/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 The parallel plates in a capacit... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The parallel plates in a capacitor, with a plate area of \(8.50 \mathrm{~cm}^{2}\) and an air-filled separation of \(3.00 \mathrm{~mm}\), are charged by a \(6.00 \mathrm{~V}\) battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of \(8.00 \mathrm{~mm}\). Neglecting fringing, find (a) the potential difference between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates.

Short Answer

Expert verified
(a) 16.01 V, (b) 4.509e-11 J, (c) 1.20e-10 J, (d) 7.49e-11 J.

Step by step solution

01

Calculate Initial Capacitance

Using the formula for capacitance of parallel plates,\[ C = \frac{\varepsilon_0 A}{d} \] where \( \varepsilon_0 \) is the permittivity of free space \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \), \( A = 8.50 \times 10^{-4} \text{ m}^2 \), and \( d = 3 \times 10^{-3} \text{ m} \). Substituting the values, we get the initial capacitance \( C_i = \frac{8.85 \times 10^{-12} \times 8.50 \times 10^{-4}}{3 \times 10^{-3}} = 2.505 \times 10^{-12} \text{ F} \).
02

Calculate Initial Stored Energy

The stored energy in a capacitor is given by \[ U = \frac{1}{2} C V^2 \]. Substitute \( C_i = 2.505 \times 10^{-12} \text{ F} \) and \( V = 6 \text{ V} \):\[ U_i = \frac{1}{2} \times 2.505 \times 10^{-12} \times 6^2 = 4.509 \times 10^{-11} \text{ J} \].
03

Find Initial Charge on Plates

The charge on the plates is given by \( Q = CV \). Thus, \( Q = 2.505 \times 10^{-12} \times 6 = 15.03 \times 10^{-12} \text{ C} \).
04

Calculate Final Capacitance

After separation, the new distance is \( d' = 8 \times 10^{-3} \text{ m} \). The final capacitance \( C_f \) is calculated using \[ C_f = \frac{\varepsilon_0 A}{d'} \]. Thus, \( C_f = \frac{8.85 \times 10^{-12} \times 8.50 \times 10^{-4}}{8 \times 10^{-3}} = 9.39 \times 10^{-13} \text{ F} \).
05

Calculate Final Potential Difference

Since charge is conserved, use \( V_f = \frac{Q}{C_f} \). Substituting the charge \( Q = 15.03 \times 10^{-12} \text{ C} \) and \( C_f = 9.39 \times 10^{-13} \text{ F} \),\[ V_f = \frac{15.03 \times 10^{-12}}{9.39 \times 10^{-13}} \approx 16.01 \text{ V} \].
06

Calculate Final Stored Energy

Use \( U_f = \frac{1}{2} C_f V_f^2 \). With \( C_f = 9.39 \times 10^{-13} \) and \( V_f = 16.01 \),\[ U_f = \frac{1}{2} \times 9.39 \times 10^{-13} \times (16.01)^2 = 1.20 \times 10^{-10} \text{ J} \].
07

Compute Work Required to Separate Plates

The work done is the change in energy: \( W = U_f - U_i \). Substitute \( U_f = 1.20 \times 10^{-10} \) and \( U_i = 4.509 \times 10^{-11} \):\[ W = 1.20 \times 10^{-10} - 4.509 \times 10^{-11} = 7.49 \times 10^{-11} \text{ J} \].

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 91Ó°ÊÓ!

Key Concepts

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

Potential Difference
The concept of potential difference is all about the voltage between two points, in this case, the two plates of a capacitor. Initially, when the plates are connected to a 6.00 V battery, the potential difference across them is simply 6.00 V.
But what happens when we disconnect the battery and change the spacing of the plates? The potential difference isn't fixed to 6 V anymore because it's not being held constant by the battery. Instead, it depends on the charge already present and the new plate spacing.

To find the final potential difference after altering the separation, use the formula:
  • Initial charge on the plates: \( Q = C_i \times V \)
  • Final potential difference: \( V_f = \frac{Q}{C_f} \)
Given that the initial capacitance and initial potential difference were known, and assuming the charge remains constant, the new potential difference was recalculated to be approximately 16.01 V once the plates were pulled further apart.
Stored Energy
Stored energy in a capacitor is the energy kept within for future use. Initially, when connected to the battery, the energy is stored as electrostatic potential energy. It is determined by the initial capacitance and the initial voltage. Calculated using:
  • Initial stored energy: \( U_i = \frac{1}{2} C_i V^2 \)
The initial energy was found to be \( 4.509 \times 10^{-11} \) J.
After changing the separation, the initial energy configuration changes because the capacitance changes while the charge stays the same.
To find the final stored energy:
  • Final energy: \( U_f = \frac{1}{2} C_f V_f^2 \)
The new capacitance and altered voltage result in a final stored energy of \( 1.20 \times 10^{-10} \) J, reflecting the system's increased potential energy due to the increased separation.
Work Done
Work done relates to the energy required to move changes within the capacitor. When the plates are pulled apart, energy is required to physically separate charged plates against the attractive electrostatic force.
The work done is calculated by measuring the difference in energy before and after the plates are separated. Simply put, you compare the initial energy and the final energy:
  • Work done: \( W = U_f - U_i \)
In this exercise, after finding both the initial and final stored energies, the work calculated required to separate the plates was \( 7.49 \times 10^{-11} \) J.
This value gives insight into the energy balance and changes occurring when the field configuration is modified by varying plate distances. It's a crucial calculation showing how energy is conserved and transformed in such a system.

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

You have two flat metal plates, each of area \(1.00 \mathrm{~m}^{2}\), with which to construct a parallel-plate capacitor. (a) If the capacitance of the device is to be \(1.00 \mathrm{~F}\), what must be the separation between the plates? (b) Could this capacitor actually be constructed?

A certain parallel-plate capacitor is filled with a dielectric for which \(\kappa=5.5 .\) The area of each plate is \(0.034 \mathrm{~m}^{2}\), and the plates are separated by \(2.0 \mathrm{~mm}\). The capacitor will fail (short out and burn up) if the electric field between the plates exceeds \(200 \mathrm{kN} / \mathrm{C}\). What is the maximum energy that can be stored in the capacitor?

A \(2.0 \mu\) F capacitor and a \(4.0 \mu \mathrm{F}\) capacitor are connected in parallel across a \(300 \mathrm{~V}\) potential difference. Calculate the total energy stored in the capacitors.

A parallel-plate capacitor has plates of area \(0.12 \mathrm{~m}^{2}\) and a separation of \(1.2 \mathrm{~cm} .\) A battery charges the plates to a potential difference of \(120 \mathrm{~V}\) and is then disconnected. A dielectric slab of thickness \(4.0 \mathrm{~mm}\) and dielectric constant \(4.8\) is then placed symmetrically between the plates. (a) What is the capacitance before the slab is inserted? (b) What is the capacitance with the slab in place? What is the free charge \(q\) (c) before and (d) after the slab is inserted? What is the magnitude of the electric field (e) in the space between the plates and dielectric and (f) in the dielectric itself? (g) With the slab in place, what is the potential difference across the plates? (h) How much external work is involved in inserting the slab?

A potential difference of \(300 \mathrm{~V}\) is applied to a series connection of two capacitors of capacitances \(C_{1}=2.00 \mu \mathrm{F}\) and \(C_{2}=8.00 \mu \mathrm{F}\). What are (a) charge \(q_{1}\) and (b) potential difference \(V_{1}\) on capacitor 1 and (c) \(q_{2}\) and (d) \(V_{2}\) on capacitor \(2 ?\) The charged capacitors are then disconnected from each other and from the battery. Then the capacitors are reconnected with plates of the same signs wired together (the battery is not used). What now are (e) \(q_{1}\), (f) \(V_{1},\left(\right.\) g) \(q_{2}\), and (h) \(V_{2}\) ? Suppose, instead, the capacitors charged in part (a) are reconnected with plates of opposite signs wired together. What now are (i) \(q_{1},(\mathrm{j}) V_{1},(\mathrm{k}) q_{2}\), and (1) \(V_{2}\) ?

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.