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    Only me!

    Okay so I'm finding some of this electric potential business a little harder than I'd anticipated. If anyone could help on a couple of things that'd be great


    1. In a vacuum tube diode, the electric potential between the electrodes is given by

    V(x) = Cx^(4/3)

    The distance between the cathode and anode is 13mm, the potential difference between the electrodes is 240V, and the constant C has the value 7.85 x 10^4 Vm^-4/3

    Obtain a formula for the electric field between the electrodes as a function of x.

    So what I did was to say

    - E x = C x^(4/3)
    E = - C x^(1/3)
    E = - (7.85 x 10^4 Vm^4/3) x^(1/3)

    But the answer is - 1.05 x10^5 Vm^(4/3) x^(1/3) :confused:

    I can only assume I wasn't allowed to integrate dx straight to x from the integral of 'Edx'


    2. A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b. The positive charge per unit length on the inner cylinder is H and there is an equal negative charge per unit lenth on the outer cylinder.

    a) Calculate the potential V(r) for
    i) r < a
    ii) a < r < b
    Take V=0 at r=b.

    b) Show that the potential of the inner cylinder with respect to the outer is

     Vab = ( H / 2 pi £o) ln (b/a)

    c) Show that the electric field at any point between the cylinders has magnitude E(r) = Vab / r ln(b/a)

    d) What is the potential difference between the two cylinders if the outer cylinder has no net charge?

    I'm just lacking a few smarts on what I should be putting in as the limits because what my brain is doing is different to what the book is doing.


    3. Two protons are separated by 1.07 x 10^-10 m, with an electron at the mid point between them. The electron has a veolicty of magnitude 1.5 x 10^6 m/s in a direction along the perpendicular bisector of the line connecting the two protons. How far from the point midway between the two protons can the electron move? Because the masses of the protons are much greater than the electron mass, the motions of the protons are very low and can be ignored.
    Note: A realistic description of the electron motion requires the use of quantum mechanics, not Newtonian mechanics.

    Okay so conceptually this is fine, I'm just not sure which equations to use?


    4. An electron with an initial speed of 6.5 x 10^6 m/s is projected along the axis midway between the deflection plates of a cathode ray tube. The uniform electric field between the plates has a magnitude of 1.1 x 10^3 V/m upward.

    The plates are 2cm apart and 6cm long.

    Right, I've done most of this question:
    I've calculated the force on the electron (1.76 x 10^-16 N downwards)
    And the acceleration of it (1.93 x 10^14 m/s^2 downwards)
    And how far below the axis the electron has moved when it reaches the end of the plates (8.24mm)

    But when I try and calculate the angle with the axis that it's moving as it leaves the plates, I'm using

    \theta = arctan (0.00824/0.06) = - 7.82 degrees

    But the answer is apparently 15.4 degrees, no idea what I've done wrong!



    And lastly,

    5. A disk with radius R has uniform surface charge density 'sigma'. By regarding the disk as a series of thin concentric rings, calculate the electric potential V at a point on the disk's axis a distance x from the centre of the disk. Assume that the potential is zero at infinity.

    I have no idea how to start this really.


    Thanks very much for any help #kiss#
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    1. What's the relationship between V and E?
    2.Mostly Gauss, which bit are you stuck on?
    3. Use an energy arguement- you know the potential due to a point charge right?
    4. You want the angle the velocity vector makes with the horizontal (it looks like you've using the position instead).
    5.Can you find the E field a distance x (on axis) from a thin ring?
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    (Original post by plmokn)
    1. What's the relationship between V and E?
    2.Mostly Gauss, which bit are you stuck on?
    3. Use an energy arguement- you know the potential due to a point charge right?
    4. You want the angle the velocity vector makes with the horizontal (it looks like you've using the position instead).
    5.Can you find the E field a distance x (on axis) from a thin ring?
    Thanks for the reply!

    1. dV = Edx, but I tried that
    2. I'm stuck on the limits. And I need to find the sum of the potentials from the inner and outer tubes at that point, but the problem with Gauss is I can't use it to find the potential due to the outer tube?
    3. Yes, I've worked out the potential at the position of the electron, I was thinking about trying to relate this to kinetic energy but I'm not sure what good that would do really...hehe
    4. Silly me, of course - got it, thanks!
    5. Yes, but like question 2, I'm unsure on the limits and how to consider the outer rings
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    :vroam:
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    (Original post by Artemidoros)
    :vroam:
    Hehe, at least you dont' have exam on it on thursday morning! :bawling:
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    1. So E=-dV/dx and you've got V as a function of x and the numerical value of C...
    2.Whats the contribution to the E field for r<b from the outer tube (use Gauss)?
    Spoiler:
    Show
    V(r<b)= V_from outer + V_from inner=INT{infin,b,-E_outer} + INT{infin,r,-E_inner}=INT{b,r,-E_inner} in hopefully understandable notation

    3.Think of how the change in potential energy comes about
    Spoiler:
    Show
    increase in V= loss of kinetic energy

    5.Once you've got the E field from one ring you can find V from one ring by integrating between infinity and x, and then integrate the contributions of V from each radius ring between 0 and R. (Or you could do it in the opposite order if it makes the integrals nicer I haven't done it).
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    (Original post by plmokn)
    1. So E=-dV/dx and you've got V as a function of x and the numerical value of C...
    2.Whats the contribution to the E field for r<b from the outer tube (use Gauss)?
    Spoiler:
    Show
    V(r<b)= V_from outer + V_from inner=INT{infin,b,-E_outer} + INT{infin,r,-E_inner}=INT{b,r,-E_inner} in hopefully understandable notation

    3.Think of how the change in potential energy comes about
    Spoiler:
    Show
    increase in V= loss of kinetic energy

    5.Once you've got the E field from one ring you can find V from one ring by integrating between infinity and x, and then integrate the contributions of V from each radius ring between 0 and R. (Or you could do it in the opposite order if it makes the integrals nicer I haven't done it).
    1. Woops, got it, forgot the factor of 4/3!
    2. Hmm, okay, I think I get this...does that mean that the potential everywhere inside the solid inner cylinder is the same? Also for proving part c), I can do it but only if I start from V = - INT(Edr) = - INT [H/(2 pi £o r^2)]dr . It works, I just don't know if I'm meant to derive that equation for E first
    3. Yes, I understand that, but surely you need to be thinking about deceleration if the electron is slowing down and coming to a stop/coming back towards the midpoint?
    5. Okay... so from one ring, where the surface charge density is @,
     V = k INT (dQ/[\sqrt(x^2+R^2)] = k INT [(@ 2 pi x dx)/(\sqrt(x^2+R^2)] = (@/2£o)[\sqrt(x^2+R^2]
    Which limits was I meant to put in here? Because none of the sensible ones give me the correct answer from this, which is very similar to the last line above:

     V = (@/2£o)[(\sqrt(x^2+R^2)) - x]

    Thanks again
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    2. Yes, there's never an E field inside a conductor in a statics problem (otherwise charges would move until the E field no longer existed) so V is constant. The equation for E comes from Gauss's law. Since you're interested in the field inside the outer coil you can completly forget about the outer coil. Then consider a cylindrical Gaussian surface height h:
    2\pi r h E= Hh/\epsilon and E comes out (by the way you've got a typo in your post it's 1/r not 1/r^2).

    3. I don't understand what you're getting at: you want the furthest distance the electron gets away. By using an energy arguement you don't have to worry about accelerations and things.

    4. I don't follow what you've done. But here's how I'd do it:
    E_{ring,radiusR}=kQcos\theta/(x^2+R^2)

=kQx/(x^2+R^2)^{3/2}

=k2\pi R \sigma x dR/(x^2+R^2)^{3/2}
    Then integrate dR from 0 to R (ignoring my sloppyness in not calling the integration variable something different). The integrate -Edx from infinity to x to find the potential.
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    (Original post by plmokn)
    2. Yes, there's never an E field inside a conductor in a statics problem (otherwise charges would move until the E field no longer existed) so V is constant. The equation for E comes from Gauss's law. Since you're interested in the field inside the outer coil you can completly forget about the outer coil. Then consider a cylindrical Gaussian surface height h:
    2\pi r h E= Hh/\epsilon and E comes out (by the way you've got a typo in your post it's 1/r not 1/r^2).

    3. I don't understand what you're getting at: you want the furthest distance the electron gets away. By using an energy arguement you don't have to worry about accelerations and things.

    4. I don't follow what you've done. But here's how I'd do it:
    E_{ring,radiusR}=kQcos\theta/(x^2+R^2)

=kQx/(x^2+R^2)^{3/2}

=k2\pi R \sigma x dR/(x^2+R^2)^{3/2}
    Then integrate dR from 0 to R (ignoring my sloppyness in not calling the integration variable something different). The integrate -Edx from infinity to x to find the potential.
    2. Heeey I think I got it!!! Wooooo, thanks!

    3. Oh, it's just that if it wasn't decelerating I don't understand what would stop it moving?

    4. Got it, thanks again!
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    3. It is decelerating, but by introducing a potential, V, we don't have to include this deceleration explicitly. It's just like finding how high a ball thrown upwards will go by setting 0.5 m v^2=mgh except the potential is a more complicated function.
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    But it says you need to use quantum theory not newtonian mechanics?

    I think I'm going to have to give up on this one!
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    I read the question as meaning "find the answer assuming Newtonian mechanics but remember it's not going to be exact", I'm not sure now though.
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    (Original post by plmokn)
    I read the question as meaning "find the answer assuming Newtonian mechanics but remember it's not going to be exact", I'm not sure now though.
    Ahh, actually that makes more sense. It doesn't really make a difference though, I still can't do it

    I think electric potentials have the potential to kill me!
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    Oh hey hey hey, I take it back - I worked it out
 
 
 
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