# Hard Physics (unit 4) question :(

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#1
The question is number 14 from June 2010 Paper... apparently the hardest on the paper (multiple choice).

http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF

Apparently the charges on top left and bottom right (+Q and -Q) cancel, but i really don't know why.

I'm self teaching so i guess my main problem is a don't really know what electric potential is. my book says it's the amount of potential energy a point charge would have at a point within an electric field.

Following this i just can't understand why the point charges cancel.

If anyone could help at all, really basic explanation, i would massively appreciate it. thanks.
0
8 years ago
#2
First: What is the distance from each charge to the point P? Call it x for the moment. (Pythagoras will help)

The potential due to the 3 positive charges at that point is
Q/4πϵox + Q/4πϵox + Q/4πϵox = 3Q/4πϵox
and due to the negative charge is
-Q/4πϵx

So the total is 2Q/4πϵox = Q/2πϵox

Now find x (It's fairly simple geometry with Pythagoras.)
0
#3
(Original post by Stonebridge)
First: What is the distance from each charge to the point P? Call it x for the moment. (Pythagoras will help)

The potential due to the 3 positive charges at that point is
Q/4πϵox + Q/4πϵox + Q/4πϵox = 3Q/4πϵox
and due to the negative charge is
-Q/4πϵx

So the total is 2Q/4πϵox = Q/2πϵox

Now find x (It's fairly simple geometry with Pythagoras.)
Got it, thanks
0
8 years ago
#4
I've just done this paper (and got it right )

The 2 positive charges opposite each other cancel out, as the force that they would exert on a charged + particle at P would be equal and opposite

Then, as the other 2 charges both exert the same magnitude of force on a + charge at P, you can say that the elec. pot., V=2(Q/4piEroot2a) as by pythagoras, They are root2a away from P, and this gives you A 0
8 years ago
#5
I've just done this paper (and got it right )

The 2 positive charges opposite each other cancel out, as the force that they would exert on a charged + particle at P would be equal and opposite

Then, as the other 2 charges both exert the same magnitude of force on a + charge at P, you can say that the elec. pot., V=2(Q/4piEroot2a) as by pythagoras, They are root2a away from P, and this gives you A Please be careful here. (Even though you got it right. The thought process could lead to an incorrect answer in a different question.)

The fact that the forces cancel out has nothing to do with the electric potential. Force is about electric field strength, not potential.
You have mixed up the concepts of electric potential and electric field strength.
This question is about potential, not force.
You got the correct answer but by an incorrect line of reasoning. In multi choice it doesn't matter. You get the mark, of course. In a long question you could lose marks for this.
It's actually the electric potential due to the positive and negative charges at opposite corners that "cancel out". The potential due to the one is plus and due to the other is minus an equal amount. The resulting potential is then just the sum of the potentials due to the other two charges.
0
8 years ago
#6
(Original post by Stonebridge)
Please be careful here. (Even though you got it right. The thought process could lead to an incorrect answer in a different question.)

The fact that the forces cancel out has nothing to do with the electric potential. Force is about electric field strength, not potential.
You have mixed up the concepts of electric potential and electric field strength.
This question is about potential, not force.
You got the correct answer but by an incorrect line of reasoning. In multi choice it doesn't matter. You get the mark, of course. In a long question you could lose marks for this.
It's actually the electric potential due to the positive and negative charges at opposite corners that "cancel out". The potential due to the one is plus and due to the other is minus an equal amount. The resulting potential is then just the sum of the potentials due to the other two charges.
So have I got the wrong 2 cancelling out?

I thought that the potential was something to do with the path a positive charge would take?
0
8 years ago
#7
So have I got the wrong 2 cancelling out?

I thought that the potential was something to do with the path a positive charge would take?
No. That's electric field.
Potential is a measure of the potential energy a charge has in a field.
0
8 years ago
#8
Oh, ok thanks 0
8 years ago
#9
This is what I did, spent like half an hour, then found out it was wrong. If you can use what I did to get past your problem though and then find the right method, please share!

Spoiler:
Show
We're using the equation V=kQ/r (where k is the 1/(4pi*epsilon naught) stuff)

1. The top left and bottom right cancel, because if you add up the electric potential from each one, they cancel- because one is positive, the other negative, and they are equidistant from P.

2. Bottom left and top right are both positive, so won't cancel. However, because they are symmetrically opposite and again both the same distant from the middle, we can effectively consider them as one contributor with double the charge.

Therefore we get V=k*2Q/r

Use pythagoras to find the length of the diagonals (2a*sqrt(2)), which is double the distance from one corner to the middle, which is r. So r=rt(2)a.

V=2kQ/rt2a, =Q/(2rt(2)*a*pi*Eo)

AW balls that's A and the answer's C. Will have to wait for someone else to have a go...

EDIT: Ignore that, for some reasons on the examiner's report it's got the answer noted as C not A... Just wasted almost two hours trying to see why. Bloody Physics, I don't have time for panicking! 0
8 years ago
#10
This is what I did, spent like half an hour, then found out it was wrong. If you can use what I did to get past your problem though and then find the right method, please share!

Spoiler:
Show
We're using the equation V=kQ/r (where k is the 1/(4pi*epsilon naught) stuff)

1. The top left and bottom right cancel, because if you add up the electric potential from each one, they cancel- because one is positive, the other negative, and they are equidistant from P.

2. Bottom left and top right are both positive, so won't cancel. However, because they are symmetrically opposite and again both the same distant from the middle, we can effectively consider them as one contributor with double the charge.

Therefore we get V=k*2Q/r

Use pythagoras to find the length of the diagonals (2a*sqrt(2)), which is double the distance from one corner to the middle, which is r. So r=rt(2)a.

V=2kQ/rt2a, =Q/(2rt(2)*a*pi*Eo)

AW balls that's A and the answer's C. Will have to wait for someone else to have a go...

The right method is in post 2.
0
8 years ago
#11
(Original post by Stonebridge)
The right method is in post 2.
I got it, AQA screwing me over again...

Also, your profile gif thing is hypnotic, love it :P
0
8 years ago
#12
(Original post by Stonebridge)
X
Your answers are extremely helpful but it took me such a long time to read them because of your incredibly distracting avatar. 0
X

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