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

1. (Original post by a.a.k)
I only smiley faces again

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Lol! Hope this helps...

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2. (Original post by CD223)
Lol! Hope this helps...

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Thnkx but
Why cant we do other formulas and how dies this relate to charge being 50%

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3. Could we be asked about how AC generators work and how electric motors work?
4. (Original post by sykik)
Could we be asked about how AC generators work and how electric motors work?
Probabky related to induced emf in someway

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5. (Original post by a.a.k)
Thnkx but
Why cant we do other formulas and how dies this relate to charge being 50%

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Because that's the only formula relating energy and voltage that doesn't contain another term that is proportional to voltage.

E = 1/2 QV but Q=VC so in reality, E=1/2 CV^2.

This means that if the charge increases by a factor of 1.5, then the voltage will increase by a factor of 1.5, and as E is proportional to voltage squared, E increased by a factor of 2.25.

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6. (Original post by CD223)

Because that's the only formula relating energy and voltage that doesn't contain another term that is proportional to voltage.

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This part makes sense .
Thnx

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7. (Original post by a.a.k)
This part makes sense .
Thnx

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No problem! Hope revision goes well

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8. (Original post by CD223)
No problem! Hope revision goes well

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Just started today. After finushing few resits

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9. (Original post by CD223)

That means that when V becomes 1.5V, E becomes 2.25E. Does that help?

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Yes that helps a lot thank you.

When I was looking at this question I was torn between using the equation you used and
. Because they both show a relationship between E and V. How did you know your equation was correct?
10. (Original post by a.a.k)
I think its 1.5E
Becuase it says charge is increased by 50% and doesnt say anything about capacitane

1) E=0.5QV
2) E=Q^2/2C
3) E=(C*V^2)0.5

So only formula without C is no. 1 nothing is squared so i am guessing answer is A.

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Another TSR member has help me with this question.

This is how (s)he concluded his/her answer.
11. (Original post by Disney0702)
Yes that helps a lot thank you.

When I was looking at this question I was torn between using the equation you used and
.Because they both show a relationship between E and V. How did you know your equation was correct?
Because Q is directly proportional to V also you need an equation for E in terms of its relationship directly to V.

As Q = VC and E = 1/2 QV, E= 1/2 CV^2 as Q depends on V directly.

This means E is proportional to V squared

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12. (Original post by SuperMushroom)
Im pretty sure that the energy stored is proportional to the voltage squared, sorry in advance if this is incorrect
No you're correct, the answer is B.

But how did you know to use and not ?

They both show a relationship between E and V, right?
13. (Original post by CD223)
Because Q is directly proportional to V also you need an equation for E in terms of its relationship directly to V.

As Q = VC and E = 1/2 QV, E= 1/2 CV^2 as Q depends on V directly.

This means E is proportional to V squared
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Ah I see. So we needed an equation that had V dependent on one thing, not two, right?
14. (Original post by Disney0702)
Ah I see. So we needed an equation that had V dependent on one thing, not two, right?
Yup!

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15. Ah thank you so much, that really cleared things up for me.

How're you feeling for PHYA4?
16. What's the difference between magnetic field and electric field, mainly looking at how charged particles interact in such fields??

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17. (Original post by Disney0702)
Ah thank you so much, that really cleared things up for me.

How're you feeling for PHYA4?
PHYA4 I definitely prefer to PHYA5. That being said, I don't feel like I've revised it for ages! Got 2 exams on Monday and Tuesday so they're the focus this weekend. Then I'm gonna blitz PHYA4 after Tuesday morning!

You?

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18. (Original post by SuperMushroom)
What's the difference between magnetic field and electric field, mainly looking at how charged particles interact in such fields??

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In electric fields:
• Charged particles experience a force equal to EQ or QV/d when in the field, in the direction of the field lines for a positive charge, or in the opposite direction for a negative charge.
• They move in a parabolic path when perpendicular to the field lines.
• The line of force or field lines point in the direction a free positive charge would move (positive to negative).
• Particles have their velocity (magnitude and direction) changed by the field when they enter.

In magnetic fields:
• Charged particles experience a force equal to BQv or when perpendicular to the field. A component of this force is experienced at an angle to field lines.
• They move in a circular path when perpendicular to field lines.
• The line of force or field lines point in the direction a free North Pole would move (North to South).
• There is no force on a particle that is stationary, or parallel to field lines.

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19. What r u guys doing for optional unit.
I do 5D turning points

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20. Hi can someone tell me if these graphs and the statements I've made at the right hand side are correct. Also do I need to know anymore graphs?

So;

For the velocity graph, the gradient is negative from a maximum when the displacement is to the right of the equilibrium as its trying to accelerate backwards towards the normal.

The gradients are different as velocity always opposes displacement?

And for acceleration its opposite as acceleration is always toward the equilibrium position so as its displaces +ve to the right its acceleration is towards the left -ve.

Sorry this is a bit muddled, but if someone could tell me if I'm correct or just make it a little easier that would be great thanks )))

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