Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 13

\(\mathrm{A}+2.0\) -nC point charge is \(3.0 \mathrm{cm}\) away from a $-3.0 \mathrm{-nC}$ point charge. (a) What are the magnitude and direction of the electric force acting on the +2.0 -nC charge? (b) What are the magnitude and direction of the electric force acting on the -3.0 -nC charge?

Short Answer

Expert verified
(b) What is the magnitude and direction of the electric force acting on the -3.0 nC charge? Answer: (a) For the +2.0 nC charge: Magnitude of the electric force is 6.0 x 10^-9 N and the direction is towards the -3.0 nC charge. (b) For the -3.0 nC charge: Magnitude of the electric force is 6.0 x 10^-9 N and the direction is towards the +2.0 nC charge.

Step by step solution

01

Coulomb's Law formula

Write down the formula for Coulomb's Law: \(F = k \frac{q_1 q_2}{r^2}\) where \(F\) is the force between the two charges, \(k\) is Coulomb's constant (\(8.99 \times 10^9 \mathrm{Nm^2/C^2}\)), \(q_1\) and \(q_2\) are the magnitudes of the two charges, and \(r\) is the distance between the charges.
02

Convert distances and charges to appropriate units

Convert the distances and charges to appropriate units (meters and coulombs). In our case, we have: \(q_1 = +2.0 \text{nC} = +2.0 \times 10^{-9} \mathrm{C}\) \(q_2 = -3.0 \text{nC} = -3.0 \times 10^{-9} \mathrm{C}\) \(r = 3.0 \mathrm{cm} = 0.03 \mathrm{m}\)
03

Calculate electric force acting on +2.0 nC charge

Plug the values from step 2 into the Coulomb's Law formula, to find the electric force acting on the +2.0 nC charge: \(F_{+2} = k \frac{(+2.0 \times 10^{-9} \mathrm{C})(-3.0 \times 10^{-9} \mathrm{C})}{(0.03 \mathrm{m})^2}\) \(F_{+2} = (8.99 \times 10^9 \mathrm{Nm^2/C^2}) \frac{(-6.0 \times 10^{-18} \mathrm{C^2})}{(0.0009 \mathrm{m^2})}\) \(F_{+2} = -6.0 \times 10^{-9} \mathrm{N}\) The force acting on the +2.0 nC charge is \(6.0 \times 10^{-9} \mathrm{N}\) towards the -3.0 nC charge (as the force is negative).
04

Calculate electric force acting on -3.0 nC charge

From Newton's Third Law, we know that the forces in opposite directions are equal in magnitudes, so: \(F_{-3} = -F_{+2}\) Therefore, the force acting on the -3.0 nC charge is also \(6.0 \times 10^{-9} \mathrm{N}\), but its direction is towards the +2.0 nC charge (as the force is positive).
05

Final Answers

(a) For the +2.0 nC charge: Magnitude of the electric force is \(6.0 \times 10^{-9} \mathrm{N}\) and the direction is towards the -3.0 nC charge. (b) For the -3.0 nC charge: Magnitude of the electric force is \(6.0 \times 10^{-9} \mathrm{N}\) and the direction is towards the +2.0 nC charge.

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.

Start your free trial

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

A parallel-plate capacitor consists of two flat metal plates of area \(A\) separated by a small distance \(d\). The plates are given equal and opposite net charges \(\pm q\) (a) Sketch the field lines and use your sketch to explain why almost all of the charge is on the inner surfaces of the plates. (b) Use Gauss's law to show that the electric field between the plates and away from the edges is $E=q /\left(\epsilon_{0} A\right)=\sigma / \epsilon_{0} \cdot(\mathrm{c})$ Does this agree with or contra- dict the result of Problem \(70 ?\) Explain. (d) Use the principle of superposition and the result of Problem 69 to arrive at this same answer. [Hint: The inner surfaces of the two plates are thin, flat sheets of charge.]
Two otherwise identical conducting spheres carry charges of $+5.0 \mu \mathrm{C}\( and \)-1.0 \mu \mathrm{C} .$ They are initially a large distance \(L\) apart. The spheres are brought together, touched together, and then returned to their original separation \(L\). What is the ratio of the magnitude of the force on either sphere after they are touched to that before they were touched?
Two point charges, \(q_{1}=+20.0 \mathrm{nC}\) and \(q_{2}=+10.0 \mathrm{nC},\) are located on the \(x\) -axis at \(x=0\) and \(x=1.00 \mathrm{m},\) respectively. Where on the \(x\) -axis is the electric field equal to zero?
Two point charges are located on the \(x\) -axis: a charge of \(+6.0 \mathrm{nC}\) at \(x=0\) and an unknown charge \(q\) at \(x=\) \(0.50 \mathrm{m} .\) No other charges are nearby. If the electric field is zero at the point $x=1.0 \mathrm{m},\( what is \)q ?$
A thin wire with positive charge evenly spread along its length is shaped into a semicircle. What is the direction of the electric field at the center of curvature of the semicircle? Explain.
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Waves Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Mechanics and Materials

Read Explanation

Measurements

Read Explanation

Dynamics

Read Explanation
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