In this problem we have a positive charge $q = 2.1 cdot 10^{-9} ~mathrm{C}$ experiencing a force of $F_e = 14 ~mathrm{N}$ at a certain point. We must calculate the magnitude of the electric field that this charge feels.

We know that electric force $F_e$ with which electric field $E$ acts on charge $q$ in the electric field is equal to:

$$
F_e = q E
$$

From the equation above we see that electric field $E$ is calculated as:

$$
begin{align*}
E &= dfrac{F_e}{q} \
tag{plug in the given values} \
E &= dfrac{ 14 ~mathrm{N} }{ 2.1 cdot 10^{-9} ~mathrm{C}}
end{align*}
$$

$$
boxed{ E= 6.67 cdot 10^{9} ~mathrm{dfrac{N}{C}} }
$$

$$
D)~~ E= 6.67 cdot 10^{9} ~mathrm{dfrac{N}{C}}
$$

In this problem we have a positive charge $q = 8.7 ~mathrm{mu C} = 8.7 cdot 10^{-6} ~mathrm{C}$ experiencing a force of $F_e = 8.1 ~mathrm{mu N} =8.1 cdot 10^{-6} ~mathrm{N}$ at a certain point in the electric field $E$. We must calculate the intensity of this electric field.

We know that electric force $F_e$ with which electric field $E$ acts on charge $q$ in the electric field $E$ is equal to:

$$
F_e = q E
$$

From the equation above we see that electric field $E$ is calculated as:

$$
begin{align*}
E &= dfrac{F_e}{q} \
tag{plug in the given values} \
E &= dfrac{ 8.1 ~mathrm{mu N} }{8.7 ~mathrm{mu C}}\
tag{convert the units } \
E &= dfrac{ 8.1cdot 10^{-6} ~mathrm{N} }{ 8.7 cdot 10^{-6} ~mathrm{C}} \
E &= 0.931 ~mathrm{dfrac{N}{C}}
end{align*}
$$

$$
boxed{ E= 9.3 cdot 10^{-1} ~mathrm{dfrac{N}{C}} }
$$

To find the direction of the electric field $E$, let’s remember that for a positive test charge $q$, direction of the electric field is same as direction of the electric force $F_e$. Since we’re told that force acts at an angle of $24^o$ N of E, we conclude that the electric field also acts at an angle of $24^o$ N of E on charge $q$.

$$
boxed{ D)~~ E= 9.3 cdot 10^{-1} ~mathrm{dfrac{N}{C}}, ~ 24^o ~text{N of E} }
$$

$$
D)~~ E= 9.3 cdot 10^{-1} ~mathrm{dfrac{N}{C}}, ~ 24^o ~text{N of E}
$$

What we need to understand is that electric field in this problem is uniform, which means that its magnitude is same in every point between the plates and equals
$E =4.8 cdot 10^3 ~mathrm{dfrac{N}{C}}$. To calculate the potential difference $Delta V$ between the plates that are $d =18 ~mathrm{cm} = 0.18 ~mathrm{m}$ apart, let’s remember that this potential difference can be calculated as a product of magnitude of the uniform electric field and distance $d$ between the plates:

$$
begin{align*}
Delta V &= E d \
tag{plug in the given values} \
Delta V &= 4.8 cdot 10^3 ~mathrm{dfrac{N}{C}} cdot 0.18 ~mathrm{m}
end{align*}
$$

$$
boxed{ Delta V = 864 ~mathrm{V} }
$$

$$
C)~~ Delta V = 0.86 ~mathrm{kV}
$$

In this problem we have a proton moving from positive plate to a negative plate of a capacitor with an electric field
$E = 125 ~mathrm{dfrac{N}{C}}$ between its plates. Distance between the plates is $d= 4.3 ~mathrm{cm}$. Charge of the proton ($q_p$) is equal to elementary charge $q_p =e = 1.6 cdot 10^{-19} ~mathrm{C}$. As we can see, charge of proton is positive and thus we need to do work $W$ to move proton from negative plate to a positive plate of the capacitor. This work $W$ needs to be calculated.

Let’s remember that work $W$ done on charge $q$ to move it from a point with electric potential $V_1$ to a point with electric potential $V_2$ is given as:

$$
W = q (V_2 – V_1) = q Delta V
$$

where $Delta V = V_2 – V_1$ is electric potential difference between the end point and starting point of travel of this charge $q$. Of course, since there is electric field $E$ between the plates of the capacitor, there is also electric potential $V$ between the plates.

As said, there is electric potential between the plates of the capacitor and more importantly, there is electric potential difference $Delta V$ between the plates of the capacitor. Knowing that electric field $E$ between the plates of the capacitor is uniform, we can apply the equation that shows connection between a uniform electric field $E$ and electric potential difference:

$$
V_2 – V_1 = Delta V = E d
$$

where $d$ is distance between points with electric potential $V_2$ and $V_1$. In our case, this is distance between the plates.

We can plug in this expression for $Delta V$ from the equation above into equation for work done on charge $q$ and have:

$$
W = q (V_2 – V_1) = q E d
$$

Given that we want to find work done on proton, we’ll plug in charge of proton $q_p = e$ into the equation above and have:

$$
W = e E d
$$

As we can see from the equation above, we have everything we need to calculate the work $W$ done on proton to move it from negative to positive plate of this capacitor. Let’s plug in the given values into the equation above and calculate the work $W$:

$$
begin{align*}
W &= e E d \
tag{plug in the values} \
W &= 1.6 cdot 10^{-19} ~mathrm{C} cdot 125 ~mathrm{dfrac{N}{C}} cdot 4.3 ~mathrm{cm} \
tag{$1mathrm{~cm} = 0.01 ~mathrm{m} $} \
W &= 1.6 cdot 10^{-19} ~mathrm{C} cdot 125 ~mathrm{dfrac{N}{C}} cdot 4.3 cdot 0.01 ~mathrm{m}
W &= 8.6 cdot 10^{-19} ~mathrm{J}
end{align*}
$$

$$
boxed{ B)~~ W = 8.6 cdot 10^{-19} ~mathrm{J} }
$$

$$
B)~~ W = 8.6 cdot 10^{-19} ~mathrm{J}
$$

What we need to understand is that electric field in Millikan apparatus is uniform, which means that its magnitude is same in every point between the plates and equals
$E$. Electric potential difference between the plates in the Millikan apparatus can be measured using a Voltmeter. This potential difference $Delta V$ is constant and it be can be calculated as a product of magnitude of the uniform electric field $E$ and distance $d$ between the plates, which is known from the construction of the apparatus:

$$
begin{align*}
Delta V &= E d \
tag{express $E$ from the equation above} \
E &= dfrac{Delta V}{d}
end{align*}
$$

As it can be seen, by measuring electric potential difference $Delta V$ and knowing distance $d$ between the plates, we see that we can find electric field between the plates in the Millikan apparatus.

$$
boxed{ text{ B)~~ from the electric potential between the plates } }
$$

$$
text{ B)~~ from the electric potential between the plates }
$$

In an oil drop experiment, we have a charged oil drop in space between the charged plates. Electric field produced by the charged acts on the charged particle with an electric force $F_e$ and counteracts its weight $Q$. In other words, we have two forces acting on the oil drop – its weight $Q$ acting downward and electric force $F_e$ acting upward. When this charged oil drop is motionless, it means that these two forces are equal in magnitude and in the opposite direction, stated in the following equation:

$$
Q = F_e
$$

Note that due to construction of Millikan apparatus, oil particle will be negatively charged. In our solution, we’ll proceed to calculate the magnitude of this negative charge.
In our case, we have an oil drop with weight $Q = 1.9 cdot 10^{-14} ~mathrm{N}$, motionless between the plates with separation $d = 63 ~mathrm{mm}$ between the plates and electric potential difference between the plates $Delta V = 0.78 ~mathrm{kV}$.

We know that the electric force $F_e$ with which electric field $E$ acts on a charged particle $q$ is equal to:

$$
F_e =|q|E
$$

where $|q|$ is magnitude of charge $q$.
We can plug in this expression for $F_e$ into the first equation in our solution and have:

$$
begin{align*}
Q &= F_e \
Q &=|q|E tag{1}
end{align*}
$$

As we can see, magnitude of the electric field $E$ between the plates is unknown, but knowing that electric field $E$ between in Millikan apparatus is uniform, we can apply the equation that shows connection between a uniform electric field $E$ and electric potential difference:

$$
V_2 – V_1 = Delta V = E d
$$

where $d$ is distance between points with electric potential $V_2$ and $V_1$. In our case, this is distance between the plates.
In other words, we can express magnitude of the electric field $E$ from the equation above as:

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

We can plug in this expression into equation $(1)$:

$$
begin{align*}
Q &=|q|E \
Q &=|q|dfrac{Delta V}{d} tag{2}
end{align*}
$$

Notice that we all of the quantities in the equation above are known, except charge $q$ of the oil drop, which we can express from the equation above:

$$
begin{align*}
Q &=|q|dfrac{Delta V}{d} \
tag{multiply equation by $d $} \
Q d &=|q|Delta V \
tag{express $|q|$ from the equation above} \
|q| &= dfrac{Qd}{Delta V} \
tag{plug in the values} \
|q|&= dfrac{1.9 cdot 10^{-14} ~mathrm{N} cdot 63 ~mathrm{mm}}{0.78 ~mathrm{V} } \
tag{$1 ~mathrm{mm} = 10^{-3} ~mathrm{m} $} \
tag{$1 ~mathrm{kV} = 10^{3} ~mathrm{V} $} \
|q|&= dfrac{1.9 cdot 10^{-14} ~mathrm{N} cdot 63 cdot 10^{-3} ~mathrm{m}}{0.78 cdot 10^3 ~mathrm{V} } \
|q|&= 1.53 cdot 10^{-18} ~mathrm{C}
end{align*}
$$

As said, oil drop in the Millikan apparatus is negatively charged and thus our final solution is:

$$
boxed{ A)~~ q = -1.5 cdot 10^{-18} ~mathrm{C} }
$$

$$
A)~~ q = -1.5 cdot 10^{-18} ~mathrm{C}
$$

We are told that capacitance $C$ of the capacitor is $C = 0.093 ~mathrm{mu F}$ and that charge $q$ on the capacitor is $q = 58 ~mathrm{ mu C}$. To calculate the potential difference $V$ across 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:

$$
begin{align*}
C &= dfrac{q}{V} \
tag{express $V$ from the equation above} \
V &= dfrac{q}{C} \
tag{Plug in the given values}\
V &= dfrac{ 58 ~mathrm{ mu C} }{ 0.093 ~mathrm{mu F} } \
%
%
tag{$1 ~mathrm{mu F} = 10^{-6} ~mathrm{F} $} \
tag{$1 ~mathrm{mu C} = 10^{-6} ~mathrm{C} $} \
%
%
V &= dfrac{ 58 cdot 10^{-6} ~mathrm{C} }{ 0.093 cdot 10^{-6} ~mathrm{F}} \
V &= 623.656 ~mathrm{V }
end{align*}
$$

$$
boxed{ C)~~ V = 6.2 cdot 10^2 ~mathrm{V } }
$$

$$
C)~~ V = 6.2 cdot 10^2 ~mathrm{V }
$$

We’re told that there are $N =18$ extra electrons on the oil drop. Weight of the oil drop is $F_g = 6.12 cdot 10^{-14} ~mathrm{N}$. Plates in the Millikan apparatus are at a distance of $d = 14.1 ~mathrm{mm}$ from each other.

To calculate the charge on the oil drop, we need to understand that any charge $q$ is equal to an integral multiple elementary charge $e$, stated as:

$$
q = N e
$$

Charge of the electron is equal to a negative value of elementary charge $q_e = – e= – 1.6 cdot 10^{-19} ~mathrm{C}$.
From the equation above we conclude that charge $q$ on the oil drop is equal to:

$$
begin{align*}
q &= N q_e \
tag{oil drop has $N = 18$ extra electrons} \
q &= 18 cdot q_e \
tag{$q_e = – e= – 1.6 cdot 10^{-19} ~mathrm{C}$}\
q &= 18 cdot (- 1.6) cdot 10^{-19} ~mathrm{C} \
q &= – 28.8 cdot 10^{-19} ~mathrm{C}
end{align*}
$$

$$
boxed{ q = – 28.8 cdot 10^{-19} ~mathrm{C} }
$$

Magnitude of the electric field $E$ through which oil drop particle is suspended in the apparatus is unknown.
What we need to understand is that electric field in Millikan apparatus is uniform, which means that its magnitude is same in every point between the plates and equals
$E$. Electric potential difference $Delta V$ between the plates in the Millikan apparatus can be calculated as a product of magnitude of the uniform electric field $E$ and distance $d$ between the plates, stated as:

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

Now let’s remember that Millikan apparatus was set up in a way that the net force acting on the charged particle is zero, which means that weight $F_g$ of the particle is equal in magnitude to the electric force $F_e$ that the particle feels, but opposite in the direction.

Electric force $F_e$ that the particle feels can be calculated as a product of electric field $E$ through which particle is suspended and charge $q$ on the particle, stated as:

$$
F_e = q E
$$

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

$$
begin{align*}
F_e &= F_g \
tag{Plug in $F_e = qE$} \
qE &= F_g \
tag{plug in $E$ from the expression above, $ E = dfrac{Delta V}{d} $} \
dfrac{q Delta V}{d} &= F_g \
tag{express $Delta V$ from the equation above} \
Delta V &= dfrac{F_g d }{q} \
tag{plug in the given values} \
Delta V &= dfrac{ 6.12 cdot 10^{-14} ~mathrm{N} cdot 14.1 ~mathrm{mm} }{- 28.8 cdot 10^{-19} ~mathrm{C}} \
tag{$ 1 ~mathrm{mm} = 1 cdot 10^{-3} ~mathrm{m} $ } \
Delta V &= dfrac{ 6.12 cdot 10^{-14} ~mathrm{N} cdot 14.1 cdot 10^{-3} ~mathrm{m} }{- 28.8 cdot 10^{-19} ~mathrm{C}}
end{align*}
$$

$$
boxed{ Delta V = -299.625 ~mathrm{V} }
$$

Note that the negative sign on potential difference means that polarization of the plates must be reversed so that this charge is balanced.

$$
q = – 28.8 cdot 10^{-19} ~mathrm{C}
$$

$$
Delta V = -299.625 ~mathrm{V}
$$