/*! 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} Q56P Two oppositely charged, identica... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two oppositely charged, identical insulating spheres, each 50.0 cm in diameter and carrying a uniformly distributed charge of magnitude 250μC, are placed 1.00 m apart center to center (Fig. P23.56). (a) If a voltmeter is connected between the nearest points (a and b) on their surfaces, what will it read? (b) Which point, a or b, is at the higher potential? How can you know this without any calculations?

Short Answer

Expert verified

a) Potential difference is 12×106V

b) Point (a) has higher potential.

Step by step solution

01

Step 1:

Given data:

R=D/2=0.25m

Q=250μ°ä

Distance between center to (br) is .075

So, the potentialVadue to both charges at (a)

localid="1664273791009" Va=kQR-kqr=kQ1R-1r=9×109×250×10-6×12.5-10.75=6.0×106V

Now the potential Vbis equal to the negative potential is

Va=-kQ1R-1r=-6.0×106V

So, the total potential difference by the voltmeter is

Vab=Va-Vb=6.0×106-(-6.0×106)=12.0×106V

Therefore, the potential difference is 12.0×106V

02

Step 2:

As the potential (a) is higher than the (b) as the charge is positive.

So electric field is in the direction of displacement, so the integration will produce a positive voltage.

Therefore, the point (a) has greater potential.

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

Question: You connect a battery, resistor, and capacitor as in Fig. 26.20a, where R = 12.0 Ω and C = 5.00 x 10-6 F. The switch S is closed at t = 0. When the current in the circuit has a magnitude of 3.00 A, the charge on the capacitor is 40.0 x 10-6 C. (a) What is the emf of the battery? (b) At what time t after the switch is closed is the charge on the capacitor equal to 40.0 x 10-6 C? (c) When the current has magnitude 3.00 A, at what rate is energy being (i) stored in the capacitor, (ii) supplied by the battery

In the circuit shown in Fig. E26.47 each capacitor initially has a charge of magnitude 3.50 nC on its plates. After the switch S is closed, what will be the current in the circuit at the instant that the capacitors have lost 80.0% of their initial stored energy?

If you peel two strips of transparent tape off the same roll and immediately let them hang near each other, they will repel each other. If you then stick the sticky side of one to the shiny side of the other and rip them apart, they will attract each other. Give a plausible explanation, involving transfer of electrons between the strips of tape, for this sequence of events.

You have a negatively charged object. How can you use it to place a net negative charge on an insulated metal sphere? To place a net positive charge on the sphere?

Question: Three negative point charges lie along a line as shown in Fig. Find the magnitude and direction of the electric field this combination of charges produces at point P, which lies 6.00 cm from the -2.00 mC charge measured perpendicular to the line connecting the three charges

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.