A potential of 12 V is applied across a parallel-plate capacitor which is constructed of square plates of side 10 cm and separation of 4.0 mm.A dielectric slab of constant $\kappa$ = 2.5 is inserted between the plates so that it completely fills the space.What is the bound surface charge density on the dielectric? A)12.3 pC/m² B)15.9 pC/m² C)39.8 pC/m² D)56.8 pC/m² E)85.4 pC/m²

Question

Quizplus · Accepted Answer

The bound surface charge density ($\sigma_b$) on the dielectric can be found using the formula $\sigma_b = P = \epsilon_0 \chi_e E$, where $P$ is the polarization density, $\epsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \, \text{F/m}$), $\chi_e$ is the electric susceptibility of the dielectric, and $E$ is the electric field in the dielectric. The electric susceptibility $\chi_e$ is related to the dielectric constant $\kappa$ by $\chi_e = \kappa - 1$. Given $\kappa = 2.5$, we find $\chi_e = 1.5$. The electric field $E$ can be found from $E = V/d$, where $V = 12 \, \text{V}$ and $d = 4.0 \, \text{mm} = 4.0 \times 10^{-3} \, \text{m}$. Thus, $E = 12 \, \text{V} / 4.0 \times 10^{-3} \, \text{m} = 3000 \, \text{V/m}$. Substituting these values into the formula for $\sigma_b$, we get $\sigma_b = 8.85 \times 10^{-12} \, \text{F/m} \times 1.5 \times 3000 \, \text{V/m} = 39.8 \times 10^{-12} \, \text{C/m}^2 = 39.8 \, \text{pC/m}^2$.

A Potential of 12 V Is Applied Across a Parallel-Plate κ\kappaκ

A Potential of 12 V Is Applied Across a Parallel-Plate $\kappa$