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    Hello i was wondering if anyone could help me with the following problem, for part a i was assuming that showing that the curl is equal to 0 would be sufficient but when it comes to evaluating this i'm starting to get stuck, the other problem is evaluating the integral for part b as it is also confusing me, part c is relatively straight forward i think as i have managed to achieve the correct result, any help on part a and b would help me a lot, Thanks see the attached image for the problem.Name:  mechanics energy problem.png
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    (Original post by Scary)
    Hello i was wondering if anyone could help me with the following problem, for part a i was assuming that showing that the curl is equal to 0 would be sufficient but when it comes to evaluating this i'm starting to get stuck,
    Showing that the curl=0 will be fine. It's not possible to comment further without seeing your working.
    the other problem is evaluating the integral for part b as it is also confusing me,
    Please put up your working so that we can see what you've done so far.
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    (Original post by atsruser)
    Showing that the curl=0 will be fine. It's not possible to comment further without seeing your working.

    Please put up your working so that we can see what you've done so far.
    Yes of course here it isName:  image.jpg
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    I'm unsure if I'm treating the E term correctly
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    (Original post by Scary)
    Yes of course here it is
    1. Your working for the curl=0 bit looks fine, if much too wordy. You wouldn't want to be writing all that in an exam, for example.

    2. The potential function: I'm not sure exactly what form or working they are expecting. However, presumably you need to perform a line integral explicitly.

    First note that for a constant function, you can't integrate from infinity - the integral will be infinite, so you have correctly chosen a sensible place as 0 potential, namely the origin.

    Then note that it doesn't matter which path from 0 to \vec{r} you choose, since the field is conservative - all paths will give the same potential. That means that you can choose an easy one, namely a straight line, which you have to describe as a parametrised vector e.g. you need a path C with position vector \vec{p}, say, with

    \vec{p}(t) = t\vec{r} and 0 \le t \le 1.

    Then your integral is:

    U(\vec{r}) = - \int_C \vec{F} \cdot \ d\vec{p} = - \int_0^1 \vec{F} \cdot \ \frac{d\vec{p}}{dt} \ dt

    This is then straightforward, as your force vector is constant - the dot product can be expressed in terms of the constant angle \theta and the parametrised vector line. Can you finish this?
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    (Original post by atsruser)
    1. Your working for the curl=0 bit looks fine, if much too wordy. You wouldn't want to be writing all that in an exam, for example.

    2. The potential function: I'm not sure exactly what form or working they are expecting. However, presumably you need to perform a line integral explicitly.

    First note that for a constant function, you can't integrate from infinity - the integral will be infinite, so you have correctly chosen a sensible place as 0 potential, namely the origin.

    Then note that it doesn't matter which path from 0 to \vec{r} you choose, since the field is conservative - all paths will give the same potential. That means that you can choose an easy one, namely a straight line, which you have to describe as a parametrised vector e.g. you need a path C with position vector \vec{p}, say, with

    \vec{p}(t) = t\vec{r} and 0 \le t \le 1.

    Then your integral is:

    U(\vec{r}) = - \int_C \vec{F} \cdot \ d\vec{p} = - \int_0^1 \vec{F} \cdot \ \frac{d\vec{p}}{dt} \ dt

    This is then straightforward, as your force vector is constant - the dot product can be expressed in terms of the constant angle \theta and the parametrised vector line. Can you finish this?
    Hi thanks for your response but this is still confusing me, im unsure why we need to write it as a parametrised vector i know that the solution is -qE_0.r its just showing this which i'm struggling with, another point i'm struggling with from your answer is how can we express the dot product in terms of the constant angle and the vector line? would it not be easier in general to write it as two integrals involving two line pieces ?
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    (Original post by Scary)
    Hi thanks for your response but this is still confusing me, im unsure why we need to write it as a parametrised vector i know that the solution is -qE_0.r
    You don't necessarily need to do it as an integral - from the context of the question, I just assumed that was what they would want. In this case, in fact, you don't need an integral at all, since everything is constant. All you need is the vector equivalent of

    work = force x distance

    which can be expressed directly in terms of the dot product, as the given answer seems to do.

    i'm struggling with from your answer is how can we express the dot product in terms of the constant angle and the vector line?
    For two vectors \vec{a}, \vec{b}, we have \vec{a} \cdot \vec{b} = ab \cos \theta

    would it not be easier in general to write it as two integrals involving two line pieces ?
    I have no idea what you mean by this, I'm afraid.
 
 
 
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