AQA Physics PHYA4 - 20th June 2016 [Exam Discussion Thread]

Original post by Music With Rocks

It kind of makes sense? haha I can't argue with the fact you got the right value

But then why was part (i) not 0.1/0.27?

I feel like the +12mm should make no difference seen as the oscillation starts from the right doesn't it, but I have no idea as you can tell

Oh yh good point, we just did part i normally. Im not sure :s-smilie:

And I only got 0.7 cos you showed the answer lol. Maybe someone else knows....

Reply 221

Cheesecake Ali

3

What do you guys predict the 6 marker to be on because it's what I struggle on tbh

Posted from TSR Mobile

Reply 222

particlestudent

13

Original post by Cheesecake Ali

What do you guys predict the 6 marker to be on because it's what I struggle on tbh

Posted from TSR Mobile

My guess is it will be on transformers :smile:

Reply 223

Music With Rocks

9

Hey I am a bit confused by another question can anyone help me? for part (a) I calculate the potential gradient as

Unparseable latex formula:

5.0jkg^-^1m

(which is given) then for part (b) I know the gravitational field strength is the negative of the potential gradient so therefore equals

Unparseable latex formula:

-5.0Nkg^-^1

However the answer at the back gives it as just

Unparseable latex formula:

5.0Nkg^-^1

Why is it not negative?

Question: physics gravity question.jpg

Reply 224

MintyMilk

3

Original post by Music With Rocks

for part (b) I know the gravitational field strength is the negative of the potential gradient so therefore equals

Unparseable latex formula:

-5.0Nkg^-^1

However the answer at the back gives it as just

Unparseable latex formula:

5.0Nkg^-^1

Why is it not negative?

Question: physics gravity question.jpg

How can a gravitational field strength be negative? If we're saying that higher field strength means higher density of field lines, how can you have a negative density of field lines?

Does it say in the book that field strength is the negative value of potential gradient?

(edited 7 years ago)

Reply 225

Music With Rocks

9

Original post by MintyMilk

How can a gravitational field strength be negative? If we're saying that higher field strength means higher density of field lines, how can you have a negative density of field lines?

Does it say in the book that field strength is the negative value of potential gradient?

Yeah it does

It is written literally just above the exercise so I assumed it applied :s-smilie:

Reply 226

username1306902

9

Original post by Music With Rocks

Yeah it does

It is written literally just above the exercise so I assumed it applied :s-smilie:

what book is that?

Reply 227

Music With Rocks

9

Original post by jarjarmonkey

what book is that?

AQA Physics A textbook, published by Oxford

It is the standard one I think, this photo is from page 58

EDIT: this one

(edited 7 years ago)

Reply 228

boyyo

6

Original post by Music With Rocks

Hey I am a bit confused by another question can anyone help me? for part (a) I calculate the potential gradient as

Unparseable latex formula:

5.0jkg^-^1m

(which is given) then for part (b) I know the gravitational field strength is the negative of the potential gradient so therefore equals

Unparseable latex formula:

-5.0Nkg^-^1

However the answer at the back gives it as just

Unparseable latex formula:

5.0Nkg^-^1

Why is it not negative?

Question: physics gravity question.jpg

I was about to help you with this, but I dont even get the first question :/
How did you calculate the potential gradient? Could you also briefly explain what the potential gradient is please? Im sorry to be a pain, I didnt even know I didnt know this, thank you for putting it up on here

Reply 229

Music With Rocks

9

Original post by boyyo

I was about to help you with this, but I dont even get the first question :/
How did you calculate the potential gradient? Could you also briefly explain what the potential gradient is please? Im sorry to be a pain, I didnt even know I didnt know this, thank you for putting it up on here

No problem at all!
The potential gradient is the change of gravitational potential per metre at that point (the biggest changes are nearer to the surface of an object)
The equation for potential gradient is

g=\Delta V/ \Delta r

So the

\Delta r

is the change in distance which we get from the fact the distance between equipotentials is 1.0km (so the closest equipotential either way is 1km away as shown by the rings on the diagram)

then [latex\] \Delta V is the change in gravitational potential, the ring we are looking at has potential of 500kjkg-1 and the ones eitherside are 495k and 505k so either way the nearest equipotential is 5k

So putting that together

Unparseable latex formula:

g= 5000 / 1000[br]g=5.0jkg^-^1m^-^1

Hope that helped :smile:

just say if you need anything clarified

(edited 7 years ago)

Reply 230

boyyo

6

Original post by Music With Rocks

No problem at all!
The potential gradient is the change of gravitational potential per metre at that point (the biggest changes are nearer to the surface of an object)
The equation for potential gradient is

g=\Delta V/ \Delta r

So the

\Delta r

is the change in distance which we get from the fact the distance between equipotentials is 1.0km (so the closest equipotential either way is 1km away as shown by the rings on the diagram)

then [latex\] \Delta V

is the change in gravitational potential, the ring we are looking at has potential of 500kjkg-1 and the ones eitherside are 495k and 505k so either way the nearest equipotential is 5k

So putting that together

Unparseable latex formula:

g= 5000 / 1000[br]g=5.0jkg^-^1m^-^1

Hope that helped :smile:

just say if you need anything clarified

Ahh yhhh ok i see. Thank you lol. That makes sense, thanks :smile:

Reply 231

Music With Rocks

9

Can anyone please help me with this question?

3. An insulated metal conductor is earthed before a negatively charged charged object is brought near to it.
a) Explain why the free electrons in the conductor move as far away from the charged object as they can.
b) The conductor is then briefly earthed. The charged object is then removed from the vicinity of the conductor. Explain why the conductor is left with an overall positive charge.

so
(a) the free electrons are negatively charged so as like charges repel they move away from the charged object
(b) 1. conductor earthed
2. Charged object discharged through object
3. conductor no longer earthed, conductor and object both neutral
4. object removed, conductor is neutral?

What have I done wrong?

Reply 232

Euclidean

OP

11

Original post by Music With Rocks

Can anyone please help me with this question?

3. An insulated metal conductor is earthed before a negatively charged charged object is brought near to it.
a) Explain why the free electrons in the conductor move as far away from the charged object as they can.
b) The conductor is then briefly earthed. The charged object is then removed from the vicinity of the conductor. Explain why the conductor is left with an overall positive charge.

so
(a) the free electrons are negatively charged so as like charges repel they move away from the charged object
(b) 1. conductor earthed
2. Charged object discharged through object
3. conductor no longer earthed, conductor and object both neutral
4. object removed, conductor is neutral?

What have I done wrong?

When the conductor is earthed, the conductor loses the negative electron charge through the ground. When it is removed from the earth again, it has a loss of negative charge so is positively charged

Reply 233

Music With Rocks

9

Original post by Euclidean

When the conductor is earthed, the conductor loses the negative electron charge through the ground. When it is removed from the earth again, it has a loss of negative charge so is positively charged

But does it not only lose electrons until it has reached a neutral charge? sorry I don't quite get it

Reply 234

Euclidean

OP

11

Original post by Music With Rocks

But does it not only lose electrons until it has reached a neutral charge? sorry I don't quite get it

The conductor is actually neutrally charged to begin with (nuclei are positively charged remember), the electrons are just delocalised which means they can move somewhat freely.

When the conductor loses electrons, it doesn't have those electrons to balance the positively charged nuclei anymore so the conductor becomes overall positively charged

Reply 235

Music With Rocks

9

Original post by Euclidean

The conductor is actually neutrally charged to begin with (nuclei are positively charged remember), the electrons are just delocalised which means they can move somewhat freely.

When the conductor loses electrons, it doesn't have those electrons to balance the positively charged nuclei anymore so the conductor becomes overall positively charged

Oh, so is this how it works (sorry for the rubbish diagrams I did)

EDIT: you have to click on the image again when it opens or else you can't see the text

Original post by Music With Rocks

Can anyone please help me with this question?

3. An insulated metal conductor is earthed before a negatively charged charged object is brought near to it.
a) Explain why the free electrons in the conductor move as far away from the charged object as they can.
b) The conductor is then briefly earthed. The charged object is then removed from the vicinity of the conductor. Explain why the conductor is left with an overall positive charge.

so
(a) the free electrons are negatively charged so as like charges repel they move away from the charged object
(b) 1. conductor earthed
2. Charged object discharged through object
3. conductor no longer earthed, conductor and object both neutral
4. object removed, conductor is neutral?

What have I done wrong?

https://www.khanacademy.org/science/physics/electric-charge-electric-force-and-voltage/charge-electric-force/v/conductors-and-insulators

6:58

(edited 7 years ago)

Reply 237

Alby1234

2

Okay so I'm getting very confused with the mark schemes here, sometimes they give their answers to 2sf and sometimes they give them to 3. Also sometimes they use the unrounded values in their calculations and sometimes they use the rounded values. I know that when it specifies to use the correct number of significant figures you should use the same amount as the the most significant figures they give in the data, but what about any other question? and should we use rounded values or keep unrounded values in the calculation until the final answer?