Question

A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates. (a) A voltmeter reads 45.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.5 V. What is the dielectric constant of this material? (b) What will the voltmeter read if the dielectric is now pulled partway out so it fills only one-third of the space between the plates?

Short answer

(a) Dielectric constant is approximately 3.91. (b) Voltmeter reads about 22.83 V.

Step-by-step solution

Understand the concept

A parallel-plate capacitor stores charge, and its voltage depends on the capacitance. The dielectric constant changes the capacitance when inserted between the plates. This affects the voltage across the capacitor.

Initial Conditions

Initially, the voltage across the capacitor is 45.0 V with no dielectric present.

Effect of Inserting Dielectric

When a dielectric is inserted, the new voltage becomes 11.5 V. The relationship involving the original voltage ($V_0$), the new voltage ($V$), and the dielectric constant ($k$) is:$$V=\frac{V_0}{k}$$.

Calculate Dielectric Constant

Rearranging the equation to solve for $k$:$$k=\frac{V_0}{V}=\frac{45.0}{11.5}$$Calculate $k$ to find the dielectric constant.

Solve for Dielectric Constant

Calculate $k=\frac{45.0}{11.5}=3.913$. Thus, the dielectric constant $k$ is approximately 3.91.

Partial Removal of Dielectric

If the dielectric fills one-third of the space, the capacitance is a combination of the dielectric-filled and air-filled sections. The equation becomes:$$V_{new}=\frac{3C\cdot V_0}{k+2}$$, where $C$ is the capacitance without dielectric, $V_0$ is the original voltage, and $k$ is the dielectric constant.

Calculate New Voltage

Substitute $k=3.913$ into the equation:$$V_{new}=\frac{3\cdot 45.0}{3.913+2}=\frac{135}{5.913}$$.Calculate $V_{new}$ to find the new voltage after partial removal of the dielectric.

Solve for New Voltage

Calculate $V_{new}=\frac{135}{5.913}\approx 22.83$. Thus, the voltmeter will read approximately 22.83 V.

Key concepts

Parallel-Plate Capacitor

A parallel-plate capacitor is an essential component in electronics that stores electrical energy. It's made of two conductive plates placed close to each other, separated by an insulating material or vacuum. These plates accumulate opposite charges when connected to a power source, like a battery. The separation between the plates prevents the current from directly flowing between them, thus enabling the capacitor to store energy. The ability to store this energy makes capacitors vital for various applications, such as filtering signals in electronic devices. The parallel-plate capacitor is also fundamental in understanding electric fields and potential differences. When connected to a voltage source, an electric field develops between the plates. The amount of charge stored and the voltage across the capacitor can be influenced by the properties of the insulator placed between the plates.

Capacitance

Capacitance is a measure of a capacitor's ability to store charge per unit voltage. This property depends not only on the physical characteristics of the capacitor itself, such as the surface area of the plates and the distance between them, but also on the properties of any material (or dielectric) placed between them.When you introduce a dielectric material between the plates of a capacitor, it enhances the capacitance. This is because the dielectric reduces the electric field within the capacitor, allowing it to store more charge for the same applied voltage. This relationship between the dielectric and capacitance is expressed by the equation:

• $C={\epsilon }_0\cdot {\epsilon }_r\cdot \frac{A}{d}$

Where $C$ is the capacitance, ${\epsilon }_0$ is the permittivity of free space, ${\epsilon }_r$ is the relative permittivity or dielectric constant, $A$ is the area of the plates, and $d$ is the distance between them.Understanding capacitance is crucial for designing and applying capacitors in various technologies, ranging from stabilizing power supplies to tuning radios.

Voltage

Voltage is the measure of electrical potential difference between two points in a circuit. In the context of a capacitor, it refers to the potential difference across the plates that allows the storage and discharge of electric energy. When it comes to a parallel-plate capacitor, the voltage is influenced by many factors, including the amount of charge stored and the presence of a dielectric. Initially, without any dielectric, the capacitor is charged to a certain voltage level. Upon inserting a dielectric, the voltage across the capacitor decreases, as the material's dielectric constant reduces the electric field, and thus the potential difference for a given amount of charge. The importance of understanding how a dielectric affects voltage lies in its applications. It allows engineers and scientists to design capacitors that can efficiently handle various voltages and store higher amounts of energy. Moreover, voltage readings from a voltmeter give insight into the effectiveness of the dielectric material used, aiding in selecting the right material for specific applications.