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

1. Anyone got a "mathsy" explanation for this answer? I just used an educated guess and multiplied by 4/3, but I wanna know why it works.

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2. (Original post by CD223)
Anyone got a "mathsy" explanation for this answer? I just used an educated guess and multiplied by 4/3, but I wanna know why it works.

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I worked out the factor ratio of the potential at the centre between the charges. Hope this satisfy your curiosity

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Edit: oops, the last bit meant to be 2/15(25)
3. (Original post by Mehrdad jafari)
I worked out the factor ratio of the potential at the centre between the charges. Hope this satisfy your curiosity

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Edit: oops, the last bit meant to be 2/15(25)
Sorry for appearing silly, but why is the Q ratio 1/5?

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4. (Original post by CD223)
Sorry for appearing silly, but why is the Q ratio 1/5?

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That's the result of factorising the equations of the potential of both charges

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5. (Original post by Mehrdad jafari)
That's the result of factorising the equations of the potential of both charges

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But the total distance between the charges is 4.0m. And potential is a scalar quantity so you add potentials?

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6. (Original post by CD223)
But the total distance between the charges is 4.0m. And potential is a scalar quantity so you add potentials?

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I thought you would add potentials of different charges( you still subtract them but with one charge being negative results in adding them)

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7. Here is the pure maths, but to do it this way in the exam takes far too long.
8. (Original post by Mehrdad jafari)
I thought you would add potentials of different charges( you still subtract them but with one charge being negative results in adding them)

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In this case you add because both are positive.

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9. (Original post by EHR223)

Here is the pure maths, but to do it this way in the exam takes far too long.
This is the only way that actually makes sense to me - thanks for sharing.

Does anyone know how one would do it with ratios?

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10. (Original post by CD223)
In this case you add because both are positive.

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Oh right, even i have difficulty with the idea of potential and how to deal with them.
So adding the potentials would mean that more work must be done to move a positive charge from infinity to that point than to move it to the middle between the charges?

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11. (Original post by CD223)
In this case you add because both are positive.

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But you would still add them if one was positive and one was negative, although one potential would be a negative value right? you never have to subtract?
12. (Original post by EHR223)
But you would still add them if one was positive and one was negative, although one potential would be a negative value right? you never have to subtract?

You always add potentials, but some might be negative.

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13. JUNE 2014 MC Q16

Anyone fancy having a stab at explaining this?

I thought energy was scalar and therefore as no work is done along equipotentials, the work done along PQRS would be twice PQ.

Why am I wrong?

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14. (Original post by CD223)
JUNE 2014 MC Q16

Anyone fancy having a stab at explaining this?

I thought energy was scalar and therefore as no work is done along equipotentials, the work done along PQRS would be twice PQ.

Why am I wrong?

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You're right, energy is a scalar but work done is a vector quantity. Energy is required to move a charge from P to Q but equal amount of energy is released ( energy is gained from some unknown place), when the charge is moved from R to S, or vice versa as the sign of the charge is not known

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15. (Original post by Mehrdad jafari)
You're right, energy is a scalar but work done is a vector quantity. Energy is required to move a charge from P to Q but equal amount of energy is released ( energy is gained from some unknown place), when the charge is moved from R to S, or vice versa as the sign of the charge is not known

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Oh I see. So energy is the "magnitude" of work done?

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16. (Original post by CD223)
Oh I see. So energy is the "magnitude" of work done?

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Yeah

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17. (Original post by CD223)
Oh I see. So energy is the "magnitude" of work done?

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You know what, actually work done is also a scalar, sorry for misleading you. But in the case of the question the overall work done across PQRS is zero because there is mo displacement in the direction of the force

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18. (Original post by Mehrdad jafari)
You know what, actually work done is also a scalar, sorry for misleading you. But in the case of the question the overall work done across PQRS is zero because there is mo displacement in the direction of the force

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I see. No worries. Thanks.

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19. (Original post by CD223)
I see. No worries. Thanks.

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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

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20. (Original post by CD223)
This is the only way that actually makes sense to me - thanks for sharing.

Does anyone know how one would do it with ratios?

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The main idea is to use only as much info as you need. You can solve this problem simply by knowing that . You don't even need to know the constant of proportionality.

So letting the constant of proportionality be and letting the two charges be , we have (where we put r=2), i.e. .

In the second arrangement,

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