What Is the Dipole Moment of a Grounded Sphere in a Uniform Electric Field?

[humanist rho]

Homework Statement

The electric potential of a grounded conducting sphere of radius r in a
uniform electric field $$E_{0}\hat{k}$$along the z direction is given by

$$V(r,\theta )=-E_{0}r\cos \theta +\frac{E_{0}R^{3}\cos \theta }{r^{2}}$$

where r is the distance from the centre of the sphere and θ is the
angle the radius vector makes with the z axis.

(a)what is the dipole moment acqured by the sphere?

Homework Equations

The Attempt at a Solution

Surface charge density,

$$\sigma (\theta )=-\varepsilon \frac{\partial V}{\partial r}% |_{r=R}=3\varepsilon E_{0}\cos \theta$$
Dipole moment,

$$P=\int \sigma (\theta ^{\prime })r^{\prime }ds^{\prime } =\int_{0}^{\pi }3\varepsilon E_{0}\cos \theta ^{\prime }r^{\prime }r^{\prime 2}\sin \theta ^{\prime }d\theta ^{\prime }d\phi ^{\prime } =6\pi \varepsilon E_{0}R^{3}\int_{0}^{\pi }\cos \theta ^{\prime }\sin \theta ^{\prime }d\theta ^{\prime } =6\pi \varepsilon E_{0}R^{3}\int_{-1}^{1}\cos \theta ^{\prime }d(\cos \theta ) =6\pi \varepsilon E_{0}R^{3}$$

But this don't match with the real answer. :(

 

[anathema]

Thank you for your question. Your attempt at solving for the dipole moment of the sphere is on the right track, but there are a few errors in your calculations. Let me walk you through the correct solution.

First, let's start with the correct surface charge density:

\sigma (\theta )=-\varepsilon \frac{\partial V}{\partial r}%
|_{r=R}=-3\varepsilon E_{0}\cos \theta

Note that the negative sign should be included in the expression for the surface charge density. Now, let's move on to calculating the dipole moment:

P=\int \sigma (\theta ^{\prime })r^{\prime }ds^{\prime }

=\int_{0}^{\pi }(-3\varepsilon E_{0}\cos \theta ^{\prime })R^{2}\sin \theta ^{\prime }d\theta ^{\prime }d\phi ^{\prime }

=6\pi \varepsilon E_{0}R^{3}\int_{0}^{\pi }\cos \theta ^{\prime }\sin \theta ^{\prime }d\theta ^{\prime }

=6\pi \varepsilon E_{0}R^{3}\int_{-1}^{1}\cos \theta ^{\prime }d(\cos \theta )

=6\pi \varepsilon E_{0}R^{3}

As you can see, the only mistake you made was in the calculation of the surface charge density. I hope this helps clarify things for you. Let me know if you have any further questions.