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

OP

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.

ImageUploadedByStudent Room1432492166.824552.jpg

ImageUploadedByStudent Room1432492166.824552.jpg

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Reply 1821

Absent Agent

21

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.

ImageUploadedByStudent Room1432492166.824552.jpg

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

ImageUploadedByStudent Room1432496286.988211.jpg

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Edit: oops, the last bit meant to be 2/15(25)

(edited 8 years ago)

Reply 1822

CD223

OP

10

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 :smile:

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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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Reply 1823

Absent Agent

21

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
ImageUploadedByStudent Room1432496947.620497.jpg

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Reply 1824

CD223

OP

10

Original post by Mehrdad jafari

That's the result of factorising the equations of the potential of both charges
ImageUploadedByStudent Room1432496947.620497.jpg

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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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Reply 1825

Absent Agent

21

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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Reply 1826

EHR223

2

Here is the pure maths, but to do it this way in the exam takes far too long.

Reply 1827

CD223

OP

10

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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Reply 1828

CD223

OP

10

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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Reply 1829

Absent Agent

21

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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(edited 8 years ago)

Reply 1830

EHR223

2

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?

Reply 1831

CD223

OP

10

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?

Yeah. Potentials are scalar addition, field strengths are vector addition.

You always add potentials, but some might be negative.

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Reply 1832

CD223

OP

10

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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Reply 1833

Absent Agent

21

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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Reply 1834

CD223

OP

10

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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Reply 1835

Absent Agent

21

Original post by CD223

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

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Yeah

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Reply 1836

Absent Agent

21

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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Reply 1837

CD223

OP

10

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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Reply 1838

Absent Agent

21

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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