Question

A parallel plate capacitor has a capacitance of \(3 \mu \mathrm{F}\). (a) Suppose another capacitor is made using the same dielectric material and with the same cross sectional area. However, the thickness of the dielectric between the plates of the capacitor is double that of the original capacitor. What is its capacitance? (b) Suppose a third capacitor is made with the same cross sectional area and thickness as the first capacitor, but from a material with twice the permittivity. What is its capacitance?

Short answer

(a) 1.5 μF; (b) 6 μF.

Step-by-step solution

Understanding Capacitance Formula

The capacitance \( C \) of a parallel plate capacitor is given by the formula \( C = \frac{\varepsilon A}{d} \), where \( \varepsilon \) is the permittivity of the dielectric, \( A \) is the area of the plates, and \( d \) is the thickness of the dielectric.

Analyzing Part (a)

For part (a), the thickness \( d \) is doubled to \( 2d \). Using the formula, the new capacitance \( C' \) becomes \( C' = \frac{\varepsilon A}{2d} \). Since the original capacitance \( C = 3 \, \mu \mathrm{F} \), substituting into the new formula, we get \( C' = \frac{3 \, \mu \mathrm{F}}{2} = 1.5 \, \mu \mathrm{F} \).

Analyzing Part (b)

For part (b), the permittivity \( \varepsilon \) is doubled to \( 2\varepsilon \), while thickness \( d \) and area \( A \) remain the same. Thus, the new capacitance \( C'' \) is \( C'' = \frac{2\varepsilon A}{d} = 2C = 2 \times 3 \, \mu \mathrm{F} = 6 \, \mu \mathrm{F} \).

Concluding the Solution

For part (a), the capacitance is reduced because the increased thickness decreases capacitance. For part (b), the capacitance is increased due to the higher permittivity material.

Key concepts

Permittivity

Permittivity is a measure of how easily a material can permit electric field lines to pass through it. It's like thinking of a highway where cars (electric field lines) travel. Some highways (materials) allow cars to zoom through easily, while others slow them down. The property of permittivity is fundamental in dielectric materials used in capacitors.
In technical terms, permittivity is symbolized by the Greek letter \(\varepsilon\). It involves two parts: the permittivity of free space, known as \(\varepsilon_0\), and the relative permittivity or dielectric constant \(\varepsilon_r\) of the material. The formula related to permittivity is:

• \(\varepsilon = \varepsilon_0 \varepsilon_r\)

Changing permittivity directly affects capacitance, which is a key point in understanding capacitors. A higher permittivity means more electric field lines can pass through the material, resulting in greater capacitance for the capacitor.

Dielectric Material

A dielectric material is an insulating substance placed between the plates of a capacitor. Its primary job is to increase the capacitor's ability to store charge. Imagine the capacitor plates as broad hands, and the dielectric material as a glove that increases the grip, or grip equivalent to storing charge efficiency.
Dielectrics increase capacitance by reducing the electric field within the capacitor, hence allowing more charge to be stored for a given voltage. Several common dielectric materials include:

• Ceramics
• Plastics
• Glass

The effectiveness of a dielectric material is determined by its dielectric constant (\(\varepsilon_r\)), which plays a crucial role in the capacitance formula. A higher dielectric constant means that the material can enhance the capacitance of the capacitor significantly more than materials with a lower value.

Capacitance Formula

The capacitance formula is the cornerstone of understanding how capacitors control and influence electric fields. For parallel plate capacitors, the capacitance \( C \) is calculated using the formula:
\[ C = \frac{\varepsilon A}{d} \]
Here's how each component works:

• \( \varepsilon \): The permittivity of the dielectric material between the plates increases the ability to store electric charge.
• \( A \): The area of the plates, where a larger area allows more charge to be stored.
• \( d \): The distance between the plates, where a smaller distance increases the capacitance.

In different scenarios, modifying any of these factors changes the capacitance. Doubling the thickness \( d \) decreases capacitance as the electric field strength weakens. On the other hand, doubling the permittivity \( \varepsilon \) or area \( A \) increases capacitance. Being able to manipulate these elements in design allows for a wide array of practical applications in electronic circuits.