Physics For Scientists And Engineers: A Strategic Approach, 4th Edition
Authors: Randall D. Knight
ISBN-13: 978-0133942651
See our solution for Question 1E from Chapter 22 from Randall Knight's Physics for Scientists and Engineers.
Problem 1E
Chapter:
Problem:
A glass rod is charged to +5.0 nC by rubbing. Have electrons been removed from the rod or protons added? Explain..
Step-by-Step Solution
Step 1
We will use the concept of quantization of charge to find the number of electrons added or removed from the rod.
Step 2: the quantization equation
According to quantization principal The charge on any body is an integral multiple of the charge on integral. Moreover, when electrons are remove from a body, the negative charge is removed from the body. That makes a body positively charge.
Step 3: Part (a)
Each body contains free electron and tightly held protons. So, when a body is charged negatively, there is an addition of electrons, and when body is negatively charged, electrons are removed from the body.
As the rod has a positive charge of $ + 8\,\,{\rm{nC}}$, there has been an removal of electrons from the rod.
Step 3: Part (b)
Number of electrons removed from the rod:
Using the equation of quantization:\[\begin{array}{l}q = ne\\\\8\,\,{\rm{nC}} = n \times 1.6 \times {10^{ - 19}}\\\\8\,\, \times {\rm{1}}{{\rm{0}}^{ - 9}} = n \times 1.6 \times {10^{ - 19}}\\\\n = \dfrac{{8\,\, \times {\rm{1}}{{\rm{0}}^{ - 9}}}}{{1.6 \times {{10}^{ - 19}}}}\\\\n = 5 \times {10^{10}}\,\,{\rm{electrons}}\end{array}\]
ANSWERS
Removal of $5 \times {10^{10}}\,\,{\rm{electrons}}$
We will use the concept of quantization of charge to find the number of electrons added or removed from the rod.
Step 2: the quantization equation
According to quantization principal The charge on any body is an integral multiple of the charge on integral. Moreover, when electrons are remove from a body, the negative charge is removed from the body. That makes a body positively charge.
Step 3: Part (a)
Each body contains free electron and tightly held protons. So, when a body is charged negatively, there is an addition of electrons, and when body is negatively charged, electrons are removed from the body.
As the rod has a positive charge of $ + 8\,\,{\rm{nC}}$, there has been an removal of electrons from the rod.
Step 3: Part (b)
Number of electrons removed from the rod:
Using the equation of quantization:\[\begin{array}{l}q = ne\\\\8\,\,{\rm{nC}} = n \times 1.6 \times {10^{ - 19}}\\\\8\,\, \times {\rm{1}}{{\rm{0}}^{ - 9}} = n \times 1.6 \times {10^{ - 19}}\\\\n = \dfrac{{8\,\, \times {\rm{1}}{{\rm{0}}^{ - 9}}}}{{1.6 \times {{10}^{ - 19}}}}\\\\n = 5 \times {10^{10}}\,\,{\rm{electrons}}\end{array}\]
ANSWERS
Removal of $5 \times {10^{10}}\,\,{\rm{electrons}}$