Electric force $F$ between charges $q_1$ and $q_2$ at a distance $r$ from each other is given as:

$$
F = dfrac{k q_1 q_2}{r^2}
$$

where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant.
We see that magnitude of the force between the two charges is proportional to those two charges and inversely proportional to the square of the distance between them.

We also see that when the two charges are like, magnitude of the force is positive because product $q_1 cdot q_2$ will give us a positive number. We interpret this as two charges repelling each other.

Notice that when two charges are opposite, magnitude of the force is negative, because product $q_1 cdot q_2$ will give as a negative number. We interpret this as two charges attracting each other.

This means that whenever you see the like charges, magnitude of the force between them will be positive and charges will try to move away from each other, whereas if you see two opposite charges, magnitude of the force between them is negative and they will try to move closer to one another.

Hint: Electric force $F$ between charges $q_1$ and $q_2$ at a distance $r$ from each other is given as:

$$
F = dfrac{k q_1 q_2}{r^2}
$$

Electric force $F$ between charges $q_1$ and $q_2$ at a distance $r$ from each other is given as:

$$
F = dfrac{k q_1 q_2}{r^2}
$$

where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant.
We see that magnitude of the force between the two charges is proportional to those two charges and inversely proportional to the square of the distance between them.

This means if distance between the two charges was tripled, force between the same two charges would decrease by a factor of $3^2$, which is $9$. We can prove this. Let $F_i$ be the electric force between charges $q_1$ and $q_2$ that are initially at a distance $r_i$. In this case, force $F_i$ between these two charges is given as:

$$
begin{equation}
F_i = dfrac{k q_1 q_2}{r_i^2}
end{equation}
$$

Let’s now move these same charges further apart to a distance 3 times larger than the initial distance. This final distance between the charges will be given as $r_f = 3 r_i$. Force $F_f$ between these two charges in the final case is given as:

$$
begin{equation}
F_f = dfrac{k q_1 q_2}{r_f^2} = dfrac{k q_1 q_2}{(3r_i)^2}
end{equation}
$$

Let’s divide equations $(2)$ and $(1)$:

$$
begin{align*}
dfrac{F_f}{F_i} &= dfrac{dfrac{k q_1 q_2}{(3r_i)^2}}{dfrac{k q_1 q_2}{r_i^2} } \
dfrac{F_f}{F_i} &= dfrac{dfrac{1}{9r_i^2}}{dfrac{1}{r_i^2} } \
dfrac{F_f}{F_i} &= dfrac{ r_i^2 }{ 9 r_i^2 } \
dfrac{F_f}{F_i} &= dfrac{ 1 }{ 9 }
end{align*}
$$

$$
boxed{F_f = dfrac{F_i}{9}}
$$

Electric force between two charges is inversely proportional to the square of the distance between them. If distance was tripled, force between charges would decrease by a factor of 9.

$$
F_f = dfrac{F_i}{9}
$$

When an electroscope is charged, this means that charge of the same sign is spread out evenly across the metal parts of the electroscope, including the leaves. This means that the leaves are charged with charge of the same sign and thus repel each other.
Keep in mind that electrostatic force between the two charges is inversely proportional to the square of the distance $r$ between the charges and proportional to both of the charges (in this case $q^2$).
Leaves are the only part of the electroscope that can freely move and because leaves repel one another, they will try to go as far as they can from one another, which they do by lifting themselves up and making the distance between the ends of the leaves as high as possible. The reason why leaves stop lifting although they still repel one another with electrostatic force is because these leaves to have same mass and thus gravitational force always pulls them downward. What essentially happens is that gravitational pull downward becomes balanced with the electrostatic repulsion between the leaves. Note that leaves in the electroscope will rise farther if charge on the electroscope is greater, but electrostatic force between the leaves will balance itself with gravitational pull, but at a higher angle.

$a)$ We can charge electroscope positively using positive rod in the following two steps. First step is to make sure that electroscope is uncharged. To discharge the electroscope, we’ll connect knob of the electroscope to the Earth (ground). Any charge on the electroscope will flow to the ground. At this moment electroscope is electrically neutral. We’ll now disconnect the knob from the ground and touch knob of the electroscope with positively charged rod. Positive charges from rod will be redistributed in a system positive rod-electroscope and electroscope will be charged positively by conduction.

$b)$ We can charge electroscope positively using negative rod by connecting knob of the electroscope to the Earth (ground) and by bringing negatively charged rod in proximity of knob of the electroscope. Negative charges will be repelled from the knob of the electroscope and they’ll flow to the ground. We can now disconnect the electroscope from the ground and remove a negative rod. Knob of the electroscope was charged positively by induction.

There is not much (if any) free electrons in an insulator, but since insulator is made from polar molecules, which means they respond to electric field, what really happens when we bring an insulator near a charged object is that molecules of the insulator re-orient themselves in a way to be attracted to the charged object. For example, if we bring a glass of water near a positive charge, since $H^+$ ions in the water are positive, they will be repelled by the positive charge and water molecules will turn around and re-orient themselves so that oxygen ion $O^{2-}$ is brought closer to the positively charged object.
This makes the water as a whole to behave like a charged object.

Electric force $F_{12}$ with which charge $q_1$ acts on charge $q_2$ which is at a distance $r$ from charge $q_1$ is calculated as:

$$
F_{12} = dfrac{k q_1 q_2}{r^2}
$$

where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant.

Analogously,
electric force $F_{21}$ with which charge $q_2$ acts on charge $q_1$ which is at a distance $r$ from charge $q_2$ is calculated as:

$$
F_{21} = dfrac{k q_2 q_1}{r^2}
$$

Notice that magnitude of both of the forces is same, which could be expected from Newton’s third law, which states that a force with which object 1 acts on object 2 is equal in magnitude, but in the opposite direction to the force with which object 2 acts on object 1.
We thus write:

$$
vec{F}_{12}= – vec{F}_{21}
$$