All Solutions
Page 584: Assessment
$$
W = q Delta V
$$
we conclude, by taking a look at the units of the quantities in the equation above, that
$1 ~mathrm{V} = 1 mathrm{dfrac{J}{C}}$, which means that correct answer for this field is $mathrm{dfrac{J}{C}}$.
Keep in mind that from the work $W$ done by the electric field, by moving a test charge $q$ through an electric potential difference $Delta V$ is equal to
$$
W = q Delta V
$$
we conclude that
correct answer for this field is work.
Electric field strength is measured in $mathrm{dfrac{N}{C}}$.
Work $W$ done by the electric field is calculated as $W = q Delta V$.
the source of the electric field $E$.
Direction of the electric field is defined so that electric field lines exit the positive charge, which means that direction of the electric field coming from a positively charged particle is in outward direction from the positively charged particle, whereas electric field lines enter the negatively charged particle, which means that direction of the electric field coming from a negatively charged particle is in inward direction, pointing to the negatively charged particle.
To sum this up, direction of the electric field is defined in the outward direction from the positive charges and inward direction to the negative charges.
out in space surrounding the source of the electric field and these lines are drawn so that direction of the electric field in any point is in the direction tangent to the electric field line. Electric field lines exit the positive charge, spread out in all directions and enter the negative charge.
out in space surrounding the source of the electric field and these lines are drawn so that direction of the electric field in any point is in the direction tangent to the electric field line. Electric field lines exit the positive charge, spread out in all directions and enter the negative charge.
1) Denser field lines represent greater magnitude of the electric field and vice versa.
2) Electric field lines point from the positively charged particle and to the negatively charged particle.
3) Field lines should not intersect.
4) Like charges repel, while unlike charges attract, which means that there is no field lines directly between two like charges, while in case of unlike charges, field lines are densest in space between unlike charges.
5) The greater the magnitude of charge, the greater the density of field lines around it.
6) In general case, field lines are curved lines because the tangent to the electric field at a given point should represent the direction of the electric field at that point.
a)~~
$$
In this case we have two like charges of equal magnitude. As said, like charges repel, so there is no electric field lines on the line directly between them. Since both charges are of the equal magnitude, density of the field lines around them is the same.
Note that the figure below represents two positive charges, but it looks almost identical to the case of two negative charges, the only difference being the different direction of the field lines.
In this case we have two unlike charges of equal magnitude. As said, since these charges attract, field lines are the densest between the two charges. We can also see that the field lines start from the positive charge and point to the negative charge. Since both charges are of the equal magnitude, density of the field lines around them is the same, but field lines point away from the positive charge and point toward the negative charge.
In general case, field lines are curved lines because the tangent to the electric field at a given point should represent the direction of the electric field at that point.
Click for further explanation.
$$
E= dfrac{kq }{r^2}
$$
where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant.
This means that electric field lines have to options where they can end. They either spread out to some finite distance and reach a negative charge to which they point to or they spread out to infinity where field gets weaker and weaker, until infinity, where field is finally zero and where electric field lines stop because magnitude of the electric field is zero.
Electric potential $V$ is measured in Volts ($mathrm{V}$), which means that potential difference $Delta V$, which is just a difference in electric potential between two points is also measured in Volts.
$$
W = q Delta V
$$
As work $W$ is done on charge $q$, its potential energy $U$ increases by the same amount which means that change in the potential energy $Delta U$ is equal to
$$
Delta U = q Delta V
$$
In an uniform electric field, potential difference $Delta V$ is equal to a product of electric field $E$ and distance $d$ between points with potentials $V_2$ and $V_1$, stated as
$$
begin{equation}
Delta V = E d
end{equation}
$$
This means that in case of unit displacement ($d = 1~mathrm{m}$) of the charge $q$ in a uniform electric field with unit magnitude bigg($E = 1~mathrm{dfrac{N}{C}}$bigg), change in the electric potential is equal to $Delta V = 1 ~mathrm{V}$.
Also, we can express potential difference $Delta V$ from the equation $Delta U = q Delta V$ as
$$
Delta V = dfrac{Delta U }{q}
$$
This means that if a unit charge ($q = 1~mathrm{C}$) changes its potential energy by 1 unit of energy ($Delta U= 1~mathrm{J}$), we have:
$$
begin{align*}
Delta V &= dfrac{Delta U }{q} tag{2} \
Delta V &= dfrac{ 1~mathrm{J} }{ 1~mathrm{C}} \
Delta V &= 1~mathrm{V}
end{align*}
$$
From the two results above we see that definition of ($1~mathrm{V}$) can be seen from
$$
Delta V = dfrac{Delta U }{q} = E d
$$
We see that $1~mathrm{V}$ is defined as a unit change in the potential energy ($Delta U= 1~mathrm{J}$) of a unit charge ($q = 1~mathrm{C}$) while it moved a unit distance ($d = 1~mathrm{m}$) through a uniform electric field with unit magnitude bigg($E = 1~mathrm{dfrac{N}{C}}$bigg).
If this sphere touches the ground or is connected to ground through a wire, charges will flow until electric potential on both of the objects (Earth and sphere) is the same. This is due to a fact that work $W$ is done by moving charge $q$ from a point with potential $V_1$ to a point with potential $V_E$ and this work equals
$$
W = q (V_2 – V_1)
$$
but when electric potential on surface of the Earth becomes equal to the electric potential on the charged sphere, the potential difference $V_2 – V_1$ will become zero, which makes the work done on moving the charges zero and thus charges won’t move from one object to the other.
Electric potential $V$ on surface of a charged sphere with charge $q$ and radius $R$ is given as:
$$
V = dfrac{kq}{R}
$$
where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant.
Charge on small charged sphere before it’s connected to ground is equal to:
$$
V_1 = dfrac{kq_1}{R_1}
$$
whereas charge on Earth $q_E$ before it’s connected to the charged sphere could be considered zero and thus electric potential on it will will equal zero
$$
V_E = dfrac{kq_E}{R_E} = 0
$$
where $R_1$ is radius of the small charged sphere, $R_E$ is radius of the Earth and $q_1$ and $q_E$ is charge on the sphere and Earth, respectfully.
$$
begin{align*}
V_{1f} &= V_{Ef} \
tag{apply the equation for electric potential on surface of a charged sphere} \
dfrac{kq_{1f}}{R_1} &= dfrac{kq_{Ef}}{R_E} \
tag{where $q_{1f}$ and $q_{Ef}$ is charge on the sphere and charge on Earth, } \
tag{after the sphere is grounded, respectfully. Cancel out $k$}\
dfrac{q_{1f}}{R_1} &= dfrac{q_{Ef}}{R_E} \
tag{express charge $q_{1f}$} \
q_{1f} &= q_{Ef} dfrac{R_1}{R_E}
end{align*}
$$
We see that charge $q_{1f}$ that’s left on the sphere after it’s grounded is proportional to the ratio $dfrac{R_1}{R_E}$, which is very small number, given that radius of the Earth $R_E$ is much larger than radius of the sphere $R_1$.
Now that we have proven that sphere loses most of its charge after it’s grounded, we conclude that it lost it’s charge due to potential difference between potential on its surface and electric potential on the ground.
$$E = frac{F}{q}$$
$$ F = k frac{ Qq}{r^2} $$
$$begin{align*}
E &= dfrac{ k frac{ Q cancel q}{r^2} }{ cancel q} \
&= k frac{Q}{r^2}
end{align*}$$
$$
F_e = q E
$$
We know that work $W$ done on moving an object a distance $d$ by acting on it with a force $F$ is equal to:
$$
W = F d
$$
In this case, we’ll act on this charge $q$ with electric force, which means that work needed to move charge $q$ through an increasing electric field $E$ by a distance $d$ is equal to:
$$
W = q E d
$$
This work is also equal to energy needed to move the charge through an increasing electric field. Notice that since work $W$ needed to move the charge is proportional to the magnitude of the electric field, we conclude that energy needed to move the charge $q$ increases as electric field $E$ increases in magnitude.
$$
W = q E d
$$
Electric force $F$ between charges $q_1$ and $q_2$ at a distance $r$ from each other is given as:
$$
begin{equation}
F = dfrac{k q_1 q_2}{r^2}
end{equation}
$$
where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant. This means that magnitude of the force between two charges only depends on the magnitude of both charges and distance between them.
Notice that since distance $r_{xy}$ between charges $x$ and $y$ is equal to the distance $r_{xz}$ between charges $x$ and $z$ and since magnitude of all the charges is the same $q_x = q_y = q_z = q$, this means that magnitude of force $F_{xy}$ between charges $x$ and $y$ is equal in magnitude of the force $F_{xz}$ between charges $x$ and $z$. Now we need to determine the direction of these two forces. Charges $x$ and $y$ repel and direction of this force is in the $D$ direction. Charges $x$ and $z$ attract and direction of this force is in the $B$ direction. Since one of the forces is trying to push charge $x$ in direction $D$, whereas the other force is trying to push it in direction $B$, net force acting on charge $x$ will be pushing charge $x$ in $C$ direction because vector sum of forces $F_{xy}$ and $F_{xz}$ acts in $C$ direction.
$$
begin{align*}
1 ~mathrm{V} &= 1 ~mathrm{dfrac{J}{C}} &tag{1} \
&text{We know that one Joule is equal to $ 1 ~mathrm{J} = 1 ~mathrm{Nm} $}\
&text{ we’ll plug it into the equation $(1)$} \
1 ~mathrm{V} &= 1 ~mathrm{dfrac{Nm}{C}} &tag{2} \
&text{We know that one Newton is equal to $1 ~mathrm{N} = 1 ~mathrm{dfrac{kg m}{s^2}}$}\
&text{ we’ll plug it into the equation $(2)$} \
1 ~mathrm{V} &= 1 ~mathrm{ dfrac{kg m}{s^2} dfrac{m}{C}} \
&text{Finally, we have:}
end{align*}
$$
$$
boxed{ 1 ~mathrm{V} = 1 ~mathrm{ dfrac{kg m^2}{C s^2} }
}
$$
1 ~mathrm{V} = 1 ~mathrm{ dfrac{kg m^2}{C s^2} }
$$
$$
C= dfrac{q}{V}
$$
To compare the capacitance $C_1$ of an aluminum sphere with diameter $d_1$ and capacitance $C_2$ of an aluminum sphere with diameter $d_2$, we’ll find an equation that will show us how capacitance of a spherical conductor depends on its diameter.
$$
V= dfrac{kq}{r}
$$
where $r$ is distance from the center of the shell, $q$ is charge on the spherical shell and $k = 9 cdot 10^9 ~mathrm{dfrac{NC^2}{m^2}}$ is Coulomb’s constant. Equation above is valid for any distance $r$ from the center of the shell, including its radius $R$. This means that the voltage $V$ of a spherical shell on surface of a charged sphere with radius $R$ is given as:
$$
begin{align*}
V &= dfrac{kq}{R} tag{2}
end{align*}
$$
Since we’re given diameters of the two aluminum spheres, we’ll express radius $R$ of a spherical shell with its diameter. Diameter is equal to two radii. In other words:
$$
begin{align*}
d &= 2 R \
tag{express $R$ from the equation above } \
R &= dfrac{d}{2} \
end{align*}
$$
We can now plug in this expression for radius $R$ into equation $(2)$ and have:
$$
begin{align*}
V &= dfrac{kq}{R} \
V &= dfrac{ kq }{dfrac{d}{2}} \
V &= dfrac{2 k q }{d} tag{3}
end{align*}
$$
$$
C= dfrac{q}{V}
$$
we can plug in the expression for voltage $V$ on surface of a spherical shell from equation $(3)$ into the equation above and have:
$$
begin{align*}
C &= dfrac{q}{V} \
tag{plug in $ V = dfrac{2 k q }{d} $} \
C &= dfrac{q}{ dfrac{2 k q }{d} } \
C &= dfrac{q d }{2kq} \
tag{cancel out $q$} \
C &= dfrac{d}{2k}
end{align*}
$$
$$
C = dfrac{d}{2k}
$$
As we can see from this equation, capacitance $C$ is proportional to diameter $d$ of the spherical conductor, which means that the larger the spherical conductor (greater diameter), the greater the capacitance. In other words, aluminum sphere with diameter $d_2 = 10 ~mathrm{cm}$ has higher capacitance $C$.
text{ Aluminum sphere with diameter $d_2 = 10 ~mathrm{cm}$ has higher capacitance.}
$$
$$C = frac{Q}{V}$$
$$Q = CV$$
We know that electric force $F_e$ with which electric field $E$ acts on charge $q$ in the electric field is equal to:
$$
F_e = q E
$$
From the equation above we see that charge $q$ is calculated as:
$$
begin{align*}
q &= dfrac{F_e}{E} \
tag{plug in the given values} \
q &= dfrac{ 1.4 cdot 10^{-8} ~mathrm{N} }{5 cdot 10^{-4} ~mathrm{dfrac{N}{C}}}
end{align*}
$$
$$
boxed{ q = 2.8 cdot 10^{-5} ~mathrm{C} }
$$
q = 2.8 cdot 10^{-5} ~mathrm{C}
$$
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:
$$
begin{align*}
E &= dfrac{F_e}{q} \
tag{plug in the given values} \
E &= dfrac{ 0.3 ~mathrm{N} }{ 1 cdot 10^{-5} ~mathrm{C}}
end{align*}
$$
$$
boxed{ E= 0.3 cdot 10^{5} ~mathrm{dfrac{N}{C}} }
$$
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 upward, we conclude that the electric field also acts upward on charge $q$.
E = 3 cdot 10^{ 4} ~mathrm{dfrac{N}{C}}
$$
Direction of the electric field is upward.
We know that electric force $F_e$ with which electric field $E$ acts on charge $q$ in the electric field is equal to:
$$
F_e = q E
$$
From the equation above we see that charge $q$ is calculated as:
$$
begin{align*}
q &= dfrac{F_e}{E} \
tag{plug in the given values} \
q &= dfrac{ 0.3 ~mathrm{N} }{ 4.5 cdot 10^{5} ~mathrm{dfrac{N}{C}} }
end{align*}
$$
$$
boxed{ q = 0.0666 cdot 10^{-5} ~mathrm{C} }
$$
q = 6.6667 cdot 10^{-7} ~mathrm{C}
$$
$a)$~ To determine the direction of the electric force on a negatively charged particle, let’s remember that electric force $F_e$ with which electric field $E$ acts on charge $q$ in the electric field is equal to:
$$
F_e = q E
$$
We see that if charge is negative, electric field and electric force are in the opposite direction. We can see this by plugging in a negative charge in the equation above:
$$
F_e = – q E
$$
Minus sign in the equation above means that the direction of $F_e$ and $E$ are opposite. Since electric field of the atmosphere is downward, electric force acting on the negatively charged particle is upward.
$$
begin{align*}
F_e &= q E \
tag{charge on the electron is $-e$} \
F_e &= -e E\
tag{plug in the given values}\
F_e &= -1.6 cdot 10^{-19} ~mathrm{C} cdot 150 ~mathrm{dfrac{N}{C}}
end{align*}
$$
$$
boxed{ F_e = – 2.4 cdot 10^{-17} ~mathrm{N} }
$$
Note that the minus sign above means that this force is pointing upward.
$$
F_g = m_e g
$$
Now let’s compare electric force $F_e$ with gravitational force $F_g$ acting on the electron:
$$
begin{align*}
dfrac{F_e}{F_g} &= dfrac{ F_e }{m_e g} \
dfrac{F_e}{F_g} &= dfrac{ – 2.4 cdot 10^{-17} ~mathrm{N} }{ 9.11 cdot 10^{-31} ~mathrm{kg} cdot 9.81 ~mathrm{dfrac{m}{s^2}} }
end{align*}
$$
$$
boxed{ dfrac{F_e}{F_g} = 2.6855 cdot 10^{12} }
$$
$$
b)~ F_e = – 2.4 cdot 10^{-17} ~mathrm{N}
$$
$$
c)~ dfrac{F_e}{F_g} = 2.6855 cdot 10^{12}
$$
We know that electric force $F_e$ with which electric field $E$ acts on charge $q$ in the electric field is equal to:
$$
F_e = q E
$$
As it can be seen, we have everything we need to calculate the force $F_e$ exerted on this test charge.
$$
begin{align*}
F_e &= q E \
tag{plug in the given values}\
F_e &= 6 cdot 10^{-6} ~mathrm{C} cdot 50 ~mathrm{dfrac{N}{C}}
end{align*}
$$
$$
boxed{ F_e = 3 cdot 10^{-4} ~mathrm{N} }
$$
F_e = 3 cdot 10^{-4} ~mathrm{N}
$$
the distance between each of the two charges is the same. This is only possible if the three charges are placed in vertices of an equilateral triangle.
Charge magnitude of charge $X$ is $q_{X}=+1.0 mathrm{~mu C}$, while charge magnitude of charge $Y$ is $q_{Y}=+2.0 mathrm{~mu C}$ and charge $Z$ is negative with unknown charge magnitude.
This is shown in the figure below.
$a)~$ We know that electric force $F_e$ with which electric field $E$ acts on charge $q$ in the electric field is equal to:
$$
F_e = q E
$$
We also know that charge on the electron is negative and equal to $q = -e = -1.6 cdot 10^{-19} ~mathrm{C}$.
From the equation for $F_e$ we can calculate electric force acting on the electron as:
$$
begin{align*}
F_e &= q E \
tag{charge on the electron is $-e$} \
F_e &= -e E\
tag{plug in the given values}\
F_e &= -1.6 cdot 10^{-19} ~mathrm{C} cdot 1 cdot 10^5 ~mathrm{dfrac{N}{C}} end{align*}
$$
Note that the minus sign in the result above means that this force points in the opposite direction to the electric field.
$$
boxed{ F_e = – 1.6 cdot 10^{-14} ~mathrm{N} }
$$
$$
a = dfrac{F}{m}
$$
In our case we have an electron with mass $m_e$ accelerating due to force $F_e$, which means that its acceleration is equal to:
$$
begin{align*}
a &= dfrac{F_e}{m_e} \
tag{plug in the given value of $m_e$} \
tag{and calculated magnitude of the electric force $F_e$} \
a &= dfrac{ – 1.6 cdot 10^{-14} ~mathrm{N} }{9.11 cdot 10^{-31} ~mathrm{kg}}
end{align*}
$$
$$
boxed{ a = – 1.7563 cdot 10^{16} ~mathrm{dfrac{m }{s^2}} }
$$
Note that the negative sign of acceleration in the result above doesn’t mean that the electron is slowing down, but means that the force that accelerates the electron is in the opposite direction to the electric field, which also makes direction of acceleration of the electron in the opposite direction to the direction of electric field.
a)~ F_e = – 1.6 cdot 10^{-14} ~mathrm{N}
$$
$$
b)~ a = – 1.7563 cdot 10^{16} ~mathrm{dfrac{m }{s^2}}
$$
We know that magnitude of the electric field $E$ coming from a charge $q$ at a distance $r$ from the source of the electric field is given as:
$$
E = dfrac{kq}{r^2}
$$
where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant. As it can be seen, we have everything we need to calculate the electric field strength asked in this problem:
$$
begin{align*}
E &= dfrac{kq}{r^2} \
tag{plug in the given values}\
E &= dfrac{ 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}} cdot 8 cdot 10^{-7 } ~mathrm{C} }{ ( 0.2 ~mathrm{m})^2 }\
E &= dfrac{ 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}} cdot 8 cdot 10^{-7 } ~mathrm{C} }{ 0.04 ~mathrm{m^2} }\
E &= 180 000 ~mathrm{dfrac{N}{C}}
end{align*}
$$
$$
boxed{ E = 1.8 cdot 10^5 ~mathrm{dfrac{N}{C}} }
$$
E = 1.8 cdot 10^5 ~mathrm{dfrac{N}{C}}
$$
$$
begin{align*}
q_n &= 82 e \
q_n &= 82 cdot 1.6 cdot 10^{-19} ~mathrm{C} \
q_n &= 1.312 cdot 10^{-17} ~mathrm{C}
end{align*}
$$
a)~
$$
To calculate the magnitude of this electric field, let’s remember that electric field $E$ coming from a charge $q$ at a distance $r$ from the source of the electric field is given as:
$$
E = dfrac{kq}{r^2}
$$
where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant. As it can be seen, we have everything we need to calculate the electric field strength asked in this problem:
$$
begin{align*}
E &= dfrac{kq}{r^2} \
tag{plug in the given values}\
E &= dfrac{ 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}} cdot 1.312 cdot 10^{-17} ~mathrm{C} }{ ( 1 cdot 10^{-10} ~mathrm{m} )^2 }\
tag{calculate the term in the brackets} \
E &= dfrac{ 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}} cdot 1.312 cdot 10^{-17} ~mathrm{C} }{ 1 cdot 10^{-20} ~mathrm{m^2} }
end{align*}
$$
$$
boxed{ E = 1.1808 cdot 10^{13} ~mathrm{dfrac{N}{C}} }
$$
Since charge on the proton is positive and so is charge on a nucleus as a whole, we conclude that direction of the electric field is in a outward direction, pointing out of the nucleus.
b)~
$$
We know that electric force $F_e$ with which electric field $E$ acts on test charge $q_{test}$ in the electric field is equal to:
$$
F_e = q_{test} E
$$
In this part of the problem, we have electron as a test charge and we know that
charge $q_e$ on the electron is negative and equal to $q_e = -e = -1.6 cdot 10^{-19} ~mathrm{C}$.
From the equation for $F_e$ we can calculate electric force acting on the electron as:
$$
begin{align*}
F_e &= q_e E \
tag{charge on the electron is $-e$} \
F_e &= -e E\
tag{plug in the given values}\
F_e &= -1.6 cdot 10^{-19} ~mathrm{C} cdot 1.1808 cdot 10^{13} ~mathrm{dfrac{N}{C}}
end{align*}
$$
Note that the minus sign in the result above means that this force points in the opposite direction to the electric field.
$$
boxed{ F_e = – 1.8893 cdot 10^{-6} ~mathrm{N} }
$$
Minus sign in the result above means that electric force $F_e$ acting on an electron is in the opposite direction to the electric force $E$ in the point where electron is located. We knew that these two quantities will be in the opposite direction from the fact that test charge is negative.
a)~ E = 1.1808 cdot 10^{13} ~mathrm{dfrac{N}{C}}
$$
Electric field points outward, away from nucleus.
$$
b)~ F_e = – 1.8893 cdot 10^{-6} ~mathrm{N}
$$
Electric force points inward, to the nucleus.
To calculate the work done on charge $q$, let’s remember that work done by the electric field doesn’t depend on the path that the charge $q$ took to go from a point with electric potential $V_1$ to a point with electric potential $V_2$, but only depends on the electric potential between these two points $Delta V = V_2 – V_1$, which is unknown. Work $W$ done on charge $q$ while it moved through a potential difference $Delta V$ is calculated as:
$$
begin{align*}
W &= q Delta V \
tag{express $Delta V$ from the equation above} \
Delta V &= dfrac{W }{q} \
tag{Plug in the given values:} \
Delta V &= dfrac{120 ~mathrm{J} }{ 2.4 ~mathrm{C} }
end{align*}
$$
$$
boxed{ Delta V= 50 ~mathrm{V} }
$$
Delta V= 50 ~mathrm{V}
$$
$Delta V = 9 ~mathrm{V}$. To calculate the work done on charge $q$, let’s remember that work done by the electric field doesn’t depend on the path that the charge $q$ took to go from a point with electric potential $V_1$ to a point with electric potential $V_2$, but only depends on the electric potential between these two points, which is $Delta V = V_2 – V_1$. Work $W$ done on charge $q$ while it moved through a potential difference $Delta V$ is calculated as:
$$
begin{align*}
W &= q Delta V \
tag{Plug in the given values:} \
W &= 0.15 ~mathrm{C} cdot 9 ~mathrm{V}
end{align*}
$$
$$
boxed{ W = 1.35 ~mathrm{J} }
$$
W = 1.35 ~mathrm{J}
$$
$q = – 1.6 cdot 10^{-19} ~mathrm{C}$ moved through a potential difference of
$Delta V = 450 ~mathrm{V}$. To calculate the work done on an electron, let’s remember that work done by the electric field doesn’t depend on the path that the charge took to go from a point with electric potential $V_1$ to a point with electric potential $V_2$, but only depends on the electric potential between these two points, which is $Delta V = V_2 – V_1$. Work $W$ done on charge $q$ while it moved through a potential difference $Delta V$ is calculated as:
$$
begin{align*}
W &= q Delta V \
tag{charge on the electron is $- 1.6 cdot 10^{-19} ~mathrm{C} $} \
W &= – e Delta V \
tag{Plug in the given values:} \
W &= – 1.6 cdot 10^{-19} ~mathrm{C} cdot 450 ~mathrm{V}
end{align*}
$$
$$
boxed{ W = – 7.2 cdot 10^{-17} ~mathrm{J} }
$$
Note that the negative value of work on an electron means that work done by the electric field $E$ is in the opposite direction to the electric force $F_e$ that does the work, which is always the case with a negatively charged test charge.
W = – 7.2 cdot 10^{-17} ~mathrm{J}
$$
$Delta V = 12 ~mathrm{V}$ in a car battery, work $W = 1200 ~mathrm{J}$ is needed. This work done on charge $q$, done by the electric field only depends on the electric potential on the battery, which is $Delta V = V_2 – V_1$ and charge $q$ that got transferred. Work $W$ done on charge $q$ while it moved through a potential difference $Delta V$ is calculated as:
$$
begin{align*}
W &= q Delta V \
tag{express charge $q$ from the equation above}\
q &= dfrac{W}{Delta V} \
tag{Plug in the given values:} \
q &= dfrac{ 1200 ~mathrm{J} }{ 12 ~mathrm{V} }
end{align*}
$$
$$
boxed{ q= 100 ~mathrm{C} }
$$
q= 100 ~mathrm{C}
$$
$E = 1.5 cdot 10^3 ~mathrm{dfrac{N}{C}}$. To calculate the potential difference $Delta V$ between the plates that are $d = 0.06 ~mathrm{m}$ 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:
$$
begin{align*}
Delta V &= E d \
tag{plug in the given values} \
Delta V &= 1.5 cdot 10^3 ~mathrm{dfrac{N}{C}} cdot 0.06 ~mathrm{m}
end{align*}
$$
$$
boxed{ Delta V = 90 ~mathrm{V} }
$$
Delta V = 90 ~mathrm{V}
$$
$$
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{ 70 ~mathrm{V} }{0.02 ~mathrm{m} }
%
end{align*}
$$
$$
boxed{ E= 3500 ~mathrm{dfrac{N}{C}} }
$$
E= 3500 ~mathrm{dfrac{N}{C}}
$$
$$
begin{align*}
C &= dfrac{q}{V} \
tag{Plug in the given values}\
C &= dfrac{ 90 ~mathrm{mu C} }{ 45 ~mathrm{V} } \
tag{$1 ~mathrm{mu C} = 10^{-6} ~mathrm{C} $} \
C &= dfrac{ 90 cdot 10^{-6} ~mathrm{C} }{ 45 ~mathrm{V} } \
C &= 2 cdot 10^{-6} ~mathrm{F}
end{align*}
$$
$$
boxed{ C = 2 ~mathrm{mu F} }
$$
C = 2 ~mathrm{mu F}
$$
$$
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{8.1 cdot 10^{-4} ~mathrm{ C}}{5.4 ~mathrm{mu F} } \
%
%
tag{$1 ~mathrm{mu F} = 10^{-6} ~mathrm{F} $} \
%
%
V &= dfrac{8.1 cdot 10^{-4} ~mathrm{ C}}{5.4 cdot 10^{-6} ~mathrm{ F} }
end{align*}
$$
$$
boxed{ V = 150 ~mathrm{V } }
$$
V = 150 ~mathrm{V }
$$
but these forces must be in the opposite direction. Since direction of the gravitational force is always downward, this means that electric force must act upward. Since the top plate in the Millikan apparatus is positively charged, this means that electric field
in Millikan apparatus points downward on a positively charged particle, but we’re told our unknown charge $q$ is negatively charged and thus this same electric field points upward on our negatively charged particle. Keep in mind that since charge $q$ is negative, direction of the electric field $E$ and electric force $F_e$ is opposite, stated as:
$$
F_e = – q E
$$
Magnitude of the mentioned electric field is
$$
E =5.6 cdot 10^3 ~mathrm{dfrac{N}{C}}
$$
a)~~
$$
Now that we know that this charged particle can be stationary, we need to find charge on it. As said, weight $F_g$ is equal in magnitude to the electric force $F_e$, stated as:
$$
begin{align*}
F_e &= F_g \
tag{$F_e = q E$} \
-q E &= F_g \
tag{express charge $q$ from the equation above} \
q &= -dfrac{F_g}{E} \
tag{plug in the given values}\
q &= -dfrac{ 4.5 cdot 10^{-15} ~mathrm{N} }{ 5.6 cdot 10^3 ~mathrm{dfrac{N}{C}} }
end{align*}
$$
$$
boxed{ a)~~ q = – 8.0357 cdot 10^{-19} ~mathrm{C} }
$$
We’ve calculated the charge on the oil drop. To calculate the number of excess electrons, we need to understand that since this charge $q$ comes from excess electrons, it will be equal to integral multiple of charge of the electron $e$, stated as:
$$
q = N q_e
$$
Charge of the electron is equal to negative value of elementary charge $q_e = -e = – 1.6 cdot 10^{-19} ~mathrm{C}$.
From the equation above we conclude:
$$
begin{align*}
q &= N q_e \
tag{express $N$ from the equation above:}\
N &= dfrac{ q}{q_e} \
tag{plug in the given values} \
N &= dfrac{ – 8.0357 cdot 10^{-19} ~mathrm{C} }{-1.6 cdot 10^{-19} ~mathrm{C}}
end{align*}
$$
$$
boxed{ b)~~ N = 5 }
$$
a)~~ q = – 8.0357 cdot 10^{-19} ~mathrm{C}
$$
$$
b)~~ N = 5
$$
$$
begin{align*}
C &= dfrac{q}{V} \
tag{express $q$ from the equation above} \
q &= {V}cdot{C} \
tag{Plug in the given values}\
q &= 45.0 Vcdot 15.0 pF \
%
%
tag{$1 ~mathrm{ pF} = 10^{-12} ~mathrm{F} $} \
%
%
q &= 45.0 Vcdot 15.0cdot 10^{-12} F
end{align*}
$$
$$
boxed{ q = 6.75cdot10^{-10} ~mathrm{C } }
$$
q = 6.75cdot10^{-10} ~mathrm{C }
$$
We know that electric force $F_e$ with which electric field $E$ acts on charge $q$ in the electric field is equal to:
$$
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{ 0.065 ~mathrm{N} }{ 37 ~mathrm{mu C}} \
tag{$1 ~mathrm{mu C} = 10^{-6} ~mathrm{C}$} \
E &= dfrac{ 0.065 ~mathrm{N} }{ 37 cdot 10^{-6} ~mathrm{C}}
end{align*}
$$
$$
boxed{ E= 1756.7567 ~mathrm{dfrac{N}{C}} }
$$
What we need to understand is that electric field in this problem is uniform, which means that its magnitude is same in every point and equals
$E = 1756.7567 ~mathrm{dfrac{N}{C}}$. To calculate the potential difference $Delta V$ between the starting point and an end point, that are $d = 25 ~mathrm{cm} = 0.25 ~mathrm{m}$ apart, let’s remember that this potential difference can be calculated as a product of magnitude of the uniform electric field and distance $d$ between the starting point and an end point:
$$
begin{align*}
Delta V &= E d \
tag{plug in the given values} \
Delta V &= 1756.7567 ~mathrm{dfrac{N}{C}} cdot 0.25
~mathrm{m}
end{align*}
$$
$$
boxed{ Delta V = 439.19 ~mathrm{V} }
$$
Delta V = 439.19 ~mathrm{V}
$$
$$
W = dfrac{1}{2 } C V^2
$$
Capacitance of the capacitor in this problem is $C = 10 ~mathrm{mu F}$ and voltage across its plates is $V = 3 cdot 10^2 ~mathrm{V}$. To find the energy stored on it, we’ll apply the equation above:
$$
begin{align*}
W &= dfrac{1}{2 } C V^2 \
tag{plug in the given values}\
W &= dfrac{1}{2} cdot 10 ~mathrm{mu F} cdot (3 cdot 10^2 ~mathrm{V})^2 \
tag{$1 ~mathrm{mu F} = 10^{-6} ~mathrm{F} $ } \
W &= dfrac{1}{2} cdot 10 cdot 10^{-6} ~mathrm{F} cdot 9 cdot 10^4 ~mathrm{V^2}
end{align*}
$$
$$
boxed{ W = 0.45 ~mathrm{J} }
$$
W = 0.45 ~mathrm{J}
$$
$a)~~$ To find the average power $P$ required to charge the capacitor, let’s remember that energy $W$ stored on the capacitor is equal to a product of time $t$ it took to charge the capacitor and average power of charging $P$, stated as:
$$
begin{align*}
E &= P t \
tag{express $P$ from the equation above}\
P &= dfrac{E}{t} \
tag{plug in the given values} \
P &= dfrac{ 0.45 ~mathrm{J} }{25 ~mathrm{s}}
end{align*}
$$
$$
boxed{P = 0.018 ~mathrm{W}}
$$
$b)~~$ If the capacitor discharges in time $t_{discharge} = 1 cdot 10^{-4} ~mathrm{s}$, it transfers all of the energy $W$ stored on it to the lamp. To calculate the power of discharge $P_{discharge}$, let’s remember that this power is equal to the rate of change of energy on the capacitor, stated as:
$$
begin{align*}
P_{discharge} &= dfrac{W}{t} \
P_{discharge} &= dfrac{ 0.45 ~mathrm{J}}{ 1 cdot 10^{-4} ~mathrm{s}}
end{align*}
$$
$$
boxed{ P_{discharge} = 4500 ~mathrm{W} }
$$
$c)~~$ Capacitor is built so that it stores electric energy and release this energy in a very short period of time. Unlike battery, it’s not built to give us a source of electricity with constant voltage, but rather to store energy and release it as fast as possible. Energy stored on the capacitor decreases exponentially with time of discharge.
a)~~ P = 0.018 ~mathrm{W}
$$
$$
b)~~ P_{discharge} = 4500 ~mathrm{W}
$$
$$
c)~~ text{ Hint: Capacitor is built to give off energy in short periods of time }
$$
a)~~
$$
Energy $W$ stored on the capacitor with capacitance $C$ and voltage $V$ across its plates is given as:
$$
W = dfrac{1}{2 } C V^2
$$
Capacitance of the capacitor in this problem is $C = 61 cdot 10^{-3} ~mathrm{F}$ and voltage across its plates is $V = 10 ~mathrm{kV} = 10^4 ~mathrm{V}$. To find the energy stored on it, we’ll apply the equation above:
$$
begin{align*}
W &= dfrac{1}{2 } C V^2 \
tag{plug in the given values}\
W &= dfrac{1}{2} cdot 61 cdot 10^{-3} ~mathrm{F} cdot (10^4 ~mathrm{V})^2 \
W &= 3 050 000 ~mathrm{J} \
W &= 3.05 cdot 10^{6} ~mathrm{J}
end{align*}
$$
$$
boxed{ W = 3.05 ~mathrm{MJ} }
$$
$$
begin{align*}
P_{discharge} &= dfrac{W}{t} \
P_{discharge} &= dfrac{3.05 ~mathrm{MJ}}{ 1 cdot 10^{-8} ~mathrm{s}} \
P_{discharge} &= dfrac{3.05 cdot 10^{6} ~mathrm{J}}{ 1 cdot 10^{-8} ~mathrm{s}}
end{align*}
$$
$$
boxed{ P_{discharge} = 3.05 cdot 10^{14} ~mathrm{W} }
$$
$$
begin{align*}
W &= P_{generator} cdot t_{charging} \
tag{express time of charging from the equation above} \
t_{charging} &= dfrac{W}{P_{generator}} \
tag{plug in the given values} \
t_{charging} &= dfrac{ 3.05 ~mathrm{MJ} }{ 1000 ~mathrm{W} } \
tag{ $ 1 ~mathrm{MJ} = 10^6 ~mathrm{J}$ } \
t_{charging} &= dfrac{ 3.05 cdot 10^6 ~mathrm{J} }{ 1000 ~mathrm{W} }
end{align*}
$$
$$
boxed{ t_{charging} = 3050 ~mathrm{s} }
$$
a)~~ W = 3.05 ~mathrm{MJ}
$$
$$
b)~~ P_{discharge} = 3.05 cdot 10^{14} ~mathrm{W}
$$
$$
c)~~ t_{charging} = 305 0 ~mathrm{s}
$$
The two parallel plates are charged and thus there is potential difference $Delta V$ between them. We know that the work $W$ needed to move charge $q$ through a potential difference $Delta V$ is equal to:
$$
W = q Delta V
$$
$$
Delta V = E d
$$
where $Delta V$ is potential difference between the plates and $d$ is distance between the plates.
We can plug in this expression for potential difference $Delta V$ into the equation for work $W$ needed to move this charge and have:
$$
W = q E d
$$
$$
begin{align*}
W &= q E d \
tag{plug in the values} \
W &= 0.25 ~mathrm{mu C} cdot 6400 ~mathrm{dfrac{N}{C}} cdot 0.4 ~mathrm{cm} \
tag{$1mathrm{~cm} = 0.01 ~mathrm{m} $} \
tag{ $ 1 ~mathrm{mu C} = 10^{-6} ~mathrm{C} $ } \
W &= 0.25 cdot 10^{-6} ~mathrm{C} cdot 6400 ~mathrm{dfrac{N}{C}} cdot 0.4 cdot 0.01 ~mathrm{m} \
W&= 6.4 cdot 10^{-6} ~mathrm{J}
end{align*}
$$
W = 6.4 ~mathrm{mu J}
$$
$$
C = dfrac{q}{V}
$$
where $q$ is charge stored on the capacitor and $V$ is voltage between its plates.
We can express charge $q$ on the capacitor from the equation above:
$$
q = CV
$$
$$
V = E d
$$
where $V$ is voltage between the plates of the capacitor, $E$ is magnitude of the electric field between the plates and $d$ is distance between the plates.
We can plug in this expression for voltage $V$ between the plates into the equation for charge $q$ on the capacitor and have:
$$
q = C E d
$$
$$
begin{align*}
q &= C E d \
tag{plug in the values} \
q &= 0.22 ~mathrm{mu F} cdot 2400 ~mathrm{dfrac{N}{C}} cdot 1.2 ~mathrm{cm} \
tag{$1mathrm{~cm} = 0.01 ~mathrm{m} $} \
tag{ $ 1 ~mathrm{mu F} = 10^{-6} ~mathrm{F} $ } \
q &= 0.22 cdot 10^{-6} ~mathrm{F} cdot 2400 ~mathrm{dfrac{N}{C}} cdot 1.2 cdot 0.01 ~mathrm{m} \
q &= 6.336 cdot 10^{-6} ~mathrm{C}
end{align*}
$$
$$
boxed{q = 6.336 ~mathrm{mu C} }
$$
q = 6.336 ~mathrm{mu C}
$$
$$
begin{align*}
C &= dfrac{q }{V} \
tag{plug in the given values} \
C &= dfrac{ 0.06 ~mathrm{mu C} }{300 ~mathrm{V} } \
tag{$1 ~mathrm{mu C} = 1 cdot 10^{-6} ~mathrm{C} $ } \
C &= dfrac{ 0.06cdot 10^{-6} ~mathrm{C} }{300 ~mathrm{V} }
end{align*}
$$
$$
boxed{C = 2 cdot 10^{-10} ~mathrm{F}}
$$
C = 2 cdot 10^{-10} ~mathrm{F}
$$
We are told that capacitance of the capacitor is $C = 0.047 ~mathrm{mu F}$ and that voltage across it its plates is $V = 120 ~mathrm{V}$. To find charge $q$ on the capacitor, we’ll use the definition of capacitance, which states that capacitance of the capacitor $C$ is calculated by dividing the charge $q$ on the capacitor with voltage $V$ across the capacitor:
$$
begin{align*}
C &= dfrac{q}{V} \
tag{express $q$ from the equation above}& \
q &= C V \
tag{Plug in the given values}\
q &= 0.047 ~mathrm{mu F} cdot 120 ~mathrm{V} \
tag{$1 ~mathrm{mu C} = 10^{-6} ~mathrm{C} $} \
q &= 4.7 cdot 10^{-8} ~mathrm{C} cdot 120 ~mathrm{V} \
q &= 5.64 cdot 10^{-6} ~mathrm{C}
end{align*}
$$
$$
boxed{q = 5.64 ~mathrm{mu C} }
$$
q = 5.64 ~mathrm{mu C}
$$
$$
begin{align*}
Delta V &= E d \
tag{Express $E$ from the equation above} \
E &= dfrac{Delta V}{d} \
tag{plug in the given values} \
E &= dfrac{ 120 ~mathrm{V} }{ 2.5 cdot 10^{-3} ~mathrm{m} }
%
end{align*}
$$
$$
boxed{ E= 48000 ~mathrm{dfrac{N}{C}} }
$$
E= 48000 ~mathrm{dfrac{N}{C}}
$$
We know that electric force $F_e$ with which electric field $E$ acts on charge $q$ in the electric field is equal to:
$$
F_e = q E
$$
As it can be seen, we have everything we need to calculate the force $F_e$ exerted on this test charge.
This means that electric force $F_e$ on the electron will be equal to:
$$
begin{align*}
F_e &= q_e E \
tag{plug in the given values}\
F_e &= -1.6 cdot 10^{-19} ~mathrm{C}cdot 48000 ~mathrm{dfrac{N}{C}}
end{align*}
$$
$$
boxed{ F_e = – 7.68cdot 10^{-15} ~mathrm{N} }
$$
Note that the negative sign in the result above means that electric force $F_e$ acting on the electron and electric field $E$ are in the opposite direction, which is always the case when electric field acts on a negatively charged particle.
F_e = – 7.68cdot 10^{-15} ~mathrm{N}
$$
$$
W = q Delta V
$$
To calculate work $W$ needed to move this charge between the plates of the capacitor, we’ll plug in the given values into the equation above:
$$
begin{align*}
W & = q Delta V \
W &= 0.01 ~mathrm{mu C} cdot 120 ~mathrm{V} \
tag{$1~mathrm{mu C} = 1 cdot 10^{-6} ~mathrm{C} $} \
W &= 0.01 cdot 10^{-6} ~mathrm{C} cdot 120 ~mathrm{V} \
W &= 1.2 cdot 10^{-6} ~mathrm{J}
end{align*}
$$
$$
boxed{W = 1.2 ~mathrm{mu J} }
$$
W = 1.2 ~mathrm{mu J}
$$
$$
y = ax + b
$$
where $a$ is slope of the linear function and $b = y_{0}$. Analogously, we have:
$$
q = aV + b
$$
As we can see from the graph, $q_{0} = 0$, which we could expect because if voltage charging the capacitor is zero, charge on the capacitor is also zero. This means that for our linear function, $b = 0$, which leaves us with just the first term of the equation:
$$
q = a V
$$
we’re asked to find what does slope $a$ represent, so we’ll express it from the equation above:
$$
a = dfrac{q}{V}
$$
Definition of capacitance $C$ states that capacitance $C$ of a capacitor is equal to charge $q$ on plates of the capacitor divided by voltage $V$ between the charges, stated as:
$$
C = dfrac{q}{V}
$$
If we compare the two equations above, we see that the right side of both of the equations is the same, which means:
$$
boxed{a = C}
$$
or in other words, slope in the graph represents capacitance.
$$
y = ax + b
$$
$$
y=acdot x+b
$$
where $y$ is variable on the $y-axis$, $x$ is variable on the $x-axis$,
$a$ is a point of intersection of the function with $y-axis$ and $a$ is slope of the line. Now let’s compare the graph that we’re given what the graph of a linear function with $b = 0$.
Let’s now have a look at the graph of a linear function
$$
y = ax + b
$$
but let $b = 0$. In this scenario, our linear function will intersect with the $y$ axis at the origin.
Since plot of our function (red line) intersects with the $y-axis$ at 0, coefficient $b = 0$ and this linear function can be written as:
$$
begin{align*}
y&=acdot x+0\
y&=acdot x
end{align*}
$$
We can express the coefficient $a$, which represents the slope of the line, from the equation above as:
$$
begin{align*}
a&=dfrac{y}{x}
end{align*}
$$
By plugging in any pair of values for $x$ and $y$ that belong to a certain point on line in the graph, we can calculate slope of this line as:
$$
begin{align*}
a&=dfrac{1}{2}=dfrac{2}{4}=dfrac{3}{6}=0.5
end{align*}
$$
$$
begin{align*}
q&=a cdot Delta V + b \
tag{plug in $b = 0 $} \
q&=a cdot Delta V + 0 \
q&=a cdot Delta V \
tag{ express the unknown quantity $a$} \
a&= dfrac{q}{Delta V} tag{1}
end{align*}
$$
$$
C=dfrac{q}{Delta V}
$$
where $q$ is charge stored on the capacitor and $Delta V$ is voltage across the plates of the capacitor, we conclude that slope of the line on the graph is actually capacitance of the capacitor $C$.
$$
begin{align*}
a&=dfrac{q}{Delta V}=C\
tag{Notice $Delta V=10 mathrm{~V}$ from the graph and plug it in}\
tag{Plug in the corresponding value of charge $q=5 mathrm{~mu C}$}\
C&=dfrac{5 mathrm{~mu C}}{10 mathrm{~V}}\
tag{$5 mathrm{~mu C}= 5 cdot 10^{-6} mathrm{~C}$}\
C&=dfrac{5 cdot 10^{-6} mathrm{~C}}{10 mathrm{~V}}\
C&= 0.5 cdot 10^{-6} mathrm{~F}\
tag{$10^{-6} mathrm{~F}=1 mathrm{~mu F}$}
end{align*}
$$
$$
boxed{C= 0.5 mathrm{~mu F}}
$$
text{Slope represents the capacitance $C$ of the capacitor }
$$
$$
C = 0.5 mathrm{~mu F}
$$
$$
q = C V
$$
where $C = 5 cdot 10^{-7}$ is capacitance of the given capacitor. To see what the area under a graph line represents, let’s choose any value of voltage $V$ across the capacitor and corresponding charge $q$ across it, calculated from the equation above. Notice the triangle with one side being the line between the origin of the graph and the chosen voltage $V$ in $V$ axis, the other side being the line between the origin of the graph and the corresponding charge $q$ in $q$ axis and the third one being part of the slope between the normal projection from chosen $V$ and corresponding $q$. Area $A$ of this triangle is calculated as:
$$
A = dfrac{1}{2} qV
$$
Since $q = C V$, we can write the equation above as:
$$
A = dfrac{1}{2 } C V^2
$$
Energy $W$ stored on the capacitor with capacitance $C$ and voltage $V$ across its plates is given as:
$$
W = dfrac{1}{2 } C V^2
$$
If we compare the two equations above, we see that the right side of both of the equations is the same, which means:
$$
boxed{A = W}
$$
or in other words, area under the graph line in the graph represents energy stored on the capacitor.
$$
A = dfrac{1}{2 } C V^2
$$
$$
W = dfrac{1}{2 } C V^2
$$
where $C = 5 cdot 10^{-7}$ is capacitance of the given capacitor and $V$ is voltage across the plates of the capacitor. If we want to charge the capacitor to $V = 25 ~mathrm{V}$, work required to charge the capacitor will be equal to:
$$
begin{align*}
W &= dfrac{1}{2 } C V^2 \
tag{plug in the given values} \
W &= dfrac{1}{2} cdot 5 cdot 10^{-7} cdot (25 ~mathrm{V})^2 \
W &= dfrac{1}{2} cdot 5 cdot 10^{-7} cdot 625 ~mathrm{V}^2
end{align*}
$$
$$
boxed{W = 1.5625 cdot 10^{-4} ~mathrm{J} }
$$
W = 1.5625 cdot 10^{-4} ~mathrm{J}
$$
$$
W = dfrac{1}{2} q V
$$
where $q$ is charge on the capacitor plates and $V$ is voltage between the plates of the capacitor.
Note that we also proved that work required to charge the capacitor is equal to the area under the graph line, which is precisely equal to the expression above.
The explanation why this work does not equal $W_0 = q V$ is simple – during charging voltage across the capacitor is not the same. Charging the capacitor means that we’ll use some battery with voltage $V_{battery}$ to put charges on the capacitor. As more and more charges get stored on the plates of the capacitor, voltage $V$ across the plates of the capacitor also increases. Work needed to move one charge $q_0$ through a potential difference $V_0$ is equal to $W_0 = q_0 V_0$, but since voltage across the plates of the capacitor increases more and more as capacitor is charging, more work is needed to move one charge to the plates of the capacitor. What happens is that by using the same battery with voltage $V_{battery}$, it gets harder for battery to store charges on the capacitor.
E = dfrac{K q}{d^2}
$$
$$
begin{equation}
E = dfrac{k q }{r^2 }
end{equation}
$$
where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant. As it can be seen from the equation above, for electric $E$ to be zero we have two options:
Charge $q$ must be zero, which means that particle is neutral and thus has no electric field or particle is charged, but electric field is still somehow equal to zero. As we can see, magnitude of the electric field decreases with square of the distance from the source of the electric field, which is this point charge. This means that as the distance from $q$ increases, electric field coming from $q$ decreases. For example, if the distance tripled, electric field would decrease by a factor of 9. We come to a conclusion that at an infinite distance from charge $q$, we’d divide $kq$ with infinity squared and thus we’d calculate zero. In other words, electric field coming from point charge $q$ is zero at an infinite distance from it.
However, since nothing can be at an infinite distance from anything, this means that electric field from any charge is present in any point of the Universe.
$$
begin{equation}
E = dfrac{k q }{r^2 }
end{equation}
$$
where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant. As it can be seen, we find that at a zero distance from the point charge, $r = 0$, magnitude of the electric field will be its limit value, equal to:
$$
begin{align*}
lim_{r to 0} E &= lim_{r to 0} dfrac{kq}{r^2} \
lim_{r to 0} E &= dfrac{kq }{0} \
lim_{r to 0} E &rightarrow infty
end{align*}
$$
We found that magnitude of the electric field of a point charge $q$ at a zero distance from it is equal to infinity. Keep in mind that although this is mathematically correct, it doesn’t necessarily mean that it’s also physically correct. Of course, this result is not physically correct because magnitude of the electric field can’t be infinite and one of the reasons why it can’t be infinite is because it would require an infinite energy to make a setup of charges so that electric field coming from them is infinite. Since there is finite amount of energy in the Universe, this result is not physically possible and we conclude that it would take infinite energy to make a point charge, which also makes the point charge not physically possible.
Electric field is inversely proportional to the square of the distance from charge $q$. Point charge is not physically possible.
This way, negative charges stored on the cloud above the lightning rod are slowly taken away from that cloud, resulting in lower potential difference between the ground and the cloud, which makes the lightning strike, if it even occurs, weaker.
Ideally, dangerous high current discharge would never occur, but if even if it does, lightning rod can safely direct this high current to the ground.
$$
begin{align*}
q_1 &= 6.563cdot 10^{-19} ~mathrm{C} \
q_2 &= 8.204cdot 10^{-19} ~mathrm{C} \
q_3 &= 11.50cdot 10^{-19} ~mathrm{C} \
q_4 &= 13.13cdot 10^{-19} ~mathrm{C} \
q_5 &= 16.48cdot 10^{-19} ~mathrm{C} \
q_6 &= 18.08cdot 10^{-19} ~mathrm{C} \
q_7 &= 19.71cdot 10^{-19} ~mathrm{C} \
q_8 &= 22.89cdot 10^{-19} ~mathrm{C} \
q_9 &= 26.13cdot 10^{-19} ~mathrm{C} \
end{align*}
$$
Millikan didn’t know this at the moment, but charge is quantized, which means that there is some least amount of charge in the nature, called elementary charge $e$ and any other charge $q$ is equal to an integral multiple of $e$, stated as:
$$
q = Ne
$$
For the first example, let’s find $q_3 – q_2$
$$
begin{align*}
q_3 – q_2 &= 11.50cdot 10^{-19} ~mathrm{C} – 8.204cdot 10^{-19} ~mathrm{C} \
q_3 – q_2 &= 3.296 cdot 10^{-19} ~mathrm{C}
end{align*}
$$
For the second example, let’s find $q_8 – q_7$
$$
begin{align*}
q_8 – q_7 &= 22.89 cdot 10^{-19} ~mathrm{C} – 19.71 cdot 10^{-19} ~mathrm{C} \
q_8 – q_7 &= 3.18cdot 10^{-19} ~mathrm{C}
end{align*}
$$
For the third example, let’s find $q_5 – q_4$
$$
begin{align*}
q_5 – q_4 &= 16.48 cdot 10^{-19} ~mathrm{C} – 13.13 cdot 10^{-19} ~mathrm{C} \
q_5 – q_4 &= 3.35 cdot 10^{-19} ~mathrm{C}
end{align*}
$$
For the fourth example, let’s find $q_9 – q_9$
$$
begin{align*}
q_9 – q_8 &= 26.13 cdot 10^{-19} ~mathrm{C} – 22.89 cdot 10^{-19} ~mathrm{C} \
q_9 – q_8 &= 3.24 cdot 10^{-19} ~mathrm{C}
end{align*}
$$
Interestingly, by subtracting two consecutive possible charges in the Millikan experiment, we find almost the same result.
For the fifth example, let’s find $q_4 – q_3$
$$
begin{align*}
q_4 – q_3 &= 13.13 cdot 10^{-19} ~mathrm{C} – 11.50 cdot 10^{-19} ~mathrm{C} \
q_4 – q_3 &= 1.63 cdot 10^{-19} ~mathrm{C} tag{1}
end{align*}
$$
For the sixth example, let’s find $q_4 – q_3$
$$
begin{align*}
q_6 – q_5 &= 18.08 cdot 10^{-19} ~mathrm{C} – 16.48 cdot 10^{-19} ~mathrm{C} \
q_6 – q_6 &= 1.60 cdot 10^{-19} ~mathrm{C} tag{2}
end{align*}
$$
For the seventh example, let’s find $q_7 – q_6$
$$
begin{align*}
q_7 – q_6 &= 19.71 cdot 10^{-19} ~mathrm{C} – 18.08 cdot 10^{-19} ~mathrm{C} \
q_7 – q_6 &= 1.63 cdot 10^{-19} ~mathrm{C} tag{3}
end{align*}
$$
For the eighth example, let’s find $q_2 – q_1$
$$
begin{align*}
q_2 – q_1 &= 8.204 cdot 10^{-19} ~mathrm{C} – 6.563 cdot 10^{-19} ~mathrm{C} \
q_2 – q_1 &= 1.641 cdot 10^{-19} ~mathrm{C} tag{4}
end{align*}
$$
$$
begin{align*}
dfrac{ q_3 – q_2 }{2} &= 1.648 cdot 10^{-19} ~mathrm{C} tag{5} \
dfrac{q_8 – q_7}{2} &= 1.59cdot 10^{-19} ~mathrm{C} tag{6} \
dfrac{q_5 – q_4 }{2} &= 1.675 cdot 10^{-19} ~mathrm{C} tag{7}\
dfrac{ q_9 – q_8}{2} &= 1.62 cdot 10^{-19} ~mathrm{C} tag{8}
end{align*}
$$
We see that the equations above give us an almost identical result and from this we might see an indication that there is some least possible amount of charge $e$ with value of about $1.6 cdot 10^{-19} ~mathrm{C}$. This way, charges in the first 4 examples differ from one another by 2 $e$, whereas charges in the next 4 examples differ by one $e$.
Let’s find the average value of the results above :
$$
begin{align*}
e &= dfrac{ 1.63 + 1.60 + 1.63 +1.641 + 1.648 + 1.59 + 1.675 + 1.62 }{8 } cdot 10^{-19} ~mathrm{C} \
e &= dfrac{ 13.034 }{8 } cdot 10^{-19} ~mathrm{C}
end{align*}
$$
$$
boxed{e = 1.62925 cdot 10^{-19} ~mathrm{C} }
$$
$$
e = 1.6022 cdot 10^{-19} ~mathrm{C}
$$
e = 1.62925 cdot 10^{-19} ~mathrm{C}
$$
