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

Original post by a.a.k

I only smiley faces again :sexface:

Lol! Hope this helps...

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Reply 2581

Original post by CD223

Lol! Hope this helps...

ImageUploadedByStudent Room1433535167.816087.jpg

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

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Reply 2582

sykik

Could we be asked about how AC generators work and how electric motors work?

Reply 2583

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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Reply 2584

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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Reply 2585

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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Reply 2586

Original post by a.a.k

This part makes sense .
Thnx

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No problem! Hope revision goes well :smile:

Reply 2587

Original post by CD223

No problem! Hope revision goes well :smile:

Just started today. After finushing few resits

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Reply 2588

Original post by CD223

[br]E = \dfrac{CV^2}{2}[br]

[br]\therefore E \propto V^2[br]

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

E = \dfrac{QV}{2}

. Because they both show a relationship between E and V. How did you know your equation was correct?

(edited 8 years ago)

Reply 2589

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

3 formulad are
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.
Whats the answer btw

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The answer is B.

Another TSR member has help me with this question.

This is how (s)he concluded his/her answer.

Reply 2590

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

E = \dfrac{QV}{2}

.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 :smile:

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 :smile:

Reply 2591

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

E = \dfrac{CV^2}{2}

and not

E = \dfrac{QV}{2}

?

They both show a relationship between E and V, right?

(edited 8 years ago)

Reply 2592

Original post by CD223

Because Q is directly proportional to V also :smile:

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 :smile:

Ah I see. So we needed an equation that had V dependent on one thing, not two, right?

Reply 2593

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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Reply 2594

Original post by CD223

Yup!

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

How're you feeling for PHYA4?

Reply 2595

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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Reply 2596

Original post by Disney0702

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

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? :smile:

Reply 2597

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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Reply 2598