# Problem 5: Common static electricity involves charges varying from nanocoulombs to microcoulombs. Part (a) How many electrons are needed to form a charge of Q1 = –10 nC? N= Part (b) How many electrons...

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Problem 5:   Common static electricity involves charges varying from nanocoulombs to microcoulombs. Part (a)  How many electrons are needed to form a charge of Q1 = –10 nC? N= Part (b)  How many electrons must be removed from a neutral object to leave a net charge of Q2 = 2.8 μC? N= Problem 6:   8.8 × 1020 electrons move through a pocket calculator during a particular day’s operation.  Par.t Calculate the total charge, in coulombs, of the electron that pass through the calculator? Q= Problem 8:   A certain lightning bolt moves 38 C of charge. How many fundamental units of charge is this , in moles? N mol= Problem 9:   A 2.5 g copper penny is given a charge of -1.3 nC. How many electrons were transferred in order to create on the penny? N e= Problem 10:   A 2.5 g copper penny is given a charge of 4.9 nC. Part B How many electrons were transferred in order to create the charge on the penny? Part C If at most one electron is removed from any single atom, what percent of the atoms are ionized by this charging process? Ne/NCu= % Problem 25:   A particle with charge q = +3e and mass m = 7.8 × 10-26 kg is injected horizontally with speed 1.4 × 106 m/s into the region between two parallel horizontal plates. The plates are 28 cm long and an unknown distance d apart. The particle is injected midway between the top and bottom plates. The top plate is negatively charged and the bottom plate is positively charged, so that there is an upward-directed electric field between the plates, of magnitude E = 37 kN/C. Ignore the weight of the particle. Part A How long, in seconds, does it take for the particle to pass through the region between the plate? Part B When the particle exits the region between the plates, what will be the magnitude of its vertical displacement from its entry height, in millimeters? Problem 28:   Suppose a speck of dust in an electrostatic precipitator has Np = 2.25 × 1017 protons in it and carries a net charge of Q = -68nC. Let qe represent the charge of an electron. Part A Enter an expression for the number of electrons Ne in the speck of dust in terms of the charge of an electron ,qe, and other variables from the problem statement. Part B How many electrons are in the speck of dust? Problem 29:   A bacterium has 3.1 × 1016 protons and a net charge of 5.4 pC. Part A How many fewer electrons are there than protons? Np - Ne = Part B If you paired them up, what fraction of the protons would have no electrons? (Np - Ne) / Np = Problem 30:   Suppose you had a 115 g piece of sulfur. What net charge, in coulombs, would you place on it if you put an extra electron on 1 in 1012 of its atoms? (Sulfur has an atomic mass of 32.1) Problem 31:   Suppose a piece of dirt in an electrostatic precipitator has 0.85 × 1017 protons in it and carries a net charge of –4.7 nC. How many electrons does it have? Problem 32:   A 51 g nugget of copper has a net charge of 1.95 μC. What fraction of the copper's electrons has been removed? Each copper atom has 29 protons, and copper has an atomic mass of 63.5. Nremoved / Ninitial =   Problem 33:   A speck of dust in an electrostatic precipitator has 1.1×1012×1012 protons in it, and it has a net charge of -4.5 nC (a very large charge for a small speck). How many electrons are in the speck of dust? Problem 34:   Plutonium has an atomic mass of 244 and each plutonium atom has 94 protons. How many coulombs of positive charge are there in 4.3 kg of plutonium?  Problem 35:   An amoeba has 1.4×1016×1016 protons and a net charge of 0.36 pC. Part (a) How many fewer electrons are there than protons? Np-Ne= Part (b) If you paired them up, what fraction of the protons would have no electrons? Problem 36:   A 45 g ball of copper has a net charge of 2.5 μCμC. Part (a) How many protons are contained in the ball? (You may find the atomic mass and the atomic number for copper (Cu) in a periodic table or other resource.) Part (b) Supposing that the ball is initially neutral, what fraction of its electrons were removed? Problem 37:   A sample of sulfur has a mass of 105 g. What is the net charge on the sample if an extra electron is added to one out of every 10121012 sulfur atoms? (The properties of sulfur may be obtained from the periodic table or some alternative source.) Problem 38:   A sample of plutonium has a mass of 4.75 kg? How many coulombs of positive charge are there in the given sample of plutonium? (The properties of plutonium may be obtained from the periodic table or other reference.)

## Answer To: Problem 5: Common static electricity involves charges varying from nanocoulombs to microcoulombs....

Nishchay answered on Sep 19 2021
Notations used in the following solutions
Q=net electrostatic charge
Ne=Number of electrons
Np=Number of protons
qe=qp=electric charge on electrons and protons (nega
tive for electrons and positive for protons)
Aelement=element atomic number
Nelement=number of atoms in the element material
Sol 5:
Part (a): Number of electrons to form a charge is
1
9
19
10
10 10
1.6 10
6.25 10 electrons
e
Q
N
q
N
N

Part (b): similarly
2
6
19
13
2.8 10
1.6 10
1.75 10 electrons
e
Q
N
q
N
N
Sol 6: The net charge is given as
20 198.8 10 1.6 10
140.8 C
eQ N q
Q
Q
Sol 8: The fundamental unit of charge is the charge on electrons or protons so for 38 C charge
the number of fundamental units of charge is
19
20
38
1.6 10
2.375 10 protons
e
Q
N
q
N
N

As 231 mole=6.022 10 fundamental units so
20
23
4
2.375 10
6.022 10
3.94 10 mole
mole
mole
N
N
Sol 9:
9
19
9
1.3 10
1.6 10
8.125 10 electrons
e
e
e
e
Q
N
q
N
N
Sol 10:
Part B: Number of electrons transferred from the penny for required charge is
9
19
10
4.9 10
1.6 10
3.0625 10 electrons
e
e
e
e
Q
N
q
N
N

Part C: By the mole concept number of atoms in copper are
23
23
22
63.5 g-mole copper = 6.022 10 atoms
2.5 6.022 10
2.5 g penny has copper atoms =
63.5
2.37 10 atomsCuN

so, the percentage of ionized atoms is
10
22
10
100
ionized atom percentage=
100 3.0625 10
2.37 10
1.29 10 %
e
Cu
N
N
Sol 25:
Part A: As no force is acting in the horizontal direction so charge particle moves with constant
velocity in the horizontal direction.
So, time taken by the particle to pass the plates is
6
7
0.28
1.4 10
2 10 s
=0.2...
SOLUTION.PDF