/*! 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 40 Point \(P\) is at a potential of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Point \(P\) is at a potential of \(500.0 \mathrm{kV}\) and point \(S\) is at a potential of \(200.0 \mathrm{kV} .\) The space between these points is evacuated. When a charge of \(+2 e\) moves from \(P\) to \(S\), by how much does its kinetic energy change?

Short Answer

Expert verified
Answer: The change in kinetic energy of the charged particle is \(9.6\times10^{-14}\,\text{J}\).

Step by step solution

01

Understand the given parameters

We are given the following parameters: - Point P's potential: \(V_P = 500.0\,\text{kV}\) - Point S's potential: \(V_S = 200.0\,\text{kV}\) - Charge moving from P to S: \(q = +2e\) The electric potential difference between the two points can be found by subtracting the potential at Point S from the potential at Point P, which gives us \(\Delta V = V_P - V_S\).
02

Calculate the electric potential difference

We can now calculate the electric potential difference: \(\Delta V = V_P - V_S = (500.0 - 200.0)\,\text{kV} = 300.0\,\text{kV} = 3 \times 10^5\,\text{V}\)
03

Calculate the potential energy change

The change in the potential energy is given by the formula: \(\Delta U = q\Delta V\) Now, substitute the values of the charge (\(q = +2e = +2 \times 1.6\times10^{-19}\,\text{C}\)) and the electric potential difference (\(\Delta V = 3 \times 10^5\,\text{V}\)) to find the potential energy change: \(\Delta U = (+2\times1.6\times10^{-19}\,\text{C})(3\times10^5\,\text{V}) = 9.6\times10^{-14}\,\text{J}\) Since the charge is positive and moving from higher potential to a lower potential, it loses potential energy.
04

Apply the work-energy principle

The work-energy principle states that the change in kinetic energy is equal to the work done on a particle. In this case, the work done on the particle is equal to the change in potential energy: \(\Delta K = \Delta U\)
05

Find the change in kinetic energy

Using the result of the potential energy change: \(\Delta K = 9.6\times10^{-14}\,\text{J}\) The kinetic energy of the particle increases by \(9.6\times10^{-14}\,\text{J}\) when it moves from point P to point S.

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!

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 large parallel plate capacitor has plate separation of \(1.00 \mathrm{cm}\) and plate area of \(314 \mathrm{cm}^{2} .\) The capacitor is connected across a voltage of \(20.0 \mathrm{V}\) and has air between the plates. How much work is done on the capacitor as the plate separation is increased to $2.00 \mathrm{cm} ?$

Draw some electric field lines and a few equipotential surfaces outside a positively charged metal cube. [Hint: What shape are the equipotential surfaces close to the cube? What shape are they far away?]

A cell membrane has a surface area of \(1.0 \times 10^{-7} \mathrm{m}^{2},\) a dielectric constant of \(5.2,\) and a thickness of \(7.5 \mathrm{nm}\) The membrane acts like the dielectric in a parallel plate capacitor; a layer of positive ions on the outer surface and a layer of negative ions on the inner surface act as the capacitor plates. The potential difference between the "plates" is \(90.0 \mathrm{mV}\). (a) How much energy is stored in this capacitor? (b) How many positive ions are there on the outside of the membrane? Assume that all the ions are singly charged (charge +e).
The plates of a 15.0 - \(\mu \mathrm{F}\) capacitor have net charges of $+0.75 \mu \mathrm{C}\( and \)-0.75 \mu \mathrm{C},$ respectively. (a) What is the potential difference between the plates? (b) Which plate is at the higher potential?
A certain capacitor stores \(450 \mathrm{J}\) of energy when it holds $8.0 \times 10^{-2} \mathrm{C}$ of charge. What is (a) the capacitance of this capacitor and (b) the potential difference across the plates?
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.