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

I made one for fun:

The electric potential at a point one third along the line connecting two charges, starting from the first charge, is x volts. The electric potential one third along the same line, starting from the second charge, is y volts. What is the electric potential at the midpoint of this line (in terms of x and y)?

Reply 1841

PotterPhysics

6

Another:

Three charges are placed at the vertices of an equilateral triangle. The electric potentials at the midpoints of the sides of the triangle are x volts, y volts and z volts. Find an expression for the electric potential at the center of the triangle.

Reply 1842

zaybun

2

Does anyone have tips for the exam?

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Reply 1843

CD223

OP

10

Original post by Mehrdad jafari

It's ok. By the way, I will email my friend tomorrow to let me know when he's got the data. Once i get them i will upload them on here

Oh brilliant, thank you!

If I was to PM you my email, could you send me it that way too? (It's just I have work all day today so won't really be on TSR!) - just don't wanna lose the post if it does just uploaded :smile:

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Reply 1844

CD223

OP

10

Original post by PotterPhysics

The main idea is to use only as much info as you need. You can solve this problem simply by knowing that

V\propto \frac{Q}{r}

. You don't even need to know the constant of proportionality.

So letting the constant of proportionality be

k

and letting the two charges be

x

, we have

V_{total}=k\frac{x}{2}+k\frac{x}{2}=25

(where we put r=2), i.e.

kx=25

.

In the second arrangement,

V_{total}=k\frac{x}{1}+k\frac{x}{3}=\frac{4}{3}kx=\frac{4}{3}(25)=33.33...=33.3\ (2\ \text{s.f.})

That makes sense. Thanks for sharing :smile:

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Reply 1845

CD223

OP

10

Original post by PotterPhysics

Another:

Three charges are placed at the vertices of an equilateral triangle. The electric potentials at the midpoints of the sides of the triangle are x volts, y volts and z volts. Find an expression for the electric potential at the center of the triangle.

How would you go about this?

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Reply 1846

Absent Agent

21

Original post by CD223

Oh brilliant, thank you!

If I was to PM you my email, could you send me it that way too? (It's just I have work all day today so won't really be on TSR!) - just don't wanna lose the post if it does just uploaded :smile:

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Yeah, no problem :smile:

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Reply 1847

Fred Cantoni

2

What are you guys generally scoring on the multiple choice section?!

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Reply 1848

PotterPhysics

6

Original post by CD223

That makes sense. Thanks for sharing :smile:

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No problem!

Original post by CD223

How would you go about this?

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Leaning on a generalization of my first solution. We let the three charges be

q_1,q_2,q_3

and let the midpoints between

(q_1,q_2),(q_2,q_3),(q_3,q_1)

be

x,y,z

respectively. Then letting (wlog) the sides of the triangle be

2

, and as before letting the constant for

V\propto \frac{Q}{r}

be

k

, we have

k(\frac{q_1}{1}+\frac{q_2}{1})=x\Leftrightarrow k(q_1+q_2)=x

, and similarly

k(q_2+q_3)=y

and

k(q_3+q_1)=z

. Hence adding up we have

2k(q_1+q_2+q_3)=x+y+z

, giving

q_1+q_2+q_3=\frac{x+y+z}{2k}

. (Call this equation 1)

Now, since the figure is an equilateral triangle, its area is

\frac{s^2\sqrt{3}}{4}

where

s

is its side length (I leave this as an exercise to the reader), therefore since our triangle has s=2 then plainly its area is

\sqrt{3}

. Then from the formula

\frac{s^3}{4R}=\frac{s^2\sqrt{3}}{4}

(where R is the circumradius, and both sides give the area of the triangle) we have

R=\frac{2}{\sqrt{3}}

. Therefore, at the center of the triangle the potential is

k\frac{q_1}{R}+k\frac{q_2}{R}+k\frac{q_3}{R}=\frac{k\sqrt{3}}{2}(q_1+q_2+q_3)

. Then substituting equation (1) into this last expression yields the answer

\frac{k\sqrt{3}}{2}(\frac{x+y+z}{2k})=\frac{\sqrt{3}}{4}(x+y+z)

.

Reply 1849

Lau14

11

Original post by zaybun

Does anyone have tips for the exam?

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Keep an eye on how much time you're spending on each half - you shouldn't be exceeding the recommended 45 minutes on the multiple choice I'd say.

In the multiple choice half, look for proportionality in questions. Also look for contradicting statements (particularly on true or false type questions) - they put these in to speed it up, but they're often missed.

Don't forget to leave time to fill in the multiple choice answer sheet! You might forget about this because in mocks you probably just circle the answer on the paper, but there is in fact a separate sheet to fill in.

Obviously never leave multiple choice questions blank - you've got a one in four chance of getting a question right even if you've got absolutely no clue (if you've got the time though, aim to make educated guesses where possible!).

Decide whether doing multiple choice or written part is best for you, and whether starting or finishing with six markers works better. Remember that you don't have to do the paper in order! (May seem obvious, but a lot of people try and do everything in order and lose a lot of time when they get stuck).

If you get stuck, move on and come back to it. Get all your guaranteed "I can definitely do this right now" marks before you try and figure out anything that's confusing you.

Reply 1850

Lau14

11

Original post by Fred Cantoni

What are you guys generally scoring on the multiple choice section?!

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Initially I was stuck getting 18 or 19 on every paper, but now in the region of 22+ :smile:

you?

Reply 1851

boombozx

1

guys I really need your help. I don't know why C is correct and the other answers are invalid Screen Shot 2558-05-25 at 13.16.14.png

Reply 1852

boombozx

1

for part ciii) why didn't they use the v that is found by v=wr r is the radius but they used v for the mid point m

Reply 1853

Corectspelling

1

Original post by boombozx

guys I really need your help. I don't know why C is correct and the other answers are invalid Screen Shot 2558-05-25 at 13.16.14.png

I think that the distractor here is D. The reason why D does not induce an E.M.F may be because there is no change in magnetic flux through the coil when it is rotated in this particular direction, wheras if it is rotated in the other direction (C), there is a change in magnetic flux; hence an EMF is induced according to Faraday's law.

Anyone?

Original post by saad97

Anyone?

Think about how the mass is related to g, and what the mass of a planet is. :smile:

ImageUploadedByStudent Room1432559379.580389.jpg

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Reply 1856

saad97

2

Original post by Jimmy20002012

Think about how the mass is related to g, and what the mass of a planet is. :smile:

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Thank you! I knew the two equations needed but forgot that V=4/3pir^3 -.-
Thanks anyway :smile:

Reply 1857

Absent Agent

21

Original post by boombozx

for part ciii) why didn't they use the v that is found by v=wr r is the radius but they used v for the mid point m

Because M is the midpoint of the radius of the disc, so you need to consider half of the radius when calculating v=wr

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Reply 1858

Fred Cantoni

2

Original post by Lau14

Initially I was stuck getting 18 or 19 on every paper, but now in the region of 22+ :smile:

you?

Sort of 20-23, was just wondering if that was good enough, or if everyone was getting 25!

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Reply 1859

Leonacatherine

2

I was doing a past paper, last quesiton on june 2011 and it said that in a transformer a high voltage introduces insulation problems and raises safety issues.
I dont understand how a high voltage can affect insulation can anyone explain?

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

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