(1) predict: if the distance between the particles doubles, how will the force change?
(2) make q1 -2 μc at 4 cm and q2 +2 μc at 6 cm.
a) what is the charge on each of the particles?
b) double the distance between the particles. what is the force on each of the particles?
c) does the force on each of the particles increase, decrease or stay the same? circle the correct answer.
d) by how many times does the force on each of the particles change?
(3) make q1 +10 μc at 0 cm and q2 -10 μc at 10 cm.
a) what is the charge on each of the particles?
b) reduce the distance between the particles by half. what is the force on each of the particles?
c) does the force on each of the particles increase, decrease or stay the same? circle the correct answer.
d) by how many times does the force on each of the particles change?
(4) make q1 -5 μc at 4 cm and q2 -4 μc at 6 cm.
a) what is the charge on each of the particles?
b) triple the distance between the particles. what is the force on each of the particles?
c) does the force on each of the particles increase, decrease or stay the same? circle the correct answer.
d) by how many times does the force on each of the particles change?
(5) the force of repulsion between two charged particles is 1000 n. if the distance between the particles is quadrupled, what will be the force between the charged particles?
(6) summarize the results of your investigations for the relationship between the force on the charged particles and the distance between them.
v. summary questions
(1) the two objects on the left attract each other with a force of f. what is the force of attraction of the objects on the right?

Question

Accepted Answer

### Question (1)
# Explanation:
## Step1: Recall Coulomb's Law
Coulomb's Law states that the force $ F $ between two charged particles is given by $ F = k\frac{|q_1q_2|}{r^2} $, where $ k $ is Coulomb's constant, $ q_1 $ and $ q_2 $ are the charges of the particles, and $ r $ is the distance between them. If the distance $ r $ doubles (i.e., new distance $ r' = 2r $), we substitute into the formula.
## Step2: Calculate the new force
Let the original force be $ F = k\frac{|q_1q_2|}{r^2} $. The new force $ F' = k\frac{|q_1q_2|}{(r')^2} = k\frac{|q_1q_2|}{(2r)^2} = k\frac{|q_1q_2|}{4r^2} = \frac{1}{4} \times k\frac{|q_1q_2|}{r^2} = \frac{F}{4} $. So the force will decrease to one - fourth of its original value.
# Answer:
The force on each particle will decrease to $\frac{1}{4}$ of its original magnitude.

### Question (2)
#### (a)
# Brief Explanations:
The charge on the first particle ($ q_1 $) is given as $- 2\ \mu\text{C}$ and the charge on the second particle ($ q_2 $) is given as $ + 2\ \mu\text{C}$.
# Answer:
$ q_1=-2\ \mu\text{C} $, $ q_2 = + 2\ \mu\text{C} $

#### (b)
# Explanation:
## Step1: Determine the initial distance
The initial position of $ q_1 $ is at $ 4\ \text{cm} $ and $ q_2 $ is at $ 6\ \text{cm} $, so the initial distance $ r = 6 - 4=2\ \text{cm}=0.02\ \text{m} $. The charges $ q_1=-2\times 10^{- 6}\ \text{C} $, $ q_2 = 2\times 10^{-6}\ \text{C} $, and $ k = 9\times 10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2 $.
## Step2: Apply Coulomb's Law
Using $ F = k\frac{|q_1q_2|}{r^2} $, we substitute the values: $ F=(9\times 10^{9})\times\frac{|(-2\times 10^{-6})\times(2\times 10^{-6})|}{(0.02)^2}=(9\times 10^{9})\times\frac{4\times 10^{-12}}{4\times 10^{-4}}=(9\times 10^{9})\times(1\times 10^{-8}) = 90\ \text{N} $
# Answer:
The force on each particle is $ 90\ \text{N} $ (the force on $ q_1 $ and $ q_2 $ has the same magnitude due to Newton's third law, and the direction is attractive since the charges are opposite).

#### (c)
# Brief Explanations:
From Coulomb's Law $ F = k\frac{|q_1q_2|}{r^2} $, when the distance $ r $ doubles, the new force $ F'=\frac{F}{4} $ (as shown in question 1). Since the new force is smaller than the original force, the force decreases.
# Answer:
decrease (circle "decrease")

#### (d)
# Explanation:
## Step1: Recall the relationship from Coulomb's Law
When the distance $ r $ doubles ( $ r' = 2r $ ), from $ F = k\frac{|q_1q_2|}{r^2} $ and $ F' = k\frac{|q_1q_2|}{(r')^2} $, we have $ F'=\frac{F}{4} $. So the force changes by a factor of $\frac{1}{4}$, which means it is multiplied by $\frac{1}{4}$ or divided by 4.
# Answer:
The force on each particle changes by a factor of $\frac{1}{4}$ (or the force is $\frac{1}{4}$ of the original force).

### Question (3)
#### (a)
# Brief Explanations:
The charge on the first particle ($ q_1 $) is given as $ + 10\ \mu\text{C} $ and the charge on the second particle ($ q_2 $) is given as $ - 10\ \mu\text{C} $.
# Answer:
$ q_1 = + 10\ \mu\text{C} $, $ q_2=-10\ \mu\text{C} $

#### (b)
# Explanation:
## Step1: Determine the initial distance
The initial position of $ q_1 $ is at $ 0\ \text{cm} $ and $ q_2 $ is at $ 10\ \text{cm} $, so the initial distance $ r = 10 - 0 = 10\ \text{cm}=0.1\ \text{m} $. The charges $ q_1 = 10\times 10^{-6}\ \text{C} $, $ q_2=-10\times 10^{-6}\ \text{C} $, and $ k = 9\times 10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2 $. The initial force $ F = k\frac{|q_1q_2|}{r^2}=(9\times 10^{9})\times\frac{|(10\times 10^{-6})\times(- 10\times 10^{-6})|}{(0.1)^2}=(9\times 10^{9})\times\frac{100\times 10^{-12}}{0.01}=(9\times 10^{9})\times(1\times 10^{-8}) = 90\ \text{N} $
## Step2: Calculate the new distance and force
The new distance $ r'=\frac{r}{2}=5\ \text{cm} = 0.05\ \text{m} $. The new force $ F'=k\frac{|q_1q_2|}{(r')^2}=(9\times 10^{9})\times\frac{|(10\times 10^{-6})\times(-10\times 10^{-6})|}{(0.05)^2}=(9\times 10^{9})\times\frac{100\times 10^{-12}}{0.0025}=(9\times 10^{9})\times(4\times 10^{-8}) = 360\ \text{N} $
# Answer:
The force on each particle is $ 360\ \text{N} $

#### (c)
# Brief Explanations:
From Coulomb's Law $ F = k\frac{|q_1q_2|}{r^2} $, when the distance $ r $ is reduced by half ($ r'=\frac{r}{2} $), the new force $ F' = 4F $ (since $ F'=k\frac{|q_1q_2|}{(r/2)^2}=4k\frac{|q_1q_2|}{r^2} = 4F $). Since the new force is larger than the original force, the force increases.
# Answer:
increase (circle "increase")

#### (d)
# Explanation:
## Step1: Recall the relationship from Coulomb's Law
When the distance $ r $ is reduced by half ($ r'=\frac{r}{2} $), from $ F = k\frac{|q_1q_2|}{r^2} $ and $ F' = k\frac{|q_1q_2|}{(r')^2} $, we substitute $ r'=\frac{r}{2} $ into the formula for $ F' $. $ F'=k\frac{|q_1q_2|}{(r/2)^2}=4k\frac{|q_1q_2|}{r^2} = 4F $. So the force changes by a factor of 4.
# Answer:
The force on each particle changes by a factor of 4.

### Question (4)
#### (a)
# Brief Explanations:
The charge on the first particle ($ q_1 $) is given as $-5\ \mu\text{C}$ and the charge on the second particle ($ q_2 $) is given as $-4\ \mu\text{C}$.
# Answer:
$ q_1=-5\ \mu\text{C} $, $ q_2=-4\ \mu\text{C} $

#### (b)
# Explanation:
## Step1: Determine the initial distance
The initial position of $ q_1 $ is at $ 4\ \text{cm} $ and $ q_2 $ is at $ 6\ \text{cm} $, so the initial distance $ r = 6 - 4 = 2\ \text{cm}=0.02\ \text{m} $. The charges $ q_1=-5\times 10^{-6}\ \text{C} $, $ q_2=-4\times 10^{-6}\ \text{C} $, and $ k = 9\times 10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2 $.
## Step2: Calculate the initial force
Using $ F = k\frac{|q_1q_2|}{r^2} $, we substitute the values: $ F=(9\times 10^{9})\times\frac{|(-5\times 10^{-6})\times(-4\times 10^{-6})|}{(0.02)^2}=(9\times 10^{9})\times\frac{20\times 10^{-12}}{4\times 10^{-4}}=(9\times 10^{9})\times(5\times 10^{-8}) = 450\ \text{N} $
## Step3: Calculate the new distance and new force
The new distance $ r' = 3r=6\ \text{cm} = 0.06\ \text{m} $ (since we triple the distance). The new force $ F'=k\frac{|q_1q_2|}{(r')^2}=(9\times 10^{9})\times\frac{|(-5\times 10^{-6})\times(-4\times 10^{-6})|}{(0.06)^2}=(9\times 10^{9})\times\frac{20\times 10^{-12}}{36\times 10^{-4}}=(9\times 10^{9})\times(\frac{5}{9}\times 10^{-8}) = 50\ \text{N} $
# Answer:
The force on each particle is $ 50\ \text{N} $

#### (c)
# Brief Explanations:
From Coulomb's Law $ F = k\frac{|q_1q_2|}{r^2} $, when the distance $ r $ triples ($ r' = 3r $), the new force $ F'=\frac{F}{9} $ (since $ F'=k\frac{|q_1q_2|}{(3r)^2}=\frac{1}{9}k\frac{|q_1q_2|}{r^2}=\frac{F}{9} $). Since the new force is smaller than the original force, the force decreases.
# Answer:
decrease (circle "decrease")

#### (d)
# Explanation:
## Step1: Recall the relationship from Coulomb's Law
When the distance $ r $ triples ($ r' = 3r $), from $ F = k\frac{|q_1q_2|}{r^2} $ and $ F' = k\frac{|q_1q_2|}{(r')^2} $, we have $ F'=\frac{F}{9} $. So the force changes by a factor of $\frac{1}{9}$.
# Answer:
The force on each particle changes by a factor of $\frac{1}{9}$.

### Question (5)
# Explanation:
## Step1: Recall Coulomb's Law
Coulomb's Law is $ F = k\frac{|q_1q_2|}{r^2} $. Let the original force be $ F = 1000\ \text{N} $ and the original distance be $ r $. The new distance $ r' = 4r $ (since the distance is quadrupled).
## Step2: Calculate the new force
The new force $ F' = k\frac{|q_1q_2|}{(r')^2}=k\frac{|q_1q_2|}{(4r)^2}=\frac{1}{16}k\frac{|q_1q_2|}{r^2}=\frac{F}{16} $. Substituting $ F = 1000\ \text{N} $, we get $ F'=\frac{1000}{16}=62.5\ \text{N} $
# Answer:
The force between the charged particles will be $ 62.5\ \text{N} $

### Question (6)
# Brief Explanations:
From the investigations using Coulomb's Law $ F = k\frac{|q_1q_2|}{r^2} $, we observe that the force $ F $ between two charged particles is inversely proportional to the square of the distance $ r $ between them (when the charges are constant). Mathematically, $ F\propto\frac{1}{r^2} $. This means that if the distance between the charged particles increases, the force between them decreases, and if the distance decreases, the force between them increases. The relationship is a non - linear inverse relationship (inverse square relationship).
# Answer:
The force between two charged particles is inversely proportional to the square of the distance between them ($ F\propto\frac{1}{r^2} $). As the distance between the charged particles increases, the force between them decreases, and as the distance decreases, the force between them increases.

### Question (V.1)
# Explanation:
## Step1: Recall Coulomb's Law
For the left - hand side objects (both with charge $ q $, distance $ d $): $ F = k\frac{|q\times q|}{d^2}=k\frac{q^2}{d^2} $
For the right - hand side objects (charges $ q $ and $ 2q $, distance $ d $): $ F' = k\frac{|q\times 2q|}{d^2}=k\frac{2q^2}{d^2} $
## Step2: Find the ratio of $ F' $ to $ F $
$\frac{F'}{F}=\frac{k\frac{2q^2}{d^2}}{k\frac{q^2}{d^2}} = 2$. So $ F' = 2F $
# Answer:
The force of attraction of the objects on the right is $ 2F $

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QUESTION IMAGE

Question

Explanation:

Question (1)

Step1: Recall Coulomb's Law

Step2: Calculate the new force

Step1: Determine the initial distance

Step2: Apply Coulomb's Law

Answer:

Question (2)

(a)