Question

A parallel-plate capacitor has capacitance \(C\) = 12.5 pF when the volume between the plates is filled with air. The plates are circular, with radius 3.00 cm. The capacitor is connected to a battery, and a charge of magnitude 25.0 pC goes onto each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted, the charge on each plate has magnitude 45.0 pC. (a) What is the dielectric constant \(K\) of the dielectric? (b) What is the potential difference between the plates before and after the dielectric has been inserted? (c) What is the electric field at a point midway between the plates before and after the dielectric has been inserted?

Short answer

(a) \( K = 1.8 \); (b) \( V = 2.0 \) V for both; (c) \( E_0 = 2.68 \times 10^{4} \) V/m, \( E = 1.49 \times 10^{4} \) V/m after dielectric.

Step-by-step solution

Determine Initial Charge and Capacitance

Initially, the charge magnitude is given as \( Q_0 = 25.0 \) pC and the capacitance is \( C_0 = 12.5 \) pF with air as the dielectric. The relationship governing this system is \( Q_0 = C_0 V_0 \) where \( V_0 \) is the initial potential difference.

Calculate Initial Potential Difference

Using \( Q_0 = C_0 V_0 \), rearrange to find \( V_0 = \frac{Q_0}{C_0} = \frac{25.0 \times 10^{-12}}{12.5 \times 10^{-12}} \) Volts. Simplifying gives \( V_0 = 2.0 \) V.

Determine Capacitance with Dielectric

With the dielectric inserted, the charge on each plate becomes \( Q = 45.0 \) pC. Since \( V \) remains constant due to the battery connection, we use \( C = \frac{Q}{V} = \frac{45.0 \times 10^{-12}}{2.0} \) Farads, resulting in \( C = 22.5 \) pF.

Calculate Dielectric Constant

The relation \( C = K C_0 \) gives \( K = \frac{C}{C_0} = \frac{22.5}{12.5} \), simplifying to \( K = 1.8 \).

Determine Electric Field Before Dielectric

The electric field \( E_0 \) is calculated using \( E_0 = \frac{V_0}{d} \). Relative distances between the plates are needed, with \( d = \frac{A}{C_0 \varepsilon_0} \). Given \( A = \pi (0.03)^2 \) and \( \varepsilon_0 = 8.85 \times 10^{-12} \) F/m, calculate \( d \) and \( E_0 = \frac{2.0}{d} \). Simplification gives \( E_0 \approx 2.68 \times 10^{4} \) V/m.

Calculate Electric Field After Dielectric

The electric field after the dielectric is inserted is \( E = \frac{E_0}{K} = \frac{2.68 \times 10^{4}}{1.8} \) V/m, resulting in \( E \approx 1.49 \times 10^{4} \) V/m.

Key concepts

Capacitance Calculation

Capacitance is a core concept in the study of capacitors, which are devices designed to store electric charge. In the case of a parallel-plate capacitor, capacitance depends on both the geometry of the plates as well as the properties of the material between them, known as the dielectric. The basic formula for capacitance is given by:\[ C = \varepsilon_0 \frac{A}{d} \]where:

• \( C \) is the capacitance
• \( \varepsilon_0 \) is the permittivity of free space
• \( A \) is the area of one of the plates
• \( d \) is the separation between the plates

Dielectric materials affect capacitance by introducing a dielectric constant, \( K \), which modifies the basic capacitance to \( C = K C_0 \). This allows capacitors to hold more charge for the same potential difference.Inserting a dielectric increases the capacitance due to the increased ability of the dielectric to polarize under an electric field, thereby allowing more charge storage at the same voltage. In our exercise, with the introduction of the dielectric, the capacitance increased from 12.5 pF to 22.5 pF, with the dielectric constant \( K \) calculated to be 1.8.

Electric Field

The electric field (\( E \)) in parallel-plate capacitors can be affected significantly by the introduction of dielectric materials. It is critical for understanding how potential differences make their way through the electric field as well as the effect of charges surrounding it. In essence, the electric field in a capacitor without any dielectric (\( E_0 \)) is established by:\[ E_0 = \frac{V_0}{d} \]where:

• \( V_0 \) is the potential difference
• \( d \) is the distance between the plates

When a dielectric is added, the electric field inside the capacitor decreases as the dielectric constant (\( K \)) works to reduce the electric field. After inserting the dielectric, the relationship becomes:\[ E = \frac{E_0}{K} \]Reducing the electric field magnitude implies that for the same voltage, the dielectric material lessens the field strength between the plates.Considering the example in the exercise, before the dielectric, the electric field was around 2.68 x 10^4 V/m. After inserting the dielectric, the electric field reduced to approximately 1.49 x 10^4 V/m. This illustrates how the dielectric material modifies the internal electric interactions of the capacitor.

Potential Difference

Potential difference, often referred to as voltage, plays a key role in the functioning of capacitors. It decides how effectively a capacitor can store electric energy. The potential difference (\( V \)) across a capacitor's plates is pivotal for determining capacitance when the charge is known. Initial calculations are made using the formula:\[ V_0 = \frac{Q_0}{C_0} \]where:

• \( Q_0 \) is the initial charge
• \( C_0 \) is the initial capacitance

After inserting a dielectric, although the charge on the capacitor increases, the voltage across the plates remains constant if the capacitor remains connected to a battery. This is why the initial analysis calculated the voltage as 2.0 V, which doesn't change even as the capacitance changes due to the dielectric.In essence, the potential difference provides a driving force that pushes charges onto the plates. However, in systems with dielectrics, while the charge magnitude changes (increasing from 25.0 pC to 45.0 pC in the example), the voltage remains steady because of the constant voltage supply from the connected battery.