Understanding Capacitance in Parallel-Plate Capacitors with Dielectrics

A parallel-plate capacitor stores electrical energy in an electric field between two conductive plates separated by an insulating material. The capacitance of a capacitor is a measure of its ability to store charge at a given potential difference. It is defined as the ratio of the charge on either plate to the potential difference between the plates.

The capacitance \(C\) of a parallel-plate capacitor with vacuum or air between the plates is given by the formula:

\( C_{air} = \frac{\epsilon_0 A}{d} \)

where:

• \( \epsilon_0 \) is the permittivity of free space (a constant).
• \( A \) is the area of overlap of the two plates.
• \( d \) is the distance between the plates.

When a dielectric material is introduced between the plates, the capacitance changes. A dielectric material is an electrical insulator that can be polarized by an applied electric field. The presence of a dielectric effectively reduces the electric field strength and hence the potential difference between the plates for the same amount of charge, leading to an increase in capacitance.

The effect of a dielectric is quantified by its dielectric constant, denoted by \( \kappa \) (kappa), which is a dimensionless quantity. The dielectric constant is the ratio of the permittivity of the dielectric material (\( \epsilon \)) to the permittivity of free space (\( \epsilon_0 \)):

\( \kappa = \frac{\epsilon}{\epsilon_0} \)

When the space between the plates of a capacitor is completely filled with a dielectric material of dielectric constant \( \kappa \), the new capacitance \( C_{dielectric} \) is given by:

\( C_{dielectric} = \frac{\epsilon A}{d} = \frac{\kappa \epsilon_0 A}{d} \)

Comparing this with the capacitance with air/vacuum, \( C_{air} \), we can see the relationship:

\( C_{dielectric} = \kappa \times C_{air} \)

In this specific problem, we are given that the capacitance with air between the plates is \( C \). So, \( C_{air} = C \). We are also given that a dielectric with a dielectric constant \( \kappa = 7 \) is now filled between the plates.

Using the relationship \( C_{dielectric} = \kappa \times C_{air} \), we can find the new capacitance:

\( C_{new} = 7 \times C \)

Thus, the value of the capacitance will become 7 times the original capacitance.

Step-by-Step Calculation

Let the initial capacitance with air be \( C_{initial} \).

Given: \( C_{initial} = C \)

Let the dielectric constant of the material filled be \( \kappa \).

Given: \( \kappa = 7 \)

The new capacitance \( C_{final} \) when the space is filled with the dielectric is given by:

\( C_{final} = \kappa \times C_{initial} \)

Substituting the given values:

\( C_{final} = 7 \times C \)

So, the new capacitance is \( 7C \).

Revision Table: Key Formulas for Parallel Plate Capacitors

Additional Information: Dielectric Effect

When an external electric field is applied across a dielectric material, the molecules within the material become polarized. This polarization creates an internal electric field within the dielectric that opposes the external field. The net electric field between the capacitor plates is reduced. Since the potential difference between the plates is directly proportional to the electric field (\( V = Ed \)), a reduced electric field means a reduced potential difference for the same charge on the plates. According to the definition of capacitance \( C = Q/V \), a smaller potential difference \( V \) for the same charge \( Q \) results in a larger capacitance \( C \).

The dielectric constant \( \kappa \) indicates how much the dielectric material strengthens the electric field (and thus the capacitance) compared to vacuum. A higher dielectric constant means the material is more effective at reducing the net electric field and increasing capacitance.