Question

Consider two concentric conducting spherical shells as shown in the figure. The inner shell has a radius \(a\) and carries a charge \(+Q\). The outer shell has a radius \(b\) and carries a charge \(-Q\). The empty space between them is half-filled by a hemispherical shell of a dielectric having permittivity \(\epsilon_1\). The remaining space between the shells is filled with air having the permittivity \(\epsilon_0\). 

The electric field at a radial distance \(r\) from the center and between the shells \((a < r < b)\) is

• A. \( \frac{Q}{2 \pi (\epsilon_0 + \epsilon_1)} \hat{r} \frac{1}{r^2} \) everywhere
• B. \( \frac{Q}{4 \pi \epsilon_0 r^2} \) on the air side and \( \frac{Q}{4 \pi \epsilon_1 r^2} \) on the dielectric side
• C. \( \frac{Q}{2 \pi \epsilon_0 r^2} \) on the air side and \( \frac{Q}{2 \pi \epsilon_1 r^2} \) on the dielectric side
• D. \( \frac{Q}{4 \pi (\epsilon_0 + \epsilon_1)} r^2 \) everywhere

Hint:

For a system with different materials in between the spherical shells, the effective permittivity is used to calculate the electric field across the regions.

Answer 1

The Correct Option is A

In this scenario, we are dealing with a spherical shell arrangement where the space between the shells is half-filled with a dielectric material and the other half is air. The electric field in this case can be calculated using Gauss's law, which gives the electric field due to a charge distribution in the form: \[ E = \frac{1}{4 \pi \epsilon} \frac{Q}{r^2}, \] where \(\epsilon\) is the permittivity of the medium in which the field is present, and \(Q\) is the total charge enclosed by a Gaussian surface. For the region between the shells, two materials are present: air (with permittivity \(\epsilon_0\)) and dielectric (with permittivity \(\epsilon_1\)). - The field in the air region (between the dielectric and the outer shell) is given by the formula:
\[ E_{\text{air}} = \frac{Q}{4 \pi \epsilon_0 r^2}. \] - The field in the dielectric region (between the inner shell and the dielectric) is given by the formula:
\[ E_{\text{dielectric}} = \frac{Q}{4 \pi \epsilon_1 r^2}. \] However, because the electric field is uniform throughout the space between the shells, we need to combine the effects of both the dielectric and the air using their effective permittivities. The total permittivity will be \(\epsilon_0 + \epsilon_1\), which accounts for the different regions. Thus, the electric field is: \[ E = \frac{Q}{2 \pi (\epsilon_0 + \epsilon_1)} \hat{r} \frac{1}{r^2}, \] which matches option (A).