AQA physics, confused about past paper question

b447m

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The question is:
An electron and a proton are 1 x 10^-10m apart, what is the electric potential energy of the electron?

The solution uses the equation Ep=(1/4piE_0)*Q1Q2/r
but havent seen this equation anywhere in my AQA textbook, the only equation for Ep that i have is V=(1/4piE_0)Q/r which doesnt give me the right answer.
Have i missed this equation somewhere or was this question directed at an old spec? Ive checked my whole textbook and i cant find it

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Sinnoh

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The solution given is correct. V is potential energy per unit charge. Hence the GCSE equation E = V × Q. It's the exact same idea here.

I've moved this to the physics forum.

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b447m

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(Original post by Sinnoh)
The solution given is correct. V is potential energy per unit charge. Hence the GCSE equation E = V × Q. It's the exact same idea here.

I've moved this to the physics forum.

i know the solution is correct but im confused as ive never seen the equation the answer sheet used before, ive only seen V=(1/4piE_0)Q/r

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Callicious

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(Original post by b447m)
i know the solution is correct but im confused as ive never seen the equation the answer sheet used before, ive only seen V=(1/4piE_0)Q/r

That equation (the

one) is 100% in the AQA syllabus and should also be on the formula sheet. Although I have to reiterate as I always do... that formula sheet isn't the limit of what you should know and you shouldn't refer to an answer sheet just to get formulae you need, you should know their derivation/what they represent in actuality.

The electric field strength for a point charge at $\vec{r} = \bar{0}$ is defined by

$\vec{E}(\vec{r}) = \frac{kQ}{r^2}\hat{r}$

You can define the work done in moving a point charge

by this field over some $d\vec{r}$ as

$dW = q\vec{E}(r)\cdot{d\vec{r}}$

Now, we know via conservation that

is equal to

for the charge

. Note that up to this point,

is the potential energy, not the potential. Anyhow, Check the work done in moving the charge from infinity to $\vec{r}$ and you'll find

$W = \int_{\infty}^{\vec{r}}\frac{kQq}{r^2}=-\frac{kQq}{r}$

which will satisfy

$W = -(V(r) - V(\infty))$

and can thus find that the potential energy (as a function of

for our nice situation) satisfies

$V(r) = \frac{kQq}{r}$

where I've lazily ignored all vector formalism at this point because I'm tired and lazy. Anyway, remove the

to get it per unit charge and you finally have the potential $\phi(r)$ for our point charge.

You get the idea though... thinking with the fundamentals of integrals and the purest definitions of how all this works you should get the final result. That's what I got taught by my lord and saviour Ms Green, at least.

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b447m

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(Original post by Callicious)
That equation (the

one) is 100% in the AQA syllabus and should also be on the formula sheet. Although I have to reiterate as I always do... that formula sheet isn't the limit of what you should know and you shouldn't refer to an answer sheet just to get formulae you need, you should know their derivation/what they represent in actuality.

The electric field strength for a point charge at $\vec{r} = \bar{0}$ is defined by

$\vec{E}(\vec{r}) = \frac{kQ}{r^2}\hat{r}$

You can define the work done in moving a point charge

by this field over some $d\vec{r}$ as

$dW = q\vec{E}(r)\cdot{d\vec{r}}$

Now, we know via conservation that

is equal to

for the charge

. Note that up to this point,

is the potential energy, not the potential. Anyhow, Check the work done in moving the charge from infinity to $\vec{r}$ and you'll find

$W = \int_{\infty}^{\vec{r}}\frac{kQq}{r^2}=-\frac{kQq}{r}$

which will satisfy

$W = -(V(r) - V(\infty))$

and can thus find that the potential energy (as a function of

for our nice situation) satisfies

$V(r) = \frac{kQq}{r}$

where I've lazily ignored all vector formalism at this point because I'm tired and lazy. Anyway, remove the

to get it per unit charge and you finally have the potential $\phi(r)$ for our point charge.

You get the idea though... thinking with the fundamentals of integrals and the purest definitions of how all this works you should get the final result. That's what I got taught by my lord and saviour Ms Green, at least.

I've looked through my whole aqa textbook and I haven't seen it derived once, it's definitely not on my formula sheet either...
I understand the derivation above but I'm not sure I'm going to need it, the past paper I did was an older one, I'm just having trouble figuring out why it was used when I've never seen it derived before? I don't know how to check whether I'm going to need it?

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Callicious

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(Original post by b447m)
I've looked through my whole aqa textbook and I haven't seen it derived once, it's definitely not on my formula sheet either...
I understand the derivation above but I'm not sure I'm going to need it, the past paper I did was an older one, I'm just having trouble figuring out why it was used when I've never seen it derived before? I don't know how to check whether I'm going to need it?

I did my a levels back in 2017 and as in 2016, so my info might be a bit old.

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Callicious

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(Original post by b447m)
I've looked through my whole aqa textbook and I haven't seen it derived once, it's definitely not on my formula sheet either...
I understand the derivation above but I'm not sure I'm going to need it, the past paper I did was an older one, I'm just having trouble figuring out why it was used when I've never seen it derived before? I don't know how to check whether I'm going to need it?

It's right there

Name: Screenshot_20211005_092748_com.google.android.apps.docs.jpg
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Size: 139.0 KB

Electric potential. Remember it's defined per unit charge.
Wait a sec, what do you think V is in that formula of yours, can you give me your definition?

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b447m

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(Original post by Callicious)
It's right there

Electric potential. Remember it's defined per unit charge.
Wait a sec, what do you think V is in that formula of yours, can you give me your definition?

I think I've gotten mixed up somewhere, i know how to use the 6th one down, I've just never seen it with Qq before (if I'm making sense)

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Callicious

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(Original post by b447m)
I think I've gotten mixed up somewhere, i know how to use the 6th one down, I've just never seen it with Qq before (if I'm making sense)

The convention is that you have the potential at a point, which is the potential energy per unit charge. If you put a charge

at that point, the potential energy will be

where the potential is implicit.

You can see it in the form

or any possible combination. At the end of the day, it's all the same thing: one of the two is producing the potential, the other one (the one you're trying to find the potential energy of) is the one you're multiplying by that potential to get the energy of.

It might be easier to see it this way...

$E_{1,2} = \textrm{On 1 From 2} = V_2(\vec{r_1})\times{q_1}$

If you're doing physics you'll get used to this kind of thing really fast. In some cases there can be many conflicting conventions. Some authors might use a left handed system for a problem, the other a right-handed system: conversions are all over the place, it's a real mess.

Last edited by Callicious; 8 months ago

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b447m

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(Original post by Callicious)
The convention is that you have the potential at a point, which is the potential energy per unit charge. If you put a charge

at that point, the potential energy will be

where the potential is implicit.

You can see it in the form

or any possible combination. At the end of the day, it's all the same thing: one of the two is producing the potential, the other one (the one you're trying to find the potential energy of) is the one you're multiplying by that potential to get the energy of.

It might be easier to see it this way...

$E_{1,2} = \textrm{On 1 From 2} = V_2(\vec{r_1})\times{q_2}$

If you're doing physics you'll get used to this kind of thing really fast. In some cases there can be many conflicting conventions. Some authors might use a left handed system for a problem, the other a right-handed system: conversions are all over the place, it's a real mess.

I understand now thanks

I was getting mixed up between electric potential and electric potential energy

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