Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A parallel-plate capacitor has capacitance C0 = 8.00 pF when there is air between the plates. The separation between the plates is 1.50 mm. (a) What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 × 104 V/m? (b) A dielectric with K=2.70 is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00 × 104 V/m?

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 μC. (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference if the plate separation is doubled? (c) How much work is required to double the separation?

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 parallel-plate air capacitor of capacitance 245 pF has a charge of magnitude 0.148 μC on each plate. The plates are 0.328 mm apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the electricfield magnitude between the plates? (d) What is the surface charge density on each plate?

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?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free