Electric potential $V$ of this field at a distance $r$ from charge $q$ will be given as:

$$
V = dfrac{kq}{r}
$$

where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant. This means that if we were to put a test charge $q_{test}$ at a distance $r$ from charge $q$, test charge would feel an electric potential equal to:

$$
V = dfrac{kq}{r}
$$

As it can be seen, electric potential coming from charge $q$ won’t depend on the test charge $q_{test}$

Electric potential energy of a system of charges is equal to work needed to assemble the system as a whole by bringing charges, one by one, from infinity to the appropriate point in the system. This way, electric potential energy represents the work that can be done by the system by converting its potential energy to work. Let’s say that we had charge $q$ and that this charge is a source of electric field $E$ around it. This electric field has electric potential $V$.
Electric potential energy $U$ of a test charge $q_{test}$ at a distance $r$ from charge $q$
will be given as:

$$
U = q_{test} V = dfrac{kqq_{test}}{r}
$$

As it can be seen, electric potential energy does depend on the test charge $q_{test}$.
From the equation above we see that electric potential energy and electric potential would be numerically equal if test charge $q_{test}$ was $1~mathrm{C}$.

Clearly, the difference between electric potential and electric potential energy is that electric potential energy represents the ability of a system to do work, whereas electric potential represents work needed to bring a unit charge ($1~mathrm{C}$) to a distance $r$ from charge $q$.

To show that $mathrm{dfrac{V}{m}}$ is same as $mathrm{dfrac{N}{C}}$, we’ll use the following procedure:

From the fact that work $W$ done by moving charge $q$ through electric potential difference $Delta V$ is equal to:

$$
W = q Delta V
$$

we conclude, by taking a look at the units of the quantities above, that

$$
mathrm{J} = mathrm{C cdot V} ~~~ rightarrow ~~~~ mathrm{V = dfrac{J}{C} }
$$

From the equation above we see that $mathrm{dfrac{V}{m}}$ is equal to:

$$
mathrm{dfrac{V}{m} = dfrac{J}{Cm } }
$$

From the fact that work $W$ done by moving an object to a distance $d$ by acting on it with a force with magnitude $F$, stated as:

$$
W = F d
$$

we conclude, by taking a look at the units of the quantities above, that

$$
mathrm{J = N m }
$$

By plugging in the result from the equation above into the equation for $mathrm{dfrac{V}{m}}$ we see that $mathrm{dfrac{V}{m}}$ equals:

$$
begin{align*}
mathrm{dfrac{V}{m}} &= mathrm{ dfrac{J}{Cm } } \
tag{plug in $mathrm{J = N m } $ } \
mathrm{dfrac{V}{m}} &= mathrm{ dfrac{N m}{Cm } } \
tag{cancel out meters} \
tag{ As it can be seen, we have proven:} \
mathrm{dfrac{V}{m}} &= mathrm{ dfrac{N }{C } }
end{align*}
$$

Hint: Work done by moving charge $q$ through electric potential difference $Delta V$ is equal to:

$$
W = q Delta V
$$

Work $W$ done by moving an object to a distance $d$ by acting on it with a force with magnitude $F$, stated as:

$$
W = F d
$$

We know that in the Millikan oil-drop experiment a whole purpose of the electric field between the plates is to keep charged oil-particles stationary by making the net force acting on the particles zero. The way we do this is by modifying the electric field between the plates so that it completely stops the fall of the oil drops due to gravity. This means that gravitational force $F_g$ acting on the charged oil drop must be of equal magnitude to the electric force $F_e = q E$, but these forces must be in the opposite direction. Since direction of the gravitational force always downward, this means that electric force must act upward. The electric potential between the plates in the Millikan apparatus can be calculated from the electric field $E$ between the plates, separated by a distance $d$ as:

$$
Delta V = E d ~~~ rightarrow ~~~~ E = dfrac{Delta V }{d}
$$

which means that the electric force between the plates is equal to:

$$
F_e = q E = q dfrac{Delta V}{d}
$$

As said, this force is equal in magnitude to the gravitational force $F_g$:

$$
F_g = q dfrac{Delta V}{d} ~~~ rightarrow ~~~~ q = dfrac{F_gd}{ Delta V}
$$

We see that to counteract the gravitational force of oil drop with charge $q$ when plates are separated by a distance $d$, electric potential difference $Delta V$ must be applied. If charge on the oil drop changes and oil drop starts to fall, since its gravitational force $F_g$ can’t change and nor can the distance $d$ between the plates, we must readjust the potential difference between the plates to counteract the gravitational force $F_g$.
Notice that if charge $q$ began to fall, this means that magnitude of the electric force $F_e = q E$ acting upward is now less than the magnitude of the gravitational force $F_g$, which means that charge $q$ is now less than it used to be. In order to push charge $q$ upward, we need to increase the electric field $E$ acting on it, which means that we need to increase the potential difference between the plates.
Notice that this works because electric force is calculated as $F_e = q E = q dfrac{Delta V}{d}$, which means that increasing the voltage between the plates increases the electric force $F_e$ acting on charge $q$.

Hint: We need to counteract gravitational force $F_g$ on the oil drop with electric force $F_e$, which is calculated as:
$$
F_e = q dfrac{Delta V}{d}
$$

We know that in the Millikan oil-drop experiment a whole purpose of the electric field between the plates is to keep charged oil-particles stationary by making the net force acting on the particles zero. The way we do this is by modifying the electric field between the plates so that it completely stops the fall of the oil drops due to gravity. This means that gravitational force $F_g$ acting on the charged oil drop must be of equal magnitude to the electric force $F_e = q E$, but these forces must be in the opposite direction. Since direction of the gravitational force always downward, this means that electric force must act upward. The electric potential between the plates in the Millikan apparatus can be calculated from the electric field $E$ between the plates, separated by a distance $d$ as:

$$
Delta V = E d ~~~ rightarrow ~~~~ E = dfrac{Delta V }{d}
$$

which means that the electric force between the plates is equal to:

$$
F_e = q E = q dfrac{Delta V}{d}
$$

As said, this force is equal in magnitude to the gravitational force $F_g$:

$$
F_g = q dfrac{Delta V}{d} ~~~ rightarrow ~~~~ q = dfrac{F_gd}{ Delta V}
$$

If however changing the electric potential has no effect on the oil drop, this means that there is no electric force acting the oil drop, which is only possible if oil drop is not charged $q= 0 ~~~rightarrow~~~ F_e = 0$, which means that the only force acting on the oil drop is gravitational force and that there is no force counteracting it (except for the force of resistance of air).
This is way oil drop falls (almost) freely.

We are told that capacitance of the capacitor is $C = 0.47 ~mathrm{mu F}$ and that voltage across it is $V = 12 ~mathrm{V}$. To find charge $q$ on the capacitor, we’ll use the definition of capacitance, which states that capacitance of the capacitor $C$ is calculated by dividing the charge $q$ on the capacitor with voltage $V$ across the capacitor:

$$
begin{align*}
C &= dfrac{q}{V} \
tag{express $q$ from the equation above}& \
q &= C V \
tag{Plug in the given values}\
q &= 0.47 ~mathrm{mu F} cdot 12 ~mathrm{V} \
tag{$1 ~mathrm{mu F} = 10^{-6} ~mathrm{F} $} \
q &= 0.47cdot 10^{-6} ~mathrm{F} cdot 12 ~mathrm{V}
end{align*}
$$

$$
boxed{q = 5.64 cdot 10^{-6} ~mathrm{C} }
$$

$$
q = 5.64 cdot 10^{-6} ~mathrm{C}
$$

If a charged object touches another charged object, negative charge will flow from the object with lower potential to an object with higher potential until electric potential on both objects is the same. This is due to a fact that work $W$ is done by moving charge $q$ from a point with potential $V_1$ to a point with potential $V_2$ and this work equals

$$
W = q (V_2 – V_1)
$$

but when electric potential on both of the spheres becomes the same after charges flow from one sphere to the other, the potential difference $V_2 – V_1$ will become zero, which makes the work done on moving the charges zero and thus charges won’t move from one sphere to the other. We conclude that after spheres touch, electric potential on both of them is same.

Keep in mind that we’re told that positively charged sphere is a large sphere, while the other sphere is negatively charged and small. When these two spheres touch, negative charge will flow from the negatively charged sphere to the positively charged sphere, but since charges flow until electric potential is same on both of the spheres, the small negatively charged sphere will become positively charged. In other words, charge on both of the spheres will be positive, but magnitude of the charge on both of the spheres won’t be the same.
Electric potential $V$ on surface of a charged sphere with charge $q$ and radius $R$ is given as:

$$
V = dfrac{kq}{R}
$$

where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant.
Charge on the large positively charged sphere before the two spheres touch is equal to:

$$
V_1 = dfrac{kq_1}{R_1}
$$

while the charge on the negatively charged sphere before the two spheres touch is equal to:

$$
V_2 = dfrac{kq_2}{R_2}
$$

where $R_1$ is radius of the large positively charged sphere, $R_2$ is radius of the small negatively charged sphere ($R_2 > R_1$) and $q_1$ and $q_2$ is charge on the large and small sphere, respectfully.

After the two spheres touch, electric potential of the two spheres $V_{1f}$ and $V_{2f}$ will be the equal:

$$
begin{align*}
V_{1f} &= V_{2f} \
tag{apply the equation for electric potential on surface of a charged sphere} \
dfrac{kq_{1f}}{R_1} &= dfrac{kq_{2f}}{R_2} \
tag{where $q_{1f}$ and $q_{2f}$ is charge on the large and small sphere, respectfully. } \
tag{cancel out $k$}\
dfrac{q_{1f}}{R_1} &= dfrac{q_{2f}}{R_2} \
tag{express charge $q_{1f}$} \
q_{1f} &= q_{2f} dfrac{R_1}{R_2}
end{align*}
$$

Since radii of the two sphere aren’t the same, charge on the two spheres after they touch won’t be of the same magnitude. Also notice that since $R_1$ and $R_2$ are positive, this means that sign of charge on the two spheres after they touch will be the same. If they aren’t the equation above won’t hold.

$$
a) ~ text{ Electric potential on the two spheres is same.}
$$

$$
b) ~ text{Charge on two spheres is of the same sign, but not equal in magnitude.}
$$

$$
text{Click for further explanation.}
$$