# Hard Physics (unit 4) question :( Watch

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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,

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.

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.

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#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/4πϵ

and due to the negative charge is

-Q/4πϵx

So the total is

Now find x (It's fairly simple geometry with Pythagoras.)

The potential due to the 3 positive charges at that point is

Q/4πϵ

_{o}x + Q/4πϵ_{o}x + Q/4πϵ_{o}x =**3**Q/4πϵ_{o}xand due to the negative charge is

-Q/4πϵx

So the total is

**2**Q/4πϵ_{o}x = Q/2πϵ_{o}xNow find x (It's fairly simple geometry with Pythagoras.)

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(Original post by

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/4πϵ

and due to the negative charge is

-Q/4πϵx

So the total is

Now find x (It's fairly simple geometry with Pythagoras.)

**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/4πϵ

_{o}x + Q/4πϵ_{o}x + Q/4πϵ_{o}x =**3**Q/4πϵ_{o}xand due to the negative charge is

-Q/4πϵx

So the total is

**2**Q/4πϵ_{o}x = Q/2πϵ_{o}xNow find x (It's fairly simple geometry with Pythagoras.)

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#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

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

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#5

(Original post by

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

**BigBadJFly**)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.

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#6

(Original post by

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

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

**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.I thought that the potential was something to do with the path a positive charge would take?

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#7

(Original post by

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?

**BigBadJFly**)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?

Potential is a measure of the potential energy a charge has in a field.

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#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!

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!

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

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*

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

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*

*E*o)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!

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#10

(Original post by

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!

**Boobadoowupski**)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

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*

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

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*

*E*o)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.

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#11

(Original post by

The right method is in post 2.

**Stonebridge**)The right method is in post 2.

Also, your profile gif thing is hypnotic, love it :P

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#12

(Original post by

X

**Stonebridge**)X

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X

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