An LIH dielectric sphere has a uniform polarization \(P\). Find an expression for the surface density of polarization charge \(\sigma_{p}\) at any point on the surface. Show that the 'depolarizing' field produced at the centre of the sphere by \(\sigma_{p}\) is \(P / 3 \varepsilon_{0}\).

Since the dielectric sphere has a uniform polarization P, the bound surface charge density (σₚ) can be expressed as the discontinuity of the normal component of the polarization across the surface, which can be computed as: \(\sigma_{p} = \hat{n} \cdot \vec{P}\) Here, \(\hat{n}\) is the unit normal vector pointing outward from the sphere surface, and \(\vec{P}\) is the polarization vector.

Now, we will calculate the electric field (E) produced by the bound surface charge (σₚ). To do this, we'll use Gauss's law, which states that the total electric flux through any closed surface is equal to the charge enclosed over the permittivity of free space (ε₀): \(\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_{0}}\) Since we're interested in finding the electric field at the center of the sphere, we'll choose a Gaussian pillbox with a small height that is centered on the surface of the sphere. The total electric flux through the pillbox is equal to the net charge enclosed, which is the bound surface charge σₚ times the area A of the pillbox: \(\vec{E} \cdot A = \frac{\sigma_{p} A}{\varepsilon_{0}}\) Here, A is the area of the pillbox.

With the electric field due to bound surface charge calculated, we can now solve for the electric field at the center of the sphere. From the Gaussian pillbox, the electric field is the same on both faces since they are equidistant from the center. Therefore, the electric field at the center is half the total field on the pillbox: \(E_{c} = \frac{1}{2}\vec{E}\) Combining the above two equations, we get: \(E_{c} = \frac{1}{2} \frac{\sigma_{p} A}{\varepsilon_{0}A}\) The area A cancels out, leaving us with: \(E_{c} = \frac{1}{2} \frac{\sigma_{p}}{\varepsilon_{0}}\)

The 'depolarizing' field is the field that opposes the applied electric field and is produced by the polarization of the dielectric sphere. It can be found by subtracting the applied electric field from the total electric field at the center of the sphere. In this case, since the two fields have opposite signs, we can write the 'depolarizing' field (E_dep) as: \(E_{dep} = \frac{1}{2} \frac{\sigma_{p}}{\varepsilon_{0}}\) Now, using the expression for bound surface charge density from Step 1: \(\sigma_{p} = \hat{n} \cdot \vec{P}\) Substituting this into the above equation, we get: \(E_{dep} = \frac{1}{2} \frac{(\hat{n} \cdot \vec{P})}{\varepsilon_{0}}\) Since \(\vec{P}\) is uniform throughout the sphere, \(\vec{P} = P\hat{n}\), where P is the magnitude of polarization. Thus, \(E_{dep} = \frac{1}{2} \frac{(P\hat{n} \cdot \hat{n})}{\varepsilon_{0}}\) With \(\hat{n} \cdot \hat{n} = 1\), we obtain the final expression for the 'depolarizing' field at the center of the sphere: \(E_{dep} = \frac{P}{3 \varepsilon_{0}}\) This is the desired result, proving that the 'depolarizing' field produced at the center of the sphere by σₚ is \(\frac{P}{3 \varepsilon_{0}}\).