A solid uniform dielectric sphere of radius $R = 5$ cm and relative permittivity $K = 3$ is placed i

Question

A solid uniform dielectric sphere of radius $R = 5$ cm and relative permittivity $K = 3$ is placed in an external electric field such that a uniform electric field of magnitude $E_{in} = 10$ V/m is established inside it. The maximum surface density of bound (polarization) charges on the sphere is expressed as $\alpha \epsilon_0$ (in SI units), and the total positive bound charge on the sphere's surface is expressed as $\beta \pi \epsilon_0 R^2$. Find the value of $(\alpha + \beta)$.

Professor PC Thomas Classes · Answer

The polarization vector $\vec{P}$ inside a linear isotropic dielectric is uniform and relates to the internal electric field by $P = \epsilon_0(K - 1)E_{in}$. Substituting the given values, $P = \epsilon_0(3 - 1)(10) = 20\epsilon_0$. The surface bound charge density is given by $\sigma_b = \vec{P} \cdot \hat{n} = P \cos	heta$. The maximum surface charge density occurs at the pole where $	heta = 0$, so $\sigma_{max} = P = 20\epsilon_0$. Thus, $\alpha = 20$. The total positive bound charge is found by integrating $\sigma_b$ over the hemisphere where it is positive ($0 \le 	heta \le \pi/2$): $q_{bound} = \int_0^{\pi/2} (P \cos	heta) (2\pi R^2 \sin	heta) d	heta = P \pi R^2 \int_0^{\pi/2} \sin(2	heta) d	heta = P \pi R^2 = 20\epsilon_0 \pi R^2$. Thus, $\beta = 20$. Therefore, the sum is $\alpha + \beta = 20 + 20 = 40$.