Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 105

A point charge \(q=-2.5 \mathrm{nC}\) is initially at rest adjacent to the negative plate of a capacitor. The charge per unit area on the plates is $4.0 \mu \mathrm{C} / \mathrm{m}^{2}\( and the space between the plates is \)6.0 \mathrm{mm} .$ (a) What is the potential difference between the plates? (b) What is the kinetic energy of the point charge just before it hits the positive plate, assuming no other forces act on it?

Short Answer

Expert verified
Question: Calculate the potential difference between the plates and the kinetic energy of the point charge just before it hits the positive plate. Answer: To calculate the potential difference between the plates, use the formula: \(V = E \cdot d\) where \(E\) is the electric field calculated in step 1 and \(d\) is the distance between the plates (given as \(6.0 \mathrm{mm}\)). To calculate the kinetic energy of the point charge just before it hits the positive plate, apply energy conservation principle. The initial potential energy (\(U_i\)) will be equal to the final kinetic energy (\(K_f\)). Use the expression: \(K_f = (-2.5 × 10^{-9} \mathrm{C}) V\)

Step by step solution

01

Calculate the electric field between the plates

Given that the charge per unit area on the plates is \(4.0 \mu \mathrm{C} / \mathrm{m}^{2}\), we can calculate the electric field between the plates of the capacitor. The electric field (\(E\)) in a parallel-plate capacitor is related to the charge density (\(\sigma\)) as: \(E = \dfrac{\sigma}{\epsilon_0}\) where \(\epsilon_0 = 8.85 × 10^{-12} \mathrm{C}^2/(\mathrm{N} \cdot \mathrm{m}^2)\) is the vacuum permittivity. By substituting the given values, we have: \(E = \dfrac{4.0 × 10^{-6} \mathrm{C/m}^2}{8.85 × 10^{-12} \mathrm{C}^2/(\mathrm{N} \cdot \mathrm{m}^2)}\)
02

Calculate the potential difference between the plates

Using the electric field calculated in step 1, we can now calculate the potential difference (\(V\)) between the plates using the formula: \(V = E \cdot d\) where \(d\) is the distance between the plates. We are told that the distance is \(6.0 \mathrm{mm}\). Thus, we have: \(V = E \cdot 6.0 × 10^{-3} \mathrm{m}\)
03

Compute the initial potential energy of the point charge

We will now determine the initial potential energy of the point charge (\(U_i\)) placed adjacent to the negative plate using its charge (\(q\)) and the potential difference between the plates (\(V\)). This is calculated as: \(U_i = qV\) We are given the point charge \(q = -2.5 \mathrm{nC}\). Substitute the values into the equation: \(U_i = (-2.5 × 10^{-9} \mathrm{C}) V\)
04

Find the kinetic energy of the point charge

We are told that there are no other forces acting on the point charge, so we can use the energy conservation principle. The total initial energy (\(U_i\)) will be equal to the final kinetic energy (\(K_f\)) of the point charge just before it hits the positive plate. Thus: \(K_f = U_i\) Now, we can substitute the values and expressions from the previous steps to solve for \(K_f\). Then, the kinetic energy of the point charge just before it hits the positive plate is: \(K_f = (-2.5 × 10^{-9} \mathrm{C}) V\) Now, you can use these steps to calculate the potential difference between the plates and the kinetic energy of the point charge just before it hits the positive plate.

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.

Start your free trial

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

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

The capacitor of Problem 79 is initially charged to a \(150-\mathrm{V}\) potential difference. The plates are then physically separated by another \(0.750 \mathrm{mm}\) in such a way that none of the charge can leak off the plates. Find (a) the new capacitance and (b) the new energy stored in the capacitor. Explain the result using conservation of energy.
(a) Calculate the capacitance per unit length of an axon of radius $5.0 \mu \mathrm{m} \text { (see Fig. } 17.14) .$ The membrane acts as an insulator between the conducting fluids inside and outside the neuron. The membrane is 6.0 nm thick and has a dielectric constant of \(7.0 .\) (Note: The membrane is thin compared with the radius of the axon, so the axon can be treated as a parallel plate capacitor.) (b) In its resting state (no signal being transmitted), the potential of the fluid inside is about \(85 \mathrm{mV}\) lower than the outside. Therefore, there must be small net charges \(\pm Q\) on either side of the membrane. Which side has positive charge? What is the magnitude of the charge density on the surfaces of the membrane?
An electron is accelerated from rest through a potential difference $\Delta V\(. If the electron reaches a speed of \)7.26 \times 10^{6} \mathrm{m} / \mathrm{s},$ what is the potential difference? Be sure to include the correct sign. (Does the electron move through an increase or a decrease in potential?)
If a capacitor has a capacitance of \(10.2 \mu \mathrm{F}\) and we wish to lower the potential difference across the plates by \(60.0 \mathrm{V},\) what magnitude of charge will we have to remove from each plate?
A parallel plate capacitor is connected to a \(12-\mathrm{V}\) battery. While the battery remains connected, the plates are pushed together so the spacing is decreased. What is the effect on (a) the potential difference between the plates? (b) the electric field between the plates? (c) the magnitude of charge on the plates?
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Astrophysics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Electricity

Read Explanation

Classical Mechanics

Read Explanation
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