Intro To Dielectrics: Videos & Practice Problems

Dielectrics enhance the capacitance of capacitors, described by the equation $C=kC₀$, where C is the new capacitance, k is the dielectric constant (always > 1), and C₀ is the original capacitance. In scenarios with constant charge, inserting a dielectric decreases voltage and potential energy. Conversely, with constant voltage, charge increases, raising potential energy. Understanding these relationships is crucial for solving capacitor problems effectively.

Intro To Dielectrics

Dielectrics are materials that, when inserted into a capacitor, always increase its capacitance. The relationship between the new capacitance (C), the dielectric constant (k), and the original capacitance (C₀) is expressed by the equation:

$$C = k \cdot C₀$$

Here, the dielectric constant (k), which is a dimensionless number greater than 1, serves to weaken the electric field within the capacitor. The electric field (E) in the presence of a dielectric can be calculated using:

$$E = \frac{E₀}{k}$$

where E₀ is the electric field in a vacuum. Understanding how dielectrics affect capacitors is crucial, especially in two scenarios: when the capacitor is disconnected from a battery and when it is connected.

In the first scenario, if a dielectric is inserted into a capacitor that is disconnected from a battery, the charge (Q) on the capacitor remains constant. As the capacitance increases due to the dielectric, the voltage (V) across the capacitor must decrease to maintain the relationship:

$$Q = C \cdot V$$

Consequently, the potential energy (U) stored in the capacitor also decreases, as it is related to capacitance:

$$U = \frac{1}{2} C V^2$$

In this case, the energy density, which is dependent on the electric field, also decreases due to the weakening effect of the dielectric.

In the second scenario, when the dielectric is inserted while the capacitor is still connected to the battery, the voltage remains constant. Here, the increase in capacitance leads to an increase in charge:

$$Q = C \cdot V$$

As the capacitance increases, the charge must also increase to maintain the constant voltage. The potential energy in this case increases as well, while the energy density still decreases due to the weakened electric field.

To solve problems involving dielectrics and capacitors, it is essential to determine whether the battery is connected or disconnected. This distinction will guide the selection of the appropriate equations and relationships between charge, capacitance, and voltage.

For example, if a capacitor with an original capacitance of 3 farads has a dielectric constant of 2 inserted while connected to a 9-volt battery, the new capacitance becomes:

$$C = 2 \cdot 3 \text{ F} = 6 \text{ F}$$

Using the constant voltage relationship:

$$Q = C \cdot V = 6 \text{ F} \cdot 9 \text{ V} = 54 \text{ C}$$

This illustrates how to approach problems involving dielectrics in capacitors, emphasizing the importance of understanding the system's configuration to apply the correct principles effectively.

A capacitor in a vacuum is charged to 64V between its plates, then disconnected. Initially, each plate has 32μC. An insulating slab of dielectric glass with k = 3 is placed between the plates. a) What is the capacitor's new capacitance? b) What is the new voltage across the capacitor?

• A. a) 2.0 μFb) 192 V
• B. a) 1.50 μFb) 21.3 V
• C. a) 1.50 μFb) 64 V
• D. a) 2.0 μFb) 21.3V

Answer: (B)

A parallel plate capacitor is formed by bringing two circular plates, of radius 0.5 cm, to a distance of 2 mm apart. The capacitor is made so that it has a dielectric of constant κ between the plates. When the charge on the capacitor is 3 nC, the voltage of the capacitor is 5000 V. What is the dielectric constant?

• A. κ=0.172
• B. κ=0.580
• C. κ=1.72
• D. κ=5.80

Answer: (C)

Partial Dielectrics

In this example, we explore the effect of dielectrics on the capacitance of capacitors in two distinct configurations. Understanding how dielectrics influence capacitance is crucial for applications in electronics and circuit design.

For part (a), we consider a capacitor where the dielectric fills the top half. This configuration can be treated as two capacitors in series: one with a dielectric and one without. The capacitance of the first capacitor, denoted as \( C_1 \), is given by the formula:

\( C_1 = \frac{2 \kappa \epsilon_0 A}{d} \)

Here, \( \kappa \) represents the dielectric constant, \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates. The second capacitor, \( C_2 \), which does not contain a dielectric, is expressed as:

\( C_2 = \frac{2 \epsilon_0 A}{d} \)

To find the equivalent capacitance \( C \) of the entire system, we use the formula for capacitors in series:

\( \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \)

Substituting the expressions for \( C_1 \) and \( C_2 \), we have:

\( \frac{1}{C} = \frac{d}{2 \kappa \epsilon_0 A} + \frac{d}{2 \epsilon_0 A} \)

Finding a common denominator leads to:

\( C = \frac{2 \kappa}{1 + \kappa} \cdot \frac{\epsilon_0 A}{d} \)

This result indicates that the effective capacitance resembles that of a capacitor with an effective dielectric constant, reflecting the influence of the dielectric material.

In part (b), we analyze a different configuration where the dielectric occupies only the left half of the capacitor. In this case, the capacitors are in parallel, sharing the same voltage across their plates. The capacitance for the first capacitor with the dielectric, \( C_1 \), is calculated as:

\( C_1 = \frac{\kappa \epsilon_0 A}{2d} \)

For the second capacitor without a dielectric, \( C_2 \), the capacitance is:

\( C_2 = \frac{\epsilon_0 A}{2d} \)

Since these capacitors are in parallel, the total capacitance \( C \) is simply the sum of \( C_1 \) and \( C_2 \):

\( C = C_1 + C_2 = \frac{\kappa + 1}{2} \cdot \frac{\epsilon_0 A}{d} \)

This expression also suggests an effective dielectric constant, demonstrating how the arrangement of dielectrics can modify the overall capacitance of a capacitor system.

Here’s what students ask on this topic:

What is the role of a dielectric in a capacitor?

A dielectric in a capacitor serves to increase the capacitance of the capacitor. This is described by the equation $C=k{C}_0$, where $C$ is the new capacitance, $k$ is the dielectric constant (always greater than 1), and $C_0$ is the original capacitance. The dielectric constant weakens the electric field inside the capacitor, which allows the capacitor to store more charge for a given voltage.

How does inserting a dielectric affect the voltage and potential energy in a capacitor with constant charge?

When a dielectric is inserted into a capacitor with constant charge, the capacitance increases. Since the charge $Q$ remains the same, the voltage $V$ across the capacitor decreases, as given by $V=Q/C$. The potential energy $U$ also decreases because it is given by $U=\frac{Q^2}{/}$. The energy density, which depends on the electric field, also decreases.

What happens to the charge and potential energy when a dielectric is inserted into a capacitor with constant voltage?

When a dielectric is inserted into a capacitor with constant voltage, the capacitance increases. Since the voltage $V$ remains constant, the charge $Q$ on the capacitor increases, as given by $Q=CV$. The potential energy $U$ also increases because it is given by $U=\frac{Q^2}{/}$. The energy density decreases due to the weakening of the electric field by the dielectric.

What is the dielectric constant and how does it affect a capacitor?

The dielectric constant, denoted as $k$ or $\kappa$, is a dimensionless number that quantifies the ability of a dielectric material to increase the capacitance of a capacitor. It is always greater than 1. The capacitance of a capacitor with a dielectric is given by $C=k{C}_0$, where $C$ is the new capacitance and $C_0$ is the original capacitance. A higher dielectric constant means a greater increase in capacitance.

How do you determine which equations to use when solving problems involving dielectrics in capacitors?

To determine which equations to use when solving problems involving dielectrics in capacitors, you need to identify whether the capacitor is connected to a battery (constant voltage) or disconnected (constant charge). For constant voltage scenarios, use $Q=CV$ and $U=\frac{Q^2}{/}$. For constant charge scenarios, use $V=\frac{Q}{/}$ and $U=\frac{Q^2}{/}$. Understanding the context of the problem will guide you to the correct equations.