/*! 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} Q58P (a) Calculate the potential ener... [FREE SOLUTION] | 91影视

91影视

(a) Calculate the potential energy of a system of two small spheres, one carrying a charge of 2.00C and the other a charge of -3.50C, with their centers separated by a distance of 0.180 m. Assume that U = 0 when the charges are infinitely separated. (b) Suppose that one sphere is held in place; the other sphere, with a mass of 1.50 g, is shot away from it. What minimum initial speed would the moving sphere need to escape completely from the attraction of the fixed sphere? (To escape, the moving sphere would have to reach a velocity of zero when it is infinitely far from the fixed sphere.)

Short Answer

Expert verified

a) The potential energy is U=-0.252]

b) Initial speed is Va=18.3m/s

Step by step solution

01

Step-by-Step Solution

Given data:

Q1=-3.50cQ2=2.0Cd=0.25mm=1.5g

02

Step 2:

So, the potential between two-point charges is

U=kQ1Qd=91092.010-6-3.510-60.25=-0.252J

Hence, the potential energy is.

Using conservation law:

Ka+Ua=Kb+UbKa=-Ua=0.252J

Here Kbis zero and (v) is also zero because the potential is inversely proportional to (r) therefore Kbis

Ka=12mv2=0.252

So,

va=20.251.510-3=18.3mls

Therefore, the initial speed is18.3mls.

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 91影视!

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) What is the potential difference Vadin the circuit of Fig. P25.62? (b) What is the terminal voltage of the 4.00-Vbattery? (c) A battery with emf and internal resistance 0.50is inserted in the circuit at d, with its negative terminal connected to the negative terminal of the 8.00-Vbattery. What is the difference of potential Vbcbetween the terminals of the 4.00-Vbattery now?

An electron at point in figure has a speed v0=1.41106m/s. Find (a) the magnetic field that will cause the electron to follow the semicircular path from to and (b) The time required for the electron to move fromAtoB.

In the circuit, in Fig. E26.47 the capacitors are initially uncharged, the battery has no internal resistance, and the ammeter is idealized. Find the ammeter reading (a) just after the switch S is closed and (b) after S has been closed for a very long time.

In the circuit shown in Fig. E26.49, C = 5.90 mF, 詯 = 28.0 V, and the emf has negligible resistance. Initially, the capacitor is uncharged and the switch S is in position 1. The switch is then moved to position 2 so that the capacitor begins to charge. (a) What will be the charge on the capacitor a long time after S is moved to position 2? (b) After S has been in position 2 for 3.00 ms, the charge on the capacitor is measured to be 110 mC What is the value of the resistance R? (c) How long after S is moved to position 2 will the charge on the capacitor be equal to 99.0% of the final value found in part (a)?

Why does an electric light bulb nearly always burn out just as you turn on the light, almost never while the light is shining?

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.