Dielectric Slab Effect on Capacitor Electric Field

TITLE: Dielectric Slab Effect on Capacitor Electric Field DESCRIPTION: Understand how inserting a dielectric slab reduces the electric field between capacitor plates due to polarization and bound charges.

Concept Overview

This question explores the fundamental effect of introducing a dielectric material into the electric field between the plates of a parallel-plate capacitor. It tests the understanding of dielectric polarization, the creation of bound charges, and how these phenomena lead to a reduction in the net electric field and, consequently, an increase in capacitance. The core idea is that the dielectric material opposes the external electric field.

Step 1: Consider a parallel-plate capacitor with charge $+Q$ on one plate and $-Q$ on the other, separated by a distance $d$. In vacuum, this creates a uniform electric field $E_0$ between the plates.

$E_0 = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}$

Here, $\sigma = Q/A$ is the surface charge density on the plates, and $\epsilon_0$ is the permittivity of free space. This field is directed from the positive plate to the negative plate.

Step 2: When a dielectric slab of relative permittivity (dielectric constant) $\kappa$ is inserted between the plates, the external electric field $E_0$ polarizes the dielectric material. The molecules within the dielectric align themselves such that their positive ends point towards the negative plate and their negative ends point towards the positive plate.

Step 3: This alignment of molecular dipoles results in the accumulation of bound charges on the surfaces of the dielectric slab. A bound positive charge $+Q_b$ appears on the surface facing the negative plate, and a bound negative charge $-Q_b$ appears on the surface facing the positive plate.

$Q_b = Q \left(1 - \frac{1}{\kappa}\right)$

These bound charges create an internal electric field $E_b$ within the dielectric, which opposes the external field $E_0$.

Step 4: The electric field $E_b$ due to the bound charges is given by:

$E_b = \frac{\sigma_b}{\epsilon_0} = \frac{Q_b}{A\epsilon_0}$

where $\sigma_b = Q_b/A$ is the surface density of bound charges.

Step 5: The net electric field $E$ inside the dielectric is the vector difference between the original field $E_0$ and the induced field $E_b$:

$E = E_0 - E_b$

Substituting the expressions for $E_0$ and $E_b$:

$E = \frac{\sigma}{\epsilon_0} - \frac{\sigma_b}{\epsilon_0} = \frac{1}{\epsilon_0} (\sigma - \sigma_b)$

Step 6: It is a known property of dielectrics that the net electric field inside the dielectric is reduced by a factor of $\kappa$ compared to the field in vacuum:

$E = \frac{E_0}{\kappa}$

This relationship can also be derived by substituting $\sigma_b = \sigma (1 - 1/\kappa)$ into the equation from Step 5:

$E = \frac{1}{\epsilon_0} \left(\sigma - \sigma \left(1 - \frac{1}{\kappa}\right)\right) = \frac{1}{\epsilon_0} \left(\sigma - \sigma + \frac{\sigma}{\kappa}\right) = \frac{\sigma}{\kappa\epsilon_0} = \frac{1}{\kappa} \left(\frac{\sigma}{\epsilon_0}\right) = \frac{E_0}{\kappa}$

Since $\kappa > 1$ for all dielectric materials, the net electric field $E$ is always less than $E_0$. This reduction in electric field is the direct consequence of the polarization of the dielectric and the creation of opposing bound charges.

Key Takeaways:

• Dielectric materials polarize when placed in an external electric field, aligning their molecular dipoles.
• This polarization leads to the formation of bound charges on the dielectric surfaces, creating an internal electric field that opposes the external field.
• The net electric field between the capacitor plates is reduced by a factor equal to the dielectric constant ($\kappa$) of the material.

Answer: The insertion of a dielectric slab reduces the electric field between capacitor plates because the dielectric material polarizes, creating bound charges that generate an internal electric field opposing the original field. The net field becomes $E = E_0/\kappa$, where $\kappa > 1$.