We can use equation for difference in electric potential:
$$Delta V=Ed$$

When we put numbers in we get:
$$Delta V=6000cdot0.05$$
$$boxed{Delta V=300,,rm V}$$

$$Delta V=300,,rm V$$

We can use equation for difference in electric potential:
$$Delta V=Edrightarrow E=frac{Delta V}{d}$$

When we put numbers in we get:
$$E=frac{400}{0.02}$$
$$boxed{E=20000,,rm N/C}$$

$$E=20000,,rm N/C$$

What we need to understand is that electric field between two parallel plates is uniform, which means that its magnitude is same in every point between the plates and equals
$E = 2.5 cdot 10^3 ~mathrm{dfrac{N}{C}}$. To calculate the potential difference $Delta V$ between the plates that are $d = 0.2 ~mathrm{m}$ apart, let’s remember that this potential difference can be calculated as a product of magnitude of the uniform electric field $E$ and distance $d$ between the plates:

$$
begin{align*}
Delta V &= E d \
Delta V &= 2.5 cdot 10^3 ~mathrm{dfrac{N}{C}} cdot 0.2 ~mathrm{m}
end{align*}
$$

$$
boxed{ Delta V = 500 ~mathrm{V} }
$$

$$
Delta V = 500 ~mathrm{V}
$$

In this problem we have difference in potential between two parallel plates, we have to calculate distance between them if we know that electric field is $4.25cdot10^{3}$ N/C.

We can use equation for difference in potential:
$$Delta V=Edrightarrow d=frac{Delta V}{E}$$

When we put numbers in we get:
$$d=frac{125 }{4.25cdot10^{3}}$$
$$boxed{d=0.0294,,rm m}$$

$$d=0.0294,,rm m$$

We can use equation for difference in potential:
$$Delta V=Edrightarrow E=frac{Delta V}{d}$$

When we put numbers in we get:
$$E=frac{275}{0.035}$$
$$boxed{E=7857,,rm N/C}$$

$$E=7857,,rm N/C$$

We can use equation for work on charge in electric field:
$$W=qDelta V$$

When we put numbers in we get:
$$W=3cdot1.5$$
$$W=4.5,,rm J$$

$$W=4.5,,rm J$$

We are told that the car battery, at voltage $V= 12 ~mathrm{V}$, can store $q = 1.44 cdot 10^6 ~mathrm{C}$ of charge. To calculate how much work can be done by the battery, we need to understand that work done by the battery will be done by moving charge $q$ through a potential difference $Delta V$ on these two quantities. Work $W$ done on charge $q$ while it moved through a potential difference $Delta V$ is calculated as:

$$
begin{align*}
W &= q Delta V \
tag{Plug in the given values:} \
W &= 1.44 cdot 10^6 ~mathrm{C}cdot 12~mathrm{V} \
W &= 17.28 cdot 10^6 ~mathrm{J}
end{align*}
$$

$$
boxed{ W = 17.28 ~mathrm{MJ} }
$$

$$
W = 17.28 ~mathrm{MJ}
$$

We are told that an electron moved through a potential difference of
$Delta V = 18000 ~mathrm{V}$. To calculate the work done on an electron, let’s remember that work done by the electric field only depends on the electric potential between these two points, which is $Delta V$ and magnitude of the charge $q$ that moved through this potential difference. Magnitude of the charge on an electron is equal to elementary charge $e = 1.6 cdot 10^{-19} ~mathrm{C}$.
Work $W$ done on charge $q$ while it moved through a potential difference $Delta V$ is calculated as:

$$
begin{align*}
W &= q Delta V \
tag{Charge on the electron is $e$} \
W &= e Delta V \
tag{Plug in the given values:} \
W &= 1.6 cdot 10^{-19} ~mathrm{C} cdot 18000 ~mathrm{V}
end{align*}
$$

$$
boxed{ W = 2.88 cdot 10^{-15} ~mathrm{J} }
$$

$$
W = 2.88 cdot 10^{-15} ~mathrm{J}
$$

What we need to understand is that electric field $E$ between two parallel plates is uniform, which means that its magnitude is same in every point between the plates. Potential difference $Delta V$ between the plates is calculated in problem 18 and equals
$Delta V = 500 ~mathrm{V}$. Distance between the plates is $d = 2.4 ~mathrm{cm} = 0.024 ~mathrm{m}$. Now let’s remember that this potential difference can be calculated as a product of magnitude of the uniform electric field $E$ and distance $d$ between the plates:

$$
begin{align*}
Delta V &= E d \
tag{Express $E$ from the equation above} \
E &= dfrac{Delta V}{d} \
tag{Plug in the given values}\
E &= dfrac{ 500 ~mathrm{V} }{0.024 ~mathrm{m} }
%
end{align*}
$$

$$
boxed{ E= 20833.3333 ~mathrm{dfrac{N}{C}} }
$$

$$
E= 20833.3333 ~mathrm{dfrac{N}{C}}
$$

We are told that magnitude of the electric field in particle accelerator is $E = 4.5 cdot 10^5 ~mathrm{dfrac{N}{C}}$. Since we’re only given magnitude of this electric field, we can safely assume that this electric field is uniform, which means that its magnitude is same in every point. To calculate the work done to move a proton for a distance $d = 25 ~mathrm{cm} = 0.25 ~mathrm{m}$, we’ll first need to find a potential difference $Delta V$ between the starting point and an end point of the protons travel. Charge of the proton is equal to elementary charge $q_p = e = 1.6 cdot 10^{-19} ~mathrm{C}$

What we need to understand is that if electric field $E$ is uniform, potential difference $Delta V$ between the two two points at a distance $d$ from each other is calculated as:

$$
begin{align*}
Delta V &= E d \
Delta V &= 4.5 cdot 10^5 ~mathrm{dfrac{N}{C}} cdot 0.25 ~mathrm{m} \
Delta V &= 112500 ~mathrm{V}
end{align*}
$$

Now that we have found the potential difference between the starting point and an end point of the protons travel, we can calculate work done on the proton if it travels distance $d$ in an electric field with the given magnitude.

To calculate the work done on an proton, let’s remember that work done by the electric field only depends on the electric potential between these two points, which is $Delta V$ and magnitude of the charge $q$ that moved through this potential difference. Magnitude of the charge on a proton is equal to elementary charge $e = 1.6 cdot 10^{-19} ~mathrm{C}$.
Work $W$ done on charge $q$ while it moved through a potential difference $Delta V$ is calculated as:

$$
begin{align*}
W &= q Delta V \
tag{Charge on the proton is $e$} \
W &= e Delta V \
tag{Plug in the given values:} \
W &= 1.6 cdot 10^{-19} ~mathrm{C} cdot 18000 ~mathrm{V}
end{align*}
$$

$$
boxed{ W = 1.8 cdot 10^{-14} ~mathrm{J} }
$$

$$
W = 1.8 cdot 10^{-14} ~mathrm{J}
$$

In this problem we have an oil drop in the Millikan apparatus. Weight of the oil drop is equal to the gravitational force it feels due to Earth and equals
$F_g = 1.9 cdot 10^{-15} ~mathrm{N}$. Magnitude of the electric field through which oil drop particle is suspended in the apparatus is
$E= 6 cdot 10^3 ~mathrm{dfrac{N}{C}}$.
To calculate charge on the oil drop particle, 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:

$$
begin{align*}
F_e &= F_g
tag{Plug in $F_e = qE$} \
qE &= F_g \
tag{Express $q$ from the equation above} \
q &= dfrac{F_g}{E} \
q &= dfrac{ 1.9 cdot 10^{-15} ~mathrm{N} }{6 cdot 10^3 ~mathrm{dfrac{N}{C}} }\
q &= 3.166 cdot 10^{-19} ~mathrm{C}
end{align*}
$$

$$
boxed{q approx 3.2 cdot 10^{-19} ~mathrm{C}}
$$

We’ve calculated the charge on the oil drop. To calculate the number of excess electrons, we need to understand that any charge $q$ is equal to an integral multiple of charge of the electron $e$, stated as:

$$
q = Ne
$$

Charge of the electron is elementary charge $e = 1.6 cdot 10^{-19} ~mathrm{C}$.
From the equation above we conclude:

$$
begin{align*}
q &= N e \
tag{express $N$ from the equation above:}\
N &= dfrac{q}{e} \
N &= dfrac{3.2 cdot 10^{-19} ~mathrm{C}}{1.6 cdot 10^{-19} ~mathrm{C}}
end{align*}
$$

$$
boxed{ N = 2}
$$

$$
q approx 3.2 cdot 10^{-19} ~mathrm{C}
$$

$$
N = 2
$$

In this problem we have an oil drop in the Millikan apparatus. Weight of the oil drop is equal to the gravitational force it feels due to Earth and equals
$F_g = 6.4 cdot 10^{-15} ~mathrm{N}$. Magnitude of the electric field $E$ through which oil drop particle is suspended in the apparatus is
unknown, but charge $q$ on the oil drop particle can be calculated from the fact that it carries one excess electron. This tells us that the particle is negatively charged and that magnitude of its charge is equal to elementary charge $e = 1.6 cdot 10^{-19} ~mathrm{C}$. This menas that charge $q$ on the particle is equal to:

$$
q = -e = -1.6 cdot 10^{-19} ~mathrm{C}
$$

To calculate the electric field needed for the particle to be motionless, 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:

$$
begin{align*}
F_e &= F_g \
tag{Plug in $F_e = qE$} \
qE &= F_g \
tag{Express $E$ from the equation above} \
E &= dfrac{F_g}{q} \
E &= dfrac{ 6.4 cdot 10^{-15} ~mathrm{N} }{ -1.6 cdot 10^{-19} ~mathrm{C} }
end{align*}
$$

$$
boxed{ E = -40 000 ~mathrm{dfrac{N}{C}} }
$$

Minus sign in the result above comes from the fact that the direction of the electric force $F_e$ acting on the oil drop particle is in the opposite direction to the electric field acting on the oil drop particle. This is due to a fact that charge on the oil drop particle is negative.

$$
E = -40 000 ~mathrm{dfrac{N}{C}}
$$

In this problem we have an oil drop in the Millikan apparatus. Weight of the oil drop is equal to the gravitational force it feels due to Earth and equals
$F_g = 1.2 cdot 10^{-14} ~mathrm{N}$.
Potential difference $Delta V$ between the plates in the Millikan apparatus is $Delta V = 240 ~mathrm{V}$ and plates are separated by a distance $d = 0.64 ~mathrm{cm} = 0.0064 ~mathrm{m}$. Given this data and knowing that electric field between the plates is uniform, we can calculate the electric field between the plates of the Millikan apparatus knowing that potential difference $Delta V$ can be calculated as:

$$
begin{align*}
Delta V &= E d \
tag{express $E$ from the equation above:} \
E &= dfrac{Delta V}{d} \
E &= dfrac{ 240 ~mathrm{V} }{0.0064 ~mathrm{m}} \
E &= 37500 ~mathrm{dfrac{N}{C}}
end{align*}
$$

Magnitude of the electric field $E$ through which oil drop particle is suspended in the apparatus is
calculated above, while charge $q$ on the oil drop particle is unknown, but it can be calculated knowing 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:

$$
begin{align*}
F_e &= F_g \
tag{Plug in $F_e = qE$} \
qE &= F_g \
tag{Express $q$ from the equation above} \
q &= dfrac{F_g}{E} \
q &= dfrac{ 1.2 cdot 10^{-14} ~mathrm{N} }{ 37500 ~mathrm{dfrac{N}{C}} } \
q &= 3.2 cdot 10^{-19} ~mathrm{C}
end{align*}
$$

We’ve calculated the charge on the oil drop. To calculate the number of electrons missing, we need to understand that any charge $q$ is equal to an integral multiple of charge of the electron $e$, stated as:

$$
q = Ne
$$

Charge of the electron is elementary charge $e = 1.6 cdot 10^{-19} ~mathrm{C}$.
From the equation above we conclude:

$$
begin{align*}
q &= N e \
tag{express $N$ from the equation above:}\
N &= dfrac{q}{e} \
N &= dfrac{3.2 cdot 10^{-19} ~mathrm{C}}{1.6 cdot 10^{-19} ~mathrm{C}}
end{align*}
$$

$$
boxed{ N = 2}
$$

$$
N = 2
$$

We are told that capacitance of the capacitor is $C = 27 ~mathrm{mu F}$ and that voltage across it is $V = 45 ~mathrm{V}$. 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:

$$
begin{align*}
C &= dfrac{q}{V} \
tag{express $q$ from the equation above}& \
q &= C V \
tag{Plug in the given values}\
q &= 27 ~mathrm{mu F} cdot 45 ~mathrm{V} \
tag{$1 ~mathrm{mu C} = 10^{-6} ~mathrm{C} $} \
q &= 27 cdot 10^{-6} ~mathrm{C} cdot 45 ~mathrm{V}
end{align*}
$$

$$
boxed{q = 1.215 cdot 10^{-3} ~mathrm{C} }
$$

$$
q = 1.215 cdot 10^{-3} ~mathrm{C}
$$

We are told that capacitance of the first capacitor is $C_1 = 3.3 ~mathrm{mu F}$ , while the capacitance of the second capacitor is $C_2 = 6.8 ~mathrm{mu F}$. Voltage across each of the capacitors is $V = 24 ~mathrm{V}$.

To compare charge $q_1$ on the first capacitor and charge $q_2$ on the second 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:

$$
C = dfrac{q}{V}
$$

We can express charge $q$ on the capacitor as:

$$
q = C V
$$

As it can be seen, charge $q$ on the capacitor is proportional to the capacitance $C$ of the capacitor and voltage $V$ across it. Since both of the capacitors are connected to same voltage, the one with greater capacitance will have more charge stored on it. This means that capacitor $C_2$ will have more charge stored on it. We can prove this by applying the equation above to both of the capacitors. We find that charge $q_1$ on the first capacitor is equal to:

$$
begin{align*}
q_1 &= C_1 V \
tag{Plug in the given values}\
q_1 &= 3.3 ~mathrm{mu F} cdot 24 ~mathrm{V} \
tag{$1 ~mathrm{mu C} = 10^{-6} ~mathrm{C} $} \
q_1 &= 3.3 cdot 10^{-6} ~mathrm{C} cdot 24 ~mathrm{V}
end{align*}
$$

$$
boxed{q_1 = 79.2 cdot 10^{-6} ~mathrm{C} }
$$

We find that charge $q_2$ on the second capacitor is equal to:

$$
begin{align*}
q_2 &= C_2 V \
tag{Plug in the given values}\
q_2 &= 6.8 ~mathrm{mu F} cdot 24 ~mathrm{V} \
tag{$1 ~mathrm{mu C} = 10^{-6} ~mathrm{C} $} \
q_2 &= 6.8 cdot 10^{-6} ~mathrm{C} cdot 24 ~mathrm{V}
end{align*}
$$

$$
boxed{ q_2 = 163.2 cdot 10^{-6} ~mathrm{C} }
$$

Clearly, $q_2 > q_1$.

Hint: Charge on the capacitor is proportional to its capacitance.

$$
q_2 = 163.2 cdot 10^{-6} ~mathrm{C}
$$

We are told that capacitance of the first capacitor is $C_1 = 3.3 ~mathrm{mu F}$ , while the capacitance of the second capacitor is $C_2 = 6.8 ~mathrm{mu F}$. Charge on each of the capacitors is $q = 3.5 cdot 10^{-4} ~mathrm{C}$.

To compare voltage $V_1$ across the first capacitor and voltage $V_2$ across the second 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:

$$
C = dfrac{q}{V}
$$

We can express voltage $V$ on the capacitor as:

$$
V = dfrac{q}{C}
$$

As it can be seen, voltage $V$ on the capacitor is inversely proportional to the capacitance $C$ of the capacitor and proportional to charge $q$ on the capacitor. Since charge on each of the capacitors is same, the one with lesser capacitance will be connected to a greater voltage and vice versa.

This means that voltage on capacitor $C_1$ will be greater. We can prove this by applying the equation above to both of the capacitors. We find that voltage $V_1$ on the first capacitor is equal to:

$$
begin{align*}
V_1 &= dfrac{q}{C_1} \
tag{Plug in the given values}\
V_1 &= dfrac{ 3.5 cdot 10^{-4} ~mathrm{C}}{ 3.3 ~mathrm{mu F} } \
tag{$1 ~mathrm{mu C} = 10^{-6} ~mathrm{C} $} \
V_1 &= dfrac{ 3.5 cdot 10^{-4} ~mathrm{C}}{ 3.3 cdot 10^{-6} ~mathrm{C} } \
end{align*}
$$

$$
boxed{V_1 = 106.0606 ~mathrm{V} }
$$

We find that voltage $V_2$ on the second capacitor is equal to:

$$
begin{align*}
V_2 &= dfrac{q}{C_2} \
tag{Plug in the given values}\
V_2 &= dfrac{ 3.5 cdot 10^{-4} ~mathrm{C}}{ 6.8 ~mathrm{mu F} } \
tag{$1 ~mathrm{mu C} = 10^{-6} ~mathrm{C} $} \
V_2 &= dfrac{ 3.5 cdot 10^{-4} ~mathrm{C}}{ 6.8cdot 10^{-6} ~mathrm{C} } \
end{align*}
$$

$$
boxed{V_2 = 51.4706 ~mathrm{V} }
$$

Clearly, $V_1 > V_2$.

Hint: Voltage across the capacitor is inversely proportional to its capacitance.

$$
V_1 = 106.0606 ~mathrm{V}
$$

We are told that capacitance of the capacitor is $C = 2.2 ~mathrm{mu F}$. Initial voltage across the capacitor is $V_i = 6 ~mathrm{V}$, whereas final voltage across the capacitor after additional charging is $V_f = 15 ~mathrm{V}$. To find the additional charge needed for voltage across the capacitor to increase from $V_i$ to $V_f$,
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:

$$
C = dfrac{q}{V}
$$

We can express charge $q$ on the capacitor as:

$$
q = C V
$$

As it can be seen, initial charge $q_i$ on the capacitor is equal to:

$$
q_i = C V_i
$$

while charge on the capacitor $q_f$ in the final state is equal to:

$$
q_f = C V_f
$$

This means that additional charge needed to increase voltage across the capacitor from its initial value to its final value can be calculated as:

$$
begin{align*}
Delta q &= q_f – q_i \
Delta q &= C V_f – C V_i \
Delta q &= C (V_f – V_i) \
Delta q &= 2.2 ~mathrm{mu F} cdot ( 15 ~mathrm{V}- 6 ~mathrm{V} ) \
tag{$1 ~mathrm{mu F} = 10^{-6} ~mathrm{F} $} \
Delta q &= 2.2 cdot 10^{-6} ~mathrm{F} cdot 9 ~mathrm{V} \
Delta q &= 19.8 cdot 10^{-6} ~mathrm{C}
end{align*}
$$

$$
boxed{ q = 19.8 ~mathrm{mu C}}
$$

$$
q = 19.8 ~mathrm{mu C}
$$

Initial voltage across the capacitor is $V_i = 12 ~mathrm{V}$, whereas final voltage across the capacitor after additional charging is $V_f = 14.5 ~mathrm{V}$.
Voltage across the capacitor increased because additional charge
$Delta q = q_f – q_i = 2.5 ~cdot 10^{-5} ~mathrm{C}$ was added to the capacitor
To find the capacitance of 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:

$$
C = dfrac{q}{V}
$$

We can express charge $q$ on the capacitor as:

$$
q = C V
$$

As it can be seen, initial charge $q_i$ on the capacitor is equal to:

$$
q_i = C V_i
$$

while charge on the capacitor $q_f$ in the final state is equal to:

$$
q_f = C V_f
$$

This means that additional charge needed to increase voltage across the capacitor from its initial value to its final value can be calculated as:

$$
begin{align*}
Delta q &= q_f – q_i \
Delta q &= C V_f – C V_i \
Delta q &= C (V_f – V_i) \
tag{express $C$ from the equation above} \
C &= dfrac{Delta q}{V_f – V_i} \
C&= dfrac{ 2.5 ~cdot 10^{-5} ~mathrm{C} }{14.5 ~mathrm{V} – 12 ~mathrm{V}} \
C &= 1 cdot 10^{-5} ~mathrm{F} \
C &= 10 cdot 10^{-6} ~mathrm{F}
end{align*}
$$

$$
boxed{ C = 10 ~mathrm{mu F}}
$$