EBK DELMAR'S STANDARD TEXTBOOK OF ELECT
6th Edition
ISBN: 9781305537125
Author: Herman
Publisher: YUZU
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Textbook Question
Chapter 3, Problem 1RQ
Why is static electricity considered to be a charge and not a current?
Expert Solution & Answer
To determine
Why is static electricity considered to be a charge and not a current?
Answer to Problem 1RQ
Static electricity refers to electrons that are immobile and not moving. Hence, it is a charge and not a current.
Explanation of Solution
Description:
The word static is used to express any quantity that is not moving or idle. Static electricity refers to electrons that are idle and immobile. Hence, it is considered to be a charge and not a current. A static charge can be either positive or negative depending upon the number of electrons. If an object has less electrons, it will have a positive charge; and if it has an excess of electrons, it will carry a negative charge.
Insulators are the only materials that hold the electrons motionless and keep them from flowing to a different position. Hence, electrostatic charges are built up on insulator materials. Electrostatic charges can be built up on a conductor only if it is electrically insulated from the surrounding objects.
Conclusion:
The reasons why static electricity is considered to be a charge and not a current has been explained.
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Students have asked these similar questions
The input reactance of an infinitesimal linear dipole of length A/60 and radius a=A/200
is given by
Xin = – 120
[In(€/a) — 1]
tan(ke)
Assuming the wire of the dipole is copper with a conductivity of 5.7 x 10' S/m,
determine at f = 1 GHz the
(a) loss resistance
(b) radiation resistance
(c) radiation efficiency
(d) VSWR when the antenna is connected to a 50-ohm line
Example
Solve the octic polynomial
2x⁸-9x⁷+20x⁶-33x⁵+46x⁴-66x³+80x²-72x+32=0
Solution
Divide by x⁴
2x⁴-9x³+20x²-33x+46-66/x + 80/x² - 72/x³ + 32/x⁴=0
Combine and bring terms
2(x⁴+16/x⁴) - 9(x³+8/x³) +20(x²+4/x²)-33(x+2/x) + 46= 0
Let use substitution
Let x+2/x =u
(x+2/x)²= u²
x²+2x*2/x + 4/x² = u²
x²+4/x²= u²-4
(x+2/x)³= x³+8/x³+3x*2/x(x+2/x)
u³= x³+8/x²+6u
x³+8/x³= u³-6u
(x²+4/x²)²= x⁴+2x²*4/x² + 16/x⁴
(u²-4)²= x⁴+16/x⁴ + 8
x⁴+16/x⁴ = (u²-4)²-8
x⁴+16/x⁴ = u⁴-8u²+8
2(u⁴-8u²+8)-9(u³-6u)+20(u²-4)-33u+46=0
Expand and simplify
2u⁴-9u³+4u²+21u-18=0
After checking
(u-1)(u-2) Are factors
Then
2u²-3u-9=0
u=3, u=-3/2
Assignment question
Solve the octic polynomial
2s⁸+s⁷+2s⁶-31s⁴-16s³-32s²-160=0 using the above example question, please explain in detail
b)
Another waveform g(t) is defined by
=0
t≥0, α>0
otherwise
g(t)= At exp(-at)
and is plotted in Figure 1 (for representative values of 4 = 1 and α = 1).
g(t)
0.4T
0.3+
0.2
0.1+
2
0
2
Figure 1
8
c)
Show that its amplitude spectrum is |G(@)| = -
A
(a²+0²)²
Describe briefly, with the aid of labelled sketches, how changing a affects
the waveform in both the time and frequency domains.
d) Deduce the Fourier transform H(@) of h(t) = g(t)+g(t+b)+g(t-b) and
calculate its DC amplitude H(0).
Chapter 3 Solutions
EBK DELMAR'S STANDARD TEXTBOOK OF ELECT
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