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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 682: Lesson Check

Exercise 3

Step 1

1 of 1

If a body is electrically neutral ,then it must have equal number of positive and negative charge ,both of them cancel each other,to give net zero charge

Exercise 4

Step 1

1 of 1

Charge quantization means that a body must have charge that cannot take any arbitrary values, but only values that are integral multiples of the fundamental charge .ie.charge of proton/electron.

Exercise 5

Step 1

1 of 2

When we charge a body ,only the charges are transferred from one body to another,but the total charge of the universe remains the same.

Result

2 of 2

No.

Exercise 6

Step 1

1 of 2

$textbf{(a)}$ The object’s mass $decreases$.

$textbf{(b)}$ The $best$ explanation is

A. To give the object a positive charge means removing some of the electrons. This will reduce its mass.

Result

2 of 2

$textbf{(a)}$ Decreases $textbf{(b)}$ A

Exercise 7

Step 1

1 of 2

From the table 19.1 , we can see that Relative Charging Material with Rubbing is greater for rabbit fur than glass, which means rabbit fur will give up electrons more easily and become more positive, but since charges are to be conserved, glass will have negative charge and rabbit fur will have positive charge.

Result

2 of 2

Rabbit fur is positive and glass is negative.

Exercise 8

Step 1

1 of 2

$textbf{(a)}$ Electrons have a negative charge and protons have a positive charge but both have the same magnitude of charge. Since the system contains a greater number of electrons than protons, the total charge of the system must be $negative$,

$textbf{(b)}$ The charge of electron is $-e$ and the charge of proton is $+e$, so the total charge of the system is

$$
55(-e) + 43(e) = -12e = -12(1.6times10^{-19};mathrm{C}) = -1.9times10^{-18};mathrm{C}.
$$

Result

2 of 2

$textbf{(a)}$ Negative $textbf{(b)}$ $-1.9times10^{-18}$ C

Exercise 9

Step 1

1 of 2

The system must contain a certain number of positively charged protons such that the total charge of the system is $+22e$. The mass of the system is the sum of the masses of its electrons and protons. Assume $p$ is the total number of protons in the system, then we must have

$$
p(e) + 150(-e) = 22e ;;Rightarrow;; p – 150 = 22,
$$

where we used the fact that each electron has a charge of $-e$ charge and each proton has a charge of $+e$. Solving for $p$ gives us $p = 172$. Since the mass of proton is $m_p = 1.67times10^{-27}$ kg and the mass of electron is $m_e = 9.11times10^{-31}$ kg, the total mass of the system is

$$
begin{align*}
172m_p + 150m_e &= 172(1.67times10^{-27};mathrm{kg}) + 150(9.11times10^{-31};mathrm{kg})\
&= 2.87times10^{-25};mathrm{kg}.
end{align*}
$$

Result

2 of 2

$2.87times10^{-25}$ kg

Exercise 10

Step 1

1 of 2

The system must contain a certain number of negatively charged electrons such that the total charge of the system is $-51e$. The mass of the system is the sum of the masses of its electrons and protons. Assume $n$ is the total number of electrons in the system, then we must have

$$
n(-e) + 320(+e) = -51e ;;Rightarrow;; -n + 320 = -51,
$$

where we used the fact that each electron has a charge of $-e$ and each proton has a charge of $+e$. Solving for $n$ gives us $n = 371$. Since the mass of proton is $m_p = 1.67times10^{-27}$ kg and the mass of electron is $m_e = 9.11times10^{-31}$ kg, the total mass of the system is

$$
begin{align*}
320m_p + 371m_e &= 320(1.67times10^{-27};mathrm{kg}) + 371(9.11times10^{-31};mathrm{kg})\
&= 5.35times10^{-25};mathrm{kg}.
end{align*}
$$

Result

2 of 2

$5.35times10^{-25}$ kg

Exercise 11

Step 1

1 of 2

Each argon atom contains 18 protons, each with a positive charge of magnitude $e$. Therefore, the total positive charge in 3 moles of argon is

$$
begin{align*}
3;mathrm{moles}&left(frac{6.022times10^{23};mathrm{atoms}}{mathrm{mole}}right)left(frac{18;mathrm{protons}}{mathrm{atom}}right)left(frac{1.60times10^{-19};mathrm{C}}{mathrm{proton}}right)\
&= 5.20times10^6;mathrm{C}.
end{align*}
$$

Result

2 of 2

$$
5.20times10^6;mathrm{C}
$$

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