Ap - Charges 2
This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA
Overview
Download & View Ap - Charges 2 as PDF for free.
More details
- Words: 636
- Pages: 3
Purple Squirrel and Blue Angel
READ FIRST: THIS WEBSITE (www.myspace.com/webassign) WILL BE OPEN TO THE PUBLIC ONLY FOR THE FIRST TWO WEBASSIGNS. CHANGES TO THE NUMBER OF WEBASSIGNS BEFORE THE WEBSITE CLOSES, IS SUBJECT TO CHANGE AT OUR DISCRETION. THIS WEBSITE WILL BE CLOSED TO PREVENT TEACHERS COMING IN. IF YOU WANT TO HAVE ACCESS TO THE WEBSITE AFTER IT “CLOSES” YOU MUST REQUEST US TO ADD YOU AS FRIENDS. 1. Object A is metallic and electrically neutral. It is charged by induction so that it acquires a charge of -3.1 10-6 C. Object B is identical to object A and is also electrically neutral. It is charged by induction so that it acquires a charge of +3.1 10-6 C. Find the difference in mass between the charged objects. Solution: Part A: We must first determine whether electrons have been added or removed.
(*Note: If electrons are added the charge will be positive and if electrons are removed the charge will be positive only if the charge is neutral) Object A: Since the charge is negative (-3.1x10-6) then we know that (3.1x10-6) electrons have been added. Object B: Since the charge is positive (+3.1x10-6) then we know that (3.1x10-6) electrons have been removed. *Now we can find the total number of electrons by subtracting Object B from Object A.
(*Note: Since electrons have been added to Object A the value will be negative since electrons have a negative charge and vise versa for Object B) Object A – Object B (-3.1x10-6 C) - (+3.1x10-6 C) = - 6.2x10-6 C *Now we must convert the total electrons into mass (kg)
(*Note: The mass of one electron is 9.11x10-31kg and the charge is 1.6x10-19C) (*Note: Since mass is a scalar quantity the negative sign is dropped) (6.2x10-6 C) x (9.11x10-31kg) / (1.6x10-19C) = 3.530125x10-17kg
Part B: Since electrons were added to Object A the mass of Object A will be greater.
2. Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of -4q. Sphere B carries a charge of -7q. Sphere C carries no net charge. Spheres A and B are touched together and then separated. Sphere C is then touched to sphere A and separated from it. Lastly, sphere C is touched to sphere B and separated from it. Solution: Part A: Identical conducting spheres equalize their charge upon touching. When sphere A and B the negative charge from A flows through B is added together and the charge is then split between A and B. Sphere C is neutral so when A and C touch the charge of A is split between A and C. When B and C touch the new amount on charge B is added together with charge C and then equally split. A and B touch -4q + -7q / 2 = -5.5 q A = -5.5q B=-5.5q A and C touch -5.5q + 0q / 2 = -2.75q A= -2.75q C= -2.75q B and C touch -5.5q + -2.75q / 2 = -4.125q B= -4.125q C= -4.125q Part B: The total charge is the sum of all charges. From the problem A is charged with -4q B is charged with -7q and C is 0q. A+B+C (-7q) + (-4q) + (0q) = -11q Part C: Since electrostatic forces are conserved the charges will also be conserved thus the charge is still -11q
THESE EXPLANATIONS BROUGHT TO YOU BY:
Purple Squirrel & Blue Angel THIS WEBSITE (www.myspace.com/webassign) WILL BE OPEN TO THE PUBLIC ONLY FOR THE FIRST TWO WEBASSIGNS. CHANGES TO THE NUMBER OF WEBASSIGNS BEFORE THE WEBSITE CLOSES, IS SUBJECT TO CHANGE AT OUR DISCRETION. THIS WEBSITE WILL BE CLOSED TO PREVENT TEACHERS COMING IN. IF YOU WANT TO HAVE ACCESS TO THE WEBSITE AFTER IT “CLOSES” YOU MUST REQUEST US TO ADD YOU AS FRIENDS.
READ FIRST: THIS WEBSITE (www.myspace.com/webassign) WILL BE OPEN TO THE PUBLIC ONLY FOR THE FIRST TWO WEBASSIGNS. CHANGES TO THE NUMBER OF WEBASSIGNS BEFORE THE WEBSITE CLOSES, IS SUBJECT TO CHANGE AT OUR DISCRETION. THIS WEBSITE WILL BE CLOSED TO PREVENT TEACHERS COMING IN. IF YOU WANT TO HAVE ACCESS TO THE WEBSITE AFTER IT “CLOSES” YOU MUST REQUEST US TO ADD YOU AS FRIENDS. 1. Object A is metallic and electrically neutral. It is charged by induction so that it acquires a charge of -3.1 10-6 C. Object B is identical to object A and is also electrically neutral. It is charged by induction so that it acquires a charge of +3.1 10-6 C. Find the difference in mass between the charged objects. Solution: Part A: We must first determine whether electrons have been added or removed.
(*Note: If electrons are added the charge will be positive and if electrons are removed the charge will be positive only if the charge is neutral) Object A: Since the charge is negative (-3.1x10-6) then we know that (3.1x10-6) electrons have been added. Object B: Since the charge is positive (+3.1x10-6) then we know that (3.1x10-6) electrons have been removed. *Now we can find the total number of electrons by subtracting Object B from Object A.
(*Note: Since electrons have been added to Object A the value will be negative since electrons have a negative charge and vise versa for Object B) Object A – Object B (-3.1x10-6 C) - (+3.1x10-6 C) = - 6.2x10-6 C *Now we must convert the total electrons into mass (kg)
(*Note: The mass of one electron is 9.11x10-31kg and the charge is 1.6x10-19C) (*Note: Since mass is a scalar quantity the negative sign is dropped) (6.2x10-6 C) x (9.11x10-31kg) / (1.6x10-19C) = 3.530125x10-17kg
Part B: Since electrons were added to Object A the mass of Object A will be greater.
2. Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of -4q. Sphere B carries a charge of -7q. Sphere C carries no net charge. Spheres A and B are touched together and then separated. Sphere C is then touched to sphere A and separated from it. Lastly, sphere C is touched to sphere B and separated from it. Solution: Part A: Identical conducting spheres equalize their charge upon touching. When sphere A and B the negative charge from A flows through B is added together and the charge is then split between A and B. Sphere C is neutral so when A and C touch the charge of A is split between A and C. When B and C touch the new amount on charge B is added together with charge C and then equally split. A and B touch -4q + -7q / 2 = -5.5 q A = -5.5q B=-5.5q A and C touch -5.5q + 0q / 2 = -2.75q A= -2.75q C= -2.75q B and C touch -5.5q + -2.75q / 2 = -4.125q B= -4.125q C= -4.125q Part B: The total charge is the sum of all charges. From the problem A is charged with -4q B is charged with -7q and C is 0q. A+B+C (-7q) + (-4q) + (0q) = -11q Part C: Since electrostatic forces are conserved the charges will also be conserved thus the charge is still -11q
THESE EXPLANATIONS BROUGHT TO YOU BY:
Purple Squirrel & Blue Angel THIS WEBSITE (www.myspace.com/webassign) WILL BE OPEN TO THE PUBLIC ONLY FOR THE FIRST TWO WEBASSIGNS. CHANGES TO THE NUMBER OF WEBASSIGNS BEFORE THE WEBSITE CLOSES, IS SUBJECT TO CHANGE AT OUR DISCRETION. THIS WEBSITE WILL BE CLOSED TO PREVENT TEACHERS COMING IN. IF YOU WANT TO HAVE ACCESS TO THE WEBSITE AFTER IT “CLOSES” YOU MUST REQUEST US TO ADD YOU AS FRIENDS.
