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    5.Two identical conducting spheres on insulating supports carry charges of magnitude Q and 2Q respectively. When separated by distance d, the electrostatic repulsive force is F. The spheres are made to touch and then restored to their original separation d. If there is no loss of charge what is the new force of repulsion?

    I don't even know what it is trying to say...

    So the two spheres are made to touch, and then the are put back in their original positions? So...nothing changes...doesn't it mean the the Force stays the same?

    it's multiple choice though, and F isn't one of the answers.

    Thanks
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    The spheres will initially be at different potentials. As they are identical spheres, they have the same capacitance. Call it C. The potentials will be V=Q/C so the sphere with 2Q will be at twice the potential of the other one.
    When they are connected, charge will move from the high potential to the lower until the pds of the spheres are the same. So pd and charge on the one goes down, and on the other goes up. Charge is conserved (it's still 3Q but equally shared) so you can calculate the new Q for the two spheres, and then the new F.
    It should be 9/8 times F
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    Ok thanks. I have a physics exam soon, so Im revising loads now lol. Ima just gonna make this thread my revision thread. So here's another.

    Two parallel metal plates separated by a distance d have a potential difference V across them. What is the magnitude of the electrostatic force acting on a charge Q placed midway between the plates?


    The answer is VQ/D

    I was just wondering coz, it's got d/2 on both sides of the particle, so how would I go about working this out?

    I got F=EQ and E = V/D (D=d/2)

    Therefore F = 2VQ/d, thats my working, so I get an extra 2, and I know its something to do with forces on either side.

    And just say that the particle Q was placed .25d from Plate A and .75d from plate B, how would I go about working this out?

    Thanks

    EDIT: I think if you just explain how to do the .25d and .75d one it will explain the 1/2d one. :P Thanks
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    Between the parallel plates there is a uniform electric field. That means the value of E (and therefore F on any charge) is the same everywhere. (Don't get confused with the potential at different points between the plates).
    For a parallel plate system the field is simply E=V/d (Often called a uniform potential gradient) and is the same everywhere. So F=Eq and E=V/d
    It would be the same at the other two points.
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    OK thanks

    Another question

    Say I have two point charges. How would I find the resultant electric potential at any point between them?

    Eg:

    -----4C--------------------------------------------------12C
    ____________________________________ _x________________
    -------------------7.3m-------------------|------2.6m------

    What would be the potential at x?

    Would it just be:

     \dfrac {4}{4\pi\epsilon_{0}\times7.3} + \dfrac {12}{4\pi\epsilon_{0}\times2.6} ?

    Like, you just add both sides to get the potential. Am I right?

    Thanks
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    If they are both positive charges yes, just add the two potentials together. (IE. both are plus).
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    Thanks

    A spring is suspended from a fixed point. A mass attached to the spring is set into vertical undamped simple harmonic motion. When the mass is at its lowest position, which one of the following has its minimum value?

    the choices are 'Potential Energy of the system' and 'Kinetic Energy of the mass'

    I thought Potential energy of system is the same as GPE, so I went for that, but I also realised that KE would also be a minimum...

    The asnwer is KE, but I dont see why PE of system cant be minimum as well

    Thanks

    EDIT: another question:



    I thought both B and C would suffice. But C is the only answer. Why wouldn't B work also? Even if u move it just a fraction, it will induce an emf.

    Thanks
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    (Original post by 2710)
    Thanks

    A spring is suspended from a fixed point. A mass attached to the spring is set into vertical undamped simple harmonic motion. When the mass is at its lowest position, which one of the following has its minimum value?

    the choices are 'Potential Energy of the system' and 'Kinetic Energy of the mass'

    I thought Potential energy of system is the same as GPE, so I went for that, but I also realised that KE would also be a minimum...

    The asnwer is KE, but I dont see why PE of system cant be minimum as well

    Thanks
    The PE of the system here means the elastic PE stored in the spring. This is greatest when the spring is stretched the most. The KE is certainly a minimum there. Think too, that, if energy is conserved, the total KE + PE must remain constant. So when KE is a minimum, PE is a maximum; and vice versa.
    EDIT: another question:



    I thought both B and C would suffice. But C is the only answer. Why wouldn't B work also? Even if u move it just a fraction, it will induce an emf.

    Thanks
    These can be a bit tricky.
    To induce an emf, the number of lines of force passing through the cross sectional area of the coil must change.
    In the diagram, the coil is positioned such that the lines of flux don't pass through that area - it's edge-on.
    Moving the coil up or down or sideways, or rotating it as in D, will not cause any flux lines to pass through the end because the end is always facing you, out of the page. As soon as you rotate it as in option C, it turns so that the field passes through the end. This changes "the flux linking the coil", and an emf is generated.
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    (Original post by Stonebridge)
    The PE of the system here means the elastic PE stored in the spring. This is greatest when the spring is stretched the most. The KE is certainly a minimum there. Think too, that, if energy is conserved, the total KE + PE must remain constant. So when KE is a minimum, PE is a maximum; and vice versa.

    These can be a bit tricky.
    To induce an emf, the number of lines of force passing through the cross sectional area of the coil must change.
    In the diagram, the coil is positioned such that the lines of flux don't pass through that area - it's edge-on.
    Moving the coil up or down or sideways, or rotating it as in D, will not cause any flux lines to pass through the end because the end is always facing you, out of the page. As soon as you rotate it as in option C, it turns so that the field passes through the end. This changes "the flux linking the coil", and an emf is generated.
    Lol, oh yeh, forgot about EPE :P

    I don't really understand the second one. let's simplify the diagram to :



    The movement of the wire is up and down, ie. Perpendicular to the field lines, which is exactly the same as option B right? in my diagram, I know for a fact it will induce and emf, no matter how small steps you move it, since it will be cutting across the field lines.

    Please can you elaborate? Thanks
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    If you move that wire up or down it cuts through/across the flux lines. (They go from N to S.) The wire is perpendicular to the field. Moving it in the direction shown produces no emf. It doesn;t cut flux.
    In the coil, the wire at the top (X) when it moves down, does not cut through the lines of force. It is parallel to the lines of force. In this case there is no induced emf.
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    Oh yeh, my Magnetic fields is a bit rusty. Thanks
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    Ok, there's something I'm thinking about.

    Gravitational Field Strength Formula goes:

     g = \dfrac{GM}{r^2}

    Now in this question it goes, find the field strength on the surface of a planet that has half of the earth's radius and a quarter of earth's mass. I worked it out, using ratios, and find that this smaller planet has the same field strength as earth. This surprised me.

    I then though, what if you half that of the new planet's radius and divide its mass by 4, you will get another planet with the same field strength. So I thought, if you continue this, won't you get to a point where you have a very small object with field strength the same as earths?

    Ie. Take me for example. I weigh 50kg.

    I worked out that for me to have field strength the same as the earths, my radius would have to be 2x10^-5 m

    So I was wondering, if you had a sphere that size, that weighed 50kg, it would exert the same gravitational field strength the same as earth :eek:

    Am I right?
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    (Original post by 2710)
    Ok, there's something I'm thinking about.

    Gravitational Field Strength Formula goes:

     g = \dfrac{GM}{r^2}

    Now in this question it goes, find the field strength on the surface of a planet that has half of the earth's radius and a quarter of earth's mass. I worked it out, using ratios, and find that this smaller planet has the same field strength as earth. This surprised me.

    I then though, what if you half that of the new planet's radius and divide its mass by 4, you will get another planet with the same field strength. So I thought, if you continue this, won't you get to a point where you have a very small object with field strength the same as earths?

    Ie. Take me for example. I weigh 50kg.

    I worked out that for me to have field strength the same as the earths, my radius would have to be 2x10^-5 m

    So I was wondering, if you had a sphere that size, that weighed 50kg, it would exert the same gravitational field strength the same as earth :eek:

    Am I right?
    Yes that is true , but remeber the smaller the object the more denser(more mass / unit volume) it is ..
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    Ok thanks

    Can I just ask, is it possible to have two oscillations in phase if their frequencies are different? I'm assuming no. Ie, would osc.1 with frequency A be in phase if osc.2 has frequency 2A and also started in phase (Ie, both start at 0) Would they be in phase?

    If no, does that mean they can't be out of phase as well? Because to be out of phase, you have to have a uniform out of phase degree right?

    Thanks (Resonance makes me:mad: )
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    is it possible to have two oscillations in phase if their frequencies are different?
    They can start with the same phase, but after any amount of time passes they will be out of phase as you say.

    does that mean they can't be out of phase as well?
    No - they are always out of phase with each other (unless one frequency is a integer multiple of the other, in which case every so often they will be instantaneosly in phase, before becoming out of phase again).

    Because to be out of phase, you have to have a uniform out of phase degree right?
    I think here uniform is referring to a pair of waves that have a constant phase offset, e.g. cos x and sin x are always 90 degrees out of phase with each other, no matter the value of x.

    A constant offset is not a requirement for something to be out of phase - this offset can be a variable also. For example, if you had  cos x and  sin(x+\phi(t)) where  \phi(t) = \frac{\pi}{2} + t then at t = 0, the waves would be in phase (as  cos x = sin (x + \frac{\pi}{2}) but at time t>0 the waves become out of phase due to the extra t term in the sin wave. The amount they are out of phase by is determined by the value of t, i.e. it increases as time goes on. The phase offset is therefore exactly equal to t, so it gets larger and larger as time goes on and is certainly not constant!

    If you can, make a sketch of the waves I described above for different values of t - hopefully that'll help to make it clearer.
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    Thanks

    (Original post by 2710)
    OK thanks

    Another question

    Say I have two point charges. How would I find the resultant electric potential at any point between them?

    Eg:

    -----4C--------------------------------------------------12C
    ____________________________________ _x________________
    -------------------7.3m-------------------|------2.6m------

    What would be the potential at x?

    Would it just be:

     \dfrac {4}{4\pi\epsilon_{0}\times7.3} + \dfrac {12}{4\pi\epsilon_{0}\times2.6} ?

    Like, you just add both sides to get the potential. Am I right?

    Thanks
    (Original post by Stonebridge)
    If they are both positive charges yes, just add the two potentials together. (IE. both are plus).
    Ok so I found a contradiction, I think...



    If I do:

    \dfrac{4}{kr} + \dfrac{6}{k(100-r)} = 0 , k = 4\pi\epsilon_{0}

    It gives me a negative r. But I am just following what you told me? Or is it because this question doesn't say 'resultant potential' but rather just 'potential'.

    Can you give me a general formula (if this is the case) for doing 'potential' questions like this? Ie, if I did:

    V1 = V2, then I get the right answer. But what if it said the potential was -3 or something (instead of 0) ?

    Thanks
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    There is no point between two positive charges where the potential is zero.
    (There is a point where the force (Field) is zero, though.)
    The sum of two positive values, as I said in the earlier post, will be positive.
    The only way there can be a zero potential point is if one of the charges is negative.
    I suspect there is a misprint in the question.

    In questions like this (when there is no mistake in it) your method is correct, but be careful of the units of charge.
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    (Original post by Stonebridge)
    There is no point between two positive charges where the potential is zero.
    (There is a point where the force (Field) is zero, though.)
    The sum of two positive values, as I said in the earlier post, will be positive.
    The only way there can be a zero potential point is if one of the charges is negative.
    I suspect there is a misprint in the question.

    In questions like this (when there is no mistake in it) your method is correct, but be careful of the units of charge.
    Mm OK, thanks. What do you mean when you say be careful of the units of charge? Do you mean the polarity of the charges?

    Also, so there is no difference between 'resultant potential' and just 'potential'? Because in some questions it says resultant.

    Thanks

    PS am I right in also saying that there is also no potential 0 with two negative charges?
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    (Original post by 2710)

    \dfrac{4}{kr} + \dfrac{6}{k(100-r)} = 0 , k = 4\pi\epsilon_{0}



    Thanks
    is the answer A ?
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    (Original post by rbnphlp)
    is the answer A ?
    Yes, I think if the 6muC charge was -ve, then it would have worked.
 
 
 
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