see.
Also, reading my above replies, did I understand the question correctly?
I understood that the point where we need our speed at, we take that point as final potential. For e.g in (bii) of the question a particle travels from the surface of one sphere to anothers (Sphere B). We're required to calculate the speed of the particle when it reaches the surface of Sphere B. So, we used the potential at the surface of sphere B as final potential and from where the particle starts moving as initial.
Also, does a particle move fastest where the potential is lowest/magnitude of strength of electric field is highest? If yes, then why?
Also, reading my above replies, did I understand the question correctly?
I understood that the point where we need our speed at, we take that point as final potential. For e.g in (bii) of the question a particle travels from the surface of one sphere to anothers (Sphere B). We're required to calculate the speed of the particle when it reaches the surface of Sphere B. So, we used the potential at the surface of sphere B as final potential and from where the particle starts moving as initial.
Also, does a particle move fastest where the potential is lowest/magnitude of strength of electric field is highest? If yes, then why?
Original post by Clark20
I see.
Also, reading my above replies, did I understand the question correctly?
I understood that the point where we need our speed at, we take that point as final potential. For e.g in (bii) of the question a particle travels from the surface of one sphere to anothers (Sphere B). We're required to calculate the speed of the particle when it reaches the surface of Sphere B. So, we used the potential at the surface of sphere B as final potential and from where the particle starts moving as initial.
Also, reading my above replies, did I understand the question correctly?
I understood that the point where we need our speed at, we take that point as final potential. For e.g in (bii) of the question a particle travels from the surface of one sphere to anothers (Sphere B). We're required to calculate the speed of the particle when it reaches the surface of Sphere B. So, we used the potential at the surface of sphere B as final potential and from where the particle starts moving as initial.
First of all, I hope that you understand that you are using conservation of energy to solve the problem where we can write the following:
Loss in electric potential energy = Gain in kinetic energy
OR
Change in various types of energy in the system = Work done on the system.
I usually prefer the second way of writing the conservation of energy.
In using the second way of writing the conservation of energy, we can write the following for the problem (the system is the alpha particle and the two spheres)
Change in Electric potential energy of the system + Change in KE of alpha particle = 0
q(Vfinal – Vinitial) + (Kfinal – Kinitial) = 0
From this equation, you should be able to see what happens to the alpha particle. Note that conservation of energy can only tell you what happens NOT why this happens or how this happens.
Original post by Clark20
...Also, does a particle move fastest where the potential is lowest/magnitude of strength of electric field is highest? If yes, then why?
If you learn calculus, you should know that when is a quantity or function is maximum. The maximum point in function y(x) is when the derivative is zero. Applying a similar argument to this problem, you can come to the conclusion that when the speed is maximum, the acceleration of the alpha particle is zero. If the acceleration of the alpha particle is zero, this implies that the resultant force acting on the alpha particle is zero, which also implies that resultant electric field strength due to the two charged spheres is zero. Since the gradient of the potential curve gives the magnitude of the electric field strength, so when the electric field strength is zero, the potential is either at the maximum or minimum. In this case, the potential is at the minimum when the electric field strength is zero.
In order to “understand” why a thing happens in this way, you can sketch out the electric field strength to know the force acting on the alpha particle to see how the dynamics of alpha particles changes.
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