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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

Expert verified
(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

01

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.
02

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.
03

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.
04

Calculate Dielectric Constant

The relation \( C = K C_0 \) gives \( K = \frac{C}{C_0} = \frac{22.5}{12.5} \), simplifying to \( K = 1.8 \).
05

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.
06

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A parallel-plate capacitor has plates with area 0.0225 m\(^2\) separated by 1.00 mm of Teflon. (a) Calculate the charge on the plates when they are charged to a potential difference of 12.0 V. (b) Use Gauss's law (Eq. 24.23) to calculate the electric field inside the Teflon. (c) Use Gauss's law to calculate the electric field if the voltage source is disconnected and the Teflon is removed.

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?

A 5.80-\(\mu\)F, parallel-plate, air capacitor has a plate separation of 5.00 mm and is charged to a potential difference of 400 V. Calculate the energy density in the region between the plates, in units of J/m\(^3\).

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is 10.0 pC. The inner cylinder has radius 0.50 mm, the outer one has radius 5.00 mm, and the length of each cylinder is 18.0 cm. (a) What is the capacitance? (b) What applied potential difference is necessary to produce these charges on the cylinders?

Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is \(\pm\) 0.50 \(\times\) 10\(^{-3}\) C/m\(^2\), the cell wall is 5.0 nm thick, and the cell- wall material is air. (a) Find the magnitude of \(\vec{E}\) S in the wall between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) A typical cell in the human body has a volume of 10\(^{-16}\) m\(^3\). Estimate the total electric-field energy stored in the wall of a cell of this size. (\(Hint\): Assume that the cell is spherical, and calculate the volume of the cell wall.) (d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of 5.4. Repeat parts (a) and (b) in this case.

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