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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

Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for 1675 s with an average light power output of 2.70 × 105 W. (a) If the conversion of electrical energy to light is 95% efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 V when the stored energy equals the value calculated in part (a). What is the capacitance?

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 air capacitor is to store charge of magnitude 240.0 pC on each plate when the potential difference between the plates is 42.0 V. (a) If the area of each plate is 6.80 cm2, what is the separation between the plates? (b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude 240.0 pC on each plate?

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 budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminum foil as plates, with a few sheets of paper between them as a dielectric. The paper has a dielectric constant of 3.0, and the thickness of one sheet of it is 0.20 mm. (a) If the sheets of paper measure 22 ×28 cm and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance? (b) Suppose for convenience she wants to use a single sheet of posterboard, with the same dielectric constant but a thickness of 12.0 mm, instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 nF of capacitance? (c) Suppose she goes high-tech and finds a sheet of Teflon of the same thickness as the posterboard to use as a dielectric. Will she need a larger or smaller area of Teflon than of posterboard? Explain.

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