Finding the Dipole Moment of a Cylinder

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SUMMARY

The discussion focuses on calculating the dipole moment of a cylinder with a uniform volume charge density ##\rho_0##, radius ##R##, and height ##H##. The initial assumption is that the dipole moment should be zero due to symmetry, but the calculation reveals a non-zero result in the radial direction. The integral for the dipole moment, expressed as $$ P=\int_V \vec r \rho(\vec r)d\tau $$, leads to $$ P=2\pi \rho_0 (\frac {R^3 H} {3}e_\hat r + 0e_{\hat z}) $$, indicating a need for clarification on the treatment of the unit vector ##\hat{e}_r## during integration.

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precise
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Homework Statement


I am trying to calculate the dipole moment of a cylinder of volume charge density ##\rho_0## of radius ##R## and height ##H## with is center coinciding with the origin. My guess is that it should be 0 because of the symmetry but I am not able to show it. Below is my calculation attempt. Thanks for any help.

Homework Equations


$$ P=\int_V \vec r \rho(\vec r)d\tau
$$

The Attempt at a Solution


$$ P=\int_V \vec r \rho(\vec r)d\tau
= \rho_0 \int_0^R \int_0^{2\pi} \int_{-H/2}^{H/2} (r e_{\hat r} + ze_{\hat z}) rdrd\theta dz = 2\pi \rho_0 (\frac {R^3 H} {3}e_\hat r + 0e_{\hat z})
$$
 
Hello.
precise said:
$$ P=\int_V \vec r \rho(\vec r)d\tau
= \rho_0 \int_0^R \int_0^{2\pi} \int_{-H/2}^{H/2} (r e_{\hat r} + ze_{\hat z}) rdrd\theta dz = 2\pi \rho_0 (\frac {R^3 H} {3}e_\hat r + 0e_{\hat z})
$$

Looks like you are treating the unit vector ##\hat{e}_r ## as a constant vector in the integration. Does it have the same direction throughout the cylinder?

What is the meaning of ##\hat{e}_r ## in your final answer? What direction does this ##\hat{e}_r ## point?

Can you express ##\hat{e}_r ## at a point in the cylinder in terms of ##\hat{e}_x ##, ##\hat{e}_y ##, and ##\theta##? Would this help?
 

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