Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A solid conducting sphere has net positive charge and radius R = 0.400 m. At a point 1.20 m from the center of the sphere, the electric potential due to the charge on the sphere is 24.0 V. Assume that V = 0 at an infinite distance from the sphere. What is the electric potential at the center of the sphere?

Short Answer

Expert verified

The electric potential at the center of the sphere is 72.0V.

Step by step solution

01

Potential Energy

The equation gives the potential energy due to the sphere;

V=14ππ0qr

Here, r denotes the distance between the sphere and the point where the potential is measured; it can be inside or outside the sphere.

02

Electric potential at the center of the sphere

The potential is inversely proportional to the distance, as shown by the equation;

V1r

The potential at the sphere's center is determined by the radius R, where the potential is constant throughout the sphere. Because the charge q is constant and the term 14π0is constant, a relationship between the states inside the sphere and the states outside the sphere can be established.

role="math" localid="1664260877266" V1V2=r2r1VinsideVouside=rRVinside=rRVouside

Putting the values;

Vinside=1.2m0.40m24.0V

Hence, the electric potential at the center of the sphere is 72.0V.

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 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


An electron at pointAinFig. E27.15has a speedv0of1.41×106m/sFind (a) the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path fromAtoB, and (b) the time required for the electron to move fromAtoB.

An electron moves at 1.40×106m/sthrough a regionin which there is a magnetic field of unspecified direction and magnitude 7.40×10-2T. (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle
between the electron velocity and the magnetic field?

A 12.4-µF capacitor is connected through a 0.895-MΩ resistor to a constant potential difference of 60.0 V. (a) Compute the charge on the capacitor at the following times after the connections are made: 0, 5.0 s, 10.0 s, 20.0 s, and 100.0 s. (b) Compute the charging currents at the same instants. (c) Graph the results of parts (a) and (b) for t between 0 and 20 s

In the circuit shown in Fig. E26.41, both capacitors are initially charged to 45.0 V. (a) How long after closing the switch S will the potential across each capacitor be reduced to 10.0 V, and (b) what will be the current at that time?

An electrical conductor designed to carry large currents has a circular cross section 2.50 mm in diameter and is 14.0 m long. The resistance between its ends is 0.104Ω. (a) What is the resistivity of the material? (b) If the electric-field magnitude in the conductor is 1.28 V/m, what is the total current? (c) If the material has 8.5×1028free electrons per cubic meter, find the average drift speed under the conditions of part (b).

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.

Sign-up for free