Log In Start studying!

Select your language

Suggested languages for you:
Deutsch (DE)
Deutsch (UK)
Deutsch (US)

Americas

Europe

Answers without the blur. Sign up and see all textbooks for free! → Illustration

16-42P

Expert-verified
Physics Principles with Applications
Found in: Page 443
Physics Principles with Applications

Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922

Answers without the blur.

Just sign up for free and you're in.

Register for Free I’ll do it later
Illustration

Short Answer

(II) The field just outside a 3.50-cm-radius metal ball is \({\bf{E = 3}}{\bf{.75 \times 1}}{{\bf{0}}^{\bf{2}}}\;{\bf{N/C}}\) and points toward the ball. What charge resides on the ball?

The charge resides on the metal ball is \( - 5.11 \times {10^{ - 11}}\;{\rm{C}}\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Understanding the electric field

The value of electric field (E) due to a charge Q at any point is determined by finding the electric force (F) per unit charge acting on a small positive test charge (q) placed at that point.

The expression for electric field is given as:

\(E = \frac{F}{q} = k\frac{Q}{{{r^2}}}\)

Here, k is the electrostatic force constant whose value is \(9.0 \times {10^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}\).

Given information:

The electric field outside the metal ball is, \(E = 3.75 \times {10^2}\;{\rm{N/C}}\)

The radius of the metal ball is, \(r = 3.50\;{\rm{cm}} = 3.50 \times 1{{\rm{0}}^{ - 2}}\;{\rm{m}}\)

Determination of the charge that resides on the ball

If E is the electric field due to charge placed inside the metal ball at its center, then the magnitude of electric field due to the charge Q placed at a distance r from it is:

\(E = k\frac{Q}{{{r^2}}}\)

So, the charge inside the metal ball is given as:

\(Q = \frac{{E{r^2}}}{k}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}Q = \frac{{\left( {3.75 \times {{10}^2}\;{\rm{N/C}}} \right) \times {{\left( {3.50 \times 1{{\rm{0}}^{ - 2}}\;{\rm{m}}} \right)}^2}}}{{\left( {9.0 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)}}\\ = 5.11 \times {10^{ - 11}}\;{\rm{C}}\end{aligned}\)

Since any charge in a metallic conductor resides on the surface of the conductor, the magnitude of charge that resides on the metal ball is \(5.11 \times {10^{ - 11}}\;{\rm{C}}\). Also, as the electric field points towards the metal ball, the charge on the metal ball must be negative.

Thus, the charge that resides on the surface of the metal ball is \( - 5.11 \times {10^{ - 11}}\;{\rm{C}}\).

Most popular questions for Physics Textbooks

(II) At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q (Fig. 16–54). Determine the magnitude and direction of the force on the charge 2Q.

Assume that the two opposite charges in Fig. 16–32a are 12.0 cm apart. Consider the magnitude of the electric field 2.5 cm from the positive charge. On which side of this charge—top, bottom, left, or right—is the electric field the strongest? The weakest? Explain.

(III) A point charge Q rests at the center of an uncharged thin spherical conducting shell. (See Fig. 16–34.) What is the electric field E as a function of r (a) for r less than the inner radius of the shell, (b) inside the shell, and (c) beyond the shell? (d) How does the shell affect the field due to Q alone? How does the charge Q affect the shell?

A point charge of mass 0.185 kg, and net charge \( + {\bf{0}}{\bf{.340 }}\mu {\bf{C}}\) hangs at rest at the end of an insulating cord above a large sheet of charge. The horizontal sheet of fixed uniform charge creates a uniform vertical electric field in the vicinity of the point charge. The tension in the cord is measured to be 5.18 N. Calculate the magnitude and direction of the electric field due to the sheet of charge (Fig. 16–67).

FIGURE 16–67 Problem 61.

(I) A proton is released in a uniform electric field, and it experiences an electric force of \({\bf{1}}{\bf{.86}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{14}}}}{\bf{ N}}\) toward the south. Find the magnitude and direction of the electric field.
Icon

Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Physics Textbooks

Work Energy and Power

Read explanation

Radiation

Read explanation

Astrophysics

Read explanation

Fluids

Read explanation

Physics of Motion

Read explanation

Famous Physicists

Read explanation

Torque and Rotational Motion

Read explanation

Fields in Physics

Read explanation

Scientific Method Physics

Read explanation

Magnetism

Read explanation

Particle Model of Matter

Read explanation

Physical Quantities and Units

Read explanation

Medical Physics

Read explanation

Further Mechanics and Thermal Physics

Read explanation

Energy Physics

Read explanation

Force

Read explanation

Waves Physics

Read explanation

Modern Physics

Read explanation

Atoms and Radioactivity

Read explanation

Dynamics

Read explanation

Electricity and Magnetism

Read explanation

Fundamentals of Physics

Read explanation

Kinematics Physics

Read explanation

Linear Momentum

Read explanation

Geometrical and Physical Optics

Read explanation

Rotational Dynamics

Read explanation

Electrostatics

Read explanation

Space Physics

Read explanation

Magnetism and Electromagnetic Induction

Read explanation

Electromagnetism

Read explanation

Conservation of Energy and Momentum

Read explanation

Turning Points in Physics

Read explanation

Measurements

Read explanation

Engineering Physics

Read explanation

Mechanics and Materials

Read explanation

Electricity

Read explanation

Nuclear Physics

Read explanation

Circular Motion and Gravitation

Read explanation

Oscillations

Read explanation

Translational Dynamics

Read explanation

Thermodynamics

Read explanation

Electric Charge Field and Potential

Read explanation

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.