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Electrostatic interaction energy between a charge rod and ring.

  1. Sep 14, 2014 #1
    1. The problem statement, all variables and given/known data

    Thin rod of the length [itex]l[/itex] is placed with one of its ends placed at the center [itex]O[/itex]of the
    thin ring of radius [itex]R[/itex] as shown, perpendicular to the plane of the ring. Rod is
    charged with total charge [itex]Q[/itex] that is distributed along the rod’s length with the
    linear charge density linearly proportional to the distance from the bottom end of
    the rod. Ring is charged uniformly with total charge [itex]q_0[/itex]. Find the electrostatic
    energy of interaction between the rod and the ring.

    2. Relevant equations


    $$W=\frac{\varepsilon_0}{2}\int_{\text{all space}}|\vec{E}|^2d\tau$$

    3. The attempt at a solution

    Okay so first the rod's electric field: $$\lambda=\alpha z$$,$$dq=\lambda dz$$

    $$d\vec{E}=dE_x\hat{x}+dE_z\hat{z}\rightarrow \int d\vec{E} = E_x\hat{x}+E_z\hat{z}$$

    The x-component of the rod's field;

    $$dE_x = |d\vec{E}|\cos(\theta)=|d\vec{E}|\frac{R}{\sqrt{z^2+R^2}}$$

    $$E_x = \int\frac{dq}{4\pi\varepsilon_0}\frac{R}{(z^2+R^2)^\frac{3}{2}}$$

    $$E_x = \frac{\alpha R}{4\pi\varepsilon_0}\int_{z=0}^{l}\frac{zdz}{(z^2+R^2)^\frac{3}{2}}$$

    $$E_x = \frac{\alpha R}{4\pi\varepsilon_0}\left[-\frac{1}{z^2+R^2}\right]_{z=0}^{l}$$


    The z-component of the rod's field;

    $$dE_z=|d\vec{E}|\cos(\theta) = |d\vec{E}|\frac{z}{\sqrt{z^2+R^2}}$$

    $$E_z= \int\frac{dq}{4\pi\varepsilon_0}\frac{z}{(z^2+R^2)^\frac{3}{2}}$$

    $$E_z =

    $$E_z = \frac{\alpha}{4\pi\varepsilon_0}\left[\ln(\sqrt{R^2+z^2}+z)-\frac{z}{\sqrt{R^2+z^2}}\right]_{z=0}^{l}$$

    $$\underline{E_z = \frac{\alpha}{4\pi\varepsilon_0}\left[\ln(\frac{\sqrt{R^2+l^2}+l}{R})-\frac{l}{\sqrt{R^2+l^2}}\right]}$$

    So the rod's total field is then:

    $$\boxed{\vec{E}_{\text{rod}}=\frac{\alpha}{4\pi\varepsilon_0}\left[1-\frac{R}{\sqrt{l^2+R^2}}\right]\hat{x}+ \frac{\alpha}{4\pi\varepsilon_0}\left[\ln(\frac{\sqrt{R^2+l^2}+l}{R})-\frac{l}{\sqrt{R^2+l^2}}\right]\hat{z}}$$

    Now for the ring's field:

    by symmetry I know the ring will only have a "z" component. Also in this case the linear charge density is uniform.

    $$dq=\lambda dL=\lambda R d\beta$$


    $$d\vec{E}=dE_z\hat{z}\rightarrow \int d\vec{E} =E_z\hat{z}$$

    $$dE_z = |d\vec{E}|\cos(\beta) = |d\vec{E}|\cos(\frac{\pi}{2}-\theta) = |d\vec{E}|\sin(\theta)$$

    $$E_z = \int\frac{dq}{4\pi\varepsilon_0}\frac{z}{(z^2+R^2)^\frac{3}{2}}$$

    $$E_z = \frac{\lambda z R}{4\pi\varepsilon_0(z^2+R^2)^{\frac{3}{2}}}\int_{\beta=0}^{2\pi}d\beta = \frac{\lambda z}{2\varepsilon_0(z^2+R^2)^{\frac{3}{2}}}$$


    $$\boxed{\vec{E}_{\text{ring}}=\frac{\lambda z R}{2\varepsilon_0(z^2+R^2)^\frac{3}{2}}\hat{z}} $$

    Okay so now assuming I haven't made any grave errors in calculating my electric fields I'll add them together and square the resulting field.



    Now I think the term that I have underbraced corresponds to the interaction of the electric fields of the two charge distributions so now in order to find the interaction energy of the two objects I need to calculate the following:

    $$W_{\text{interaction}}=\varepsilon_0\int_{\text{all space}}\vec{E}_{\text{rod}}\cdot\vec{E}_{\text{ring}}d\tau$$

    $$\vec{E}_{\text{rod}}\cdot\vec{E}_{\text{ring}} = \left[\frac{\alpha}{4\pi\varepsilon_0}\left[1-\frac{R}{\sqrt{l^2+R^2}}\right]\hat{x}+ \frac{\alpha}{4\pi\varepsilon_0}\left[\ln(\frac{\sqrt{R^2+l^2}+l}{R})-\frac{l}{\sqrt{R^2+l^2}}\right]\hat{z}\right]\cdot\left[0\hat{x}+\frac{\lambda z R}{2\varepsilon_0(z^2+R^2)^\frac{3}{2}}\hat{z}\right]$$

    $$\varepsilon_0\int_{\text{all space}}\frac{\alpha}{4\pi\varepsilon_0}\left[\ln(\frac{\sqrt{R^2+l^2}+l}{R})-\frac{l}{\sqrt{R^2+l^2}}\right]\frac{\lambda z R}{2\varepsilon_0(z^2+R^2)^\frac{3}{2}}d\tau$$

    So this is where I'm stuck and I'm not even completely sure the field for my rod is correct. If I integrate over an infinite cylinder I get answer of 0 since [itex]-\infty<z<\infty[/itex] and [itex]\frac{z}{(z^2+R^2)^\frac{3}{2}}[/itex] is an odd function.
  2. jcsd
  3. Sep 14, 2014 #2
    Actually I think I figured out my problem: I need to calculate the electric fields for all space and not just where the individual objects are. Hopefully that will resolve my issue.
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