Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

Answers without the blur.

Just sign up for free and you're in.

Short Answer

How far apart must two point charges of \({\rm{75}}{\rm{.0 nC}}\) (typical of static electricity) be to have a force of \({\rm{1}}{\rm{.00 N}}\) between them?

The separation between the charges is \(7.11{\rm{ }}mm\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Electrostatic force:

When a charge enters the field of another charge, it experiences a force known as an electrostatic force.

Step 2: Separation between charges

The electrostatic force between two point charges is given as,

\(F = \frac{{K{q_1}{q_2}}}{{{r^2}}}\) ….. (1)

Here, \(F\) is the electrostatic force, \(K\) is the electrostatic force constant, \({q_1}\)is the charge on the first body, \({q_2}\) is the charge on the second body, and \(r\) is the separation between the charges.

Consider the given data as below.

The electrostatic force, \(F = 1.00{\rm{ }}N\)

The electrostatic force constant, \(K=9\times 10^{9} N.m^{2}/C^{2}\)

The charge on the first body, \({q_1} = 75.0{\rm{ }}nC\)

The charge on the second body, \({q_2} = 75.0{\rm{ }}nC\)

Rearranging equation (1) in order to get an expression for separation between the charges.

\(r = \sqrt {\frac{{K{q_1}{q_2}}}{F}} \) ….. (2)

Substituting all known values into equation (2).

\(\begin{aligned} r &= \sqrt {\frac{{\left( {9 \times {{10}^9}{\rm{ }}N \times {m^2}/{C^2}} \right) \times \left( {75.0{\rm{ }}nC} \right) \times \left( {75.0{\rm{ }}nC} \right)}}{{\left( {1.00{\rm{ }}N} \right)}}} \\ &= \sqrt {\frac{{\left( {9 \times {{10}^9}{\rm{ }}N \times {m^2}/{C^2}} \right) \times \left( {75.0{\rm{ }}nC} \right) \times \left( {\frac{{{{10}^{ - 9}}{\rm{ }}C}}{{1{\rm{ }}nC}}} \right) \times \left( {75.0{\rm{ }}nC} \right) \times \left( {\frac{{{{10}^{ - 9}}{\rm{ }}C}}{{1{\rm{ }}nC}}} \right)}}{{\left( {1.00{\rm{ }}N} \right)}}} \\ &= 7.11 \times {10^{ - 3}}{\rm{ }}m \times \left( {\frac{{1000{\rm{ }}mm}}{{1{\rm{ }}m}}} \right)\\ &= 7.11{\rm{ }}m\end{aligned}\)

Hence, the separation between the charges is \(7.11{\rm{ }}m\).

Find Study Materials for

Create Study Materials

Select your language

College Physics (Urone)

Answers without the blur.

Short Answer

How far apart must two point charges of \({\rm{75}}{\rm{.0 nC}}\) (typical of static electricity) be to have a force of \({\rm{1}}{\rm{.00 N}}\) between them?

Step by Step Solution

Step 1: Electrostatic force:

Step 2: Separation between charges

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

College Physics (Urone)

Answers without the blur.

Short Answer

How far apart must two point charges of \({\rm{75}}{\rm{.0 nC}}\) (typical of static electricity) be to have a force of \({\rm{1}}{\rm{.00 N}}\) between them?

Step by Step Solution

Step 1: Electrostatic force:

Step 2: Separation between charges

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics