- #1
Wavefunction
- 99
- 4
Homework Statement
Thin rod of the length l is placed with one of its ends placed at the center Oof the
thin ring of radius R as shown, perpendicular to the plane of the ring. Rod is
charged with total charge Q 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 q_0. Find the electrostatic
energy of interaction between the rod and the ring.
Homework Equations
$$|d\vec{E}|=\frac{dq}{4\pi\varepsilon_0r^2}$$
$$W=\frac{\varepsilon_0}{2}\int_{\text{all space}}|\vec{E}|^2d\tau$$
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;
$$dE_x = |d\vec{E}|\cos(\theta)=|d\vec{E}|\frac{R}{\sqrt{z^2+R^2}}$$
$$E_x = \int\frac{dq}{4\pi\varepsilon_0}\frac{R}{(z^2+R^2)^\frac{3}{2}}$$
$$E_x = \frac{\alpha R}{4\pi\varepsilon_0}\int_{z=0}^{l}\frac{zdz}{(z^2+R^2)^\frac{3}{2}}$$
$$E_x = \frac{\alpha R}{4\pi\varepsilon_0}\left[-\frac{1}{z^2+R^2}\right]_{z=0}^{l}$$
$$\underline{E_x=\frac{\alpha}{4\pi\varepsilon_0}\left[1-\frac{R}{\sqrt{l^2+R^2}}\right]}$$
The z-component of the rod's field;
$$dE_z=|d\vec{E}|\cos(\theta) = |d\vec{E}|\frac{z}{\sqrt{z^2+R^2}}$$
$$E_z= \int\frac{dq}{4\pi\varepsilon_0}\frac{z}{(z^2+R^2)^\frac{3}{2}}$$
$$E_z =
\frac{\alpha}{4\pi\varepsilon_0}\int_{z=0}^{l}\frac{z^2dz}{(z^2+R^2)^\frac{3}{2}}$$
$$E_z = \frac{\alpha}{4\pi\varepsilon_0}\left[\ln(\sqrt{R^2+z^2}+z)-\frac{z}{\sqrt{R^2+z^2}}\right]_{z=0}^{l}$$
$$\underline{E_z = \frac{\alpha}{4\pi\varepsilon_0}\left[\ln(\frac{\sqrt{R^2+l^2}+l}{R})-\frac{l}{\sqrt{R^2+l^2}}\right]}$$
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.
$$dq=\lambda dL=\lambda R d\beta$$
$$\beta=\frac{\pi}{2}-\theta$$
$$d\vec{E}=dE_z\hat{z}\rightarrow \int d\vec{E} =E_z\hat{z}$$
$$dE_z = |d\vec{E}|\cos(\beta) = |d\vec{E}|\cos(\frac{\pi}{2}-\theta) = |d\vec{E}|\sin(\theta)$$
$$E_z = \int\frac{dq}{4\pi\varepsilon_0}\frac{z}{(z^2+R^2)^\frac{3}{2}}$$
$$E_z = \frac{\lambda z R}{4\pi\varepsilon_0(z^2+R^2)^{\frac{3}{2}}}\int_{\beta=0}^{2\pi}d\beta = \frac{\lambda z}{2\varepsilon_0(z^2+R^2)^{\frac{3}{2}}}$$
So
$$\boxed{\vec{E}_{\text{ring}}=\frac{\lambda z R}{2\varepsilon_0(z^2+R^2)^\frac{3}{2}}\hat{z}} $$
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.
$$\vec{E}_T=\vec{E}_{\text{rod}}+\vec{E}_{\text{ring}}$$
$$|\vec{E}_T|^2=|\vec{E}_{\text{rod}}|^2+|\vec{E}_{\text{ring}}|^2
+2\underbrace{\vec{E}_{\text{rod}}\cdot\vec{E}_{\text{ring}}}$$
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 -\infty<z<\infty and \frac{z}{(z^2+R^2)^\frac{3}{2}} is an odd function.