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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) E0=2.68×104 V/m, E=1.49×104 V/m after dielectric.

Step by step solution

01

Determine Initial Charge and Capacitance

Initially, the charge magnitude is given as Q0=25.0 pC and the capacitance is C0=12.5 pF with air as the dielectric. The relationship governing this system is Q0=C0V0 where V0 is the initial potential difference.
02

Calculate Initial Potential Difference

Using Q0=C0V0, rearrange to find V0=Q0C0=25.0×101212.5×1012 Volts. Simplifying gives V0=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=QV=45.0×10122.0 Farads, resulting in C=22.5 pF.
04

Calculate Dielectric Constant

The relation C=KC0 gives K=CC0=22.512.5, simplifying to K=1.8.
05

Determine Electric Field Before Dielectric

The electric field E0 is calculated using E0=V0d. Relative distances between the plates are needed, with d=AC0ε0. Given A=π(0.03)2 and ε0=8.85×1012 F/m, calculate d and E0=2.0d. Simplification gives E02.68×104 V/m.
06

Calculate Electric Field After Dielectric

The electric field after the dielectric is inserted is E=E0K=2.68×1041.8 V/m, resulting in E1.49×104 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=ε0Adwhere:
  • C is the capacitance
  • ε0 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=KC0. 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 (E0) is established by:E0=V0dwhere:
  • V0 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=E0KReducing 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:V0=Q0C0where:
  • Q0 is the initial charge
  • C0 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

In one type of computer keyboard, each key holds a small metal plate that serves as one plate of a parallel-plate, airfilled capacitor. When the key is depressed, the plate separation decreases and the capacitance increases. Electronic circuitry detects the change in capacitance and thus detects that the key has been pressed. In one particular keyboard, the area of each metal plate is 42.0 mm2, and the separation between the plates is 0.700 mm before the key is depressed. (a) Calculate the capacitance before the key is depressed. (b) If the circuitry can detect a change in capacitance of 0.250 pF, how far must the key be depressed before the circuitry detects its depression?

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of 1.60 × 107 V>m. The capacitor is to have a capacitance of 1.25 × 109 F and must be able to withstand a maximum potential difference of 5500 V. What is the minimum area the plates of the capacitor may have?

When a 360-nF air capacitor (11 nF = 109 F) is connected to a power supply, the energy stored in the capacitor is 1.85 × 105 J. While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by 2.32 × 105 J. (a) What is the potential difference between the capacitor plates? (b) What is the dielectric constant of the slab?

A 20.0-μF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to those of an uncharged 10.0-μF capacitor. Compute (a) the original charge of the system, (b) the final potential difference across each capacitor, (c) the final energy of the system, and (d) the decrease in energy when the capacitors are connected.

A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.15 mm, what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?

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