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Fundamentals Of Physics
Found in: Page 714
Fundamentals Of Physics

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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Short Answer

Sphere 1 with radius has positive charge . Sphere 2 with radius is far from sphere 1 and initially uncharged. After the separated spheres are connected with a wire thin enough to retain only negligible charge, (a) is potential of sphere 1 greater than, less than, or equal to potential of sphere 2? What fraction of ends up on (b) sphere 1 and (c) sphere 2? (d) What is the ratio of the surface charge densities of the spheres?

  1. The potential of sphere 1 and 2 is .
  2. The charge on the sphere 1 in the fraction of is .
  3. The charge on the sphere 2 in the fraction of is .
  4. The ratio of the surface charge densities of the spheres is .
See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given data:

The radius of sphere 1 s .

The radius of sphere 2 is .

Step 2: Understanding the concept:

The electric potential due to the point charge is,

Here, is the coulomb’s constant, is the charge, and is the distance from the charge.

The potential of sphere 1 is,

….. (1)

Here, is the charge on sphere 1.

The potential of sphere 2 is,

….. (2)

Here, is the charge on sphere 2.

The total electric charge is,

….. (3)

Step 3: (a) Find out if potential V1 of sphere 1 greater than, less than, or equal to potential V2 of sphere 2:

The charges must be the same for both spheres when both spheres are connected by a thin wire. The electric potential is directly proportional to the charge. Hence, the electric potential on both spheres must be the same.

V1=V2

Here, V1 is the electric potential on sphere 1 and V2 is the electric potential on sphere 2. Hence, the electric potential of sphere 1 is equal to the electric potential on sphere 2.

Step 4: (b) Calculate what fraction of q ends up on sphere 1:

The relation between the radius of the sphere 1 and sphere 2 is,

The relation between the potential of the sphere 1 and sphere 2 is,

….. (4)

Substitute for in the equation (3) and solve for .

Therefore, the charge on the sphere 1 in the fraction of is .

Step 5: (c) Calculate what fraction of q ends up on sphere 2:

Substitute for in the equation (4) and solve for .

Hence, the charge on the sphere 2 in the fraction of is .

Step 6: (d) Calculate the ratio   of the surface charge densities of the spheres:

The surface charge density of sphere 1 is,

The surface charge density of sphere 2 is,

Take the ratio of the surface charge densities of sphere 1 and 2.

Substitute for and for in the above equation.

Hence, the ratio of the surface charge densities of the spheres is .

Most popular questions for Physics Textbooks

What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.15 m whose potential is 200 V (with V = 0 at infinity)?

A nonconducting sphere has radius R = 2.31 cm and uniformly distributed charge q = +3.50 fC. Take the electric potential at the sphere’s center to be V = 0 . What is V at radial distance (a) r = 1.45 cm and (b) r = R . (Hint: See Module 23-6.)

A metal sphere of radius 15 cm has a net charge of 3.0×10-8 C . (a) What is the electric field at the sphere’s surface? (b) If V = 0 at infinity, what is the electric potential at the sphere’s surface? (c) At what distance from the sphere’s surface has the electric potential decreased by 500 V ?

(a) what is the electric potential energy of two electrons separated by 2.00 nm? (b) If the separation increases, does the potential energy increase or decrease?

Question: An electron is placed in an x-y plane where the electric potential depends on x and y as shown, for the coordinate axes, in Fig. 24-51 (the potential does not depend on z). The scale of the vertical axis is set by Vs =500 V. In unit-vector notation, what is the electric force on the electron?
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