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Ackbach
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I had some trouble reconciling signs while attempting to teach electricity to my students here. The topic is electric potential, potential energy, electric force, electric field, and work. After much thought - several hours - I have finally come up with a presentation that solves the issues. Fundamentally, the issue was confusing a constant electric field with a point charge.
The goal of this post is to arrive at the correct sign information for the potential due to a point charge:
$$V=\frac{kq}{r}.$$
Background: suppose a ball rolls down a hill, with a starting height of $y_{i}$ and a stopping height of $y_{f}$. These heights are measured relative to a position of low potential energy. Then the work done on the ball by the gravitational force is given by $$W_{i\to f}=-\Delta U=-(U_{f}-U_{i})=-(mgy_{f}-mgy_{i})=mg(y_{i}-y_{f}).$$
Since $y_{i}>y_{f}$, the work done by the gravitational force on the ball is positive. This makes sense, because, as Young and Freedman say on page 195 of University Physics, 9th Edition, "When the body moves up, $y$ increases, the work done by the gravitational force is negative, and the gravitational potential energy increases ($\Delta U>0$). When the body moves down, $y$ decreases, the gravitational force does positive work, and the gravitational potential energy decreases ($\Delta U<0$)." So far so good.
Moving on to the electrical case. Suppose I have a uniform electric field of field strength $E$ pointed down. This could happen in the middle of a capacitor, e.g. By the definition of the electric field, the lines go away from positive charge, and towards negative charge. Hence, I have positive charges up top (say, on the top plate of a capacitor), and negative charges down below. Now suppose I have a positive test charge $q_{0}$ that is in this field pointed down. Because the test charge is positive, it's going to want to go towards the negative plate, so the electric field will be trying to move the test charge down - with the field. The electric force on the test charge $q_{0}$ is $F=q_{0}E$, a constant. Hence, the work done by the field on the test charge in moving the test charge from $a$ to $b$ is given by
$$W_{a\to b}=\int_{r_{a}}^{r_{b}}F\,dx=Fd=q_{0}Ed,$$
where $d$ is the distance from point $a$ to point $b$, and equals $|r_{a}-r_{b}|$.
If we measure distance $y$ from the negative plate, which is typical, then the potential energy is $U=q_{0}Ey$. The work done by the electric field in moving the test charge from point $a$ to point $b$ is
$$W_{a\to b}=-\Delta U=-(U_{b}-U_{a})=-(q_{0}Ey_{b}-q_{0}Ey_{a})
=q_{0}E(y_{a}-y_{b}).$$
Again, Young and Freedman on page 732:
"When $y_{a}$ is greater than $y_{b}$ ..., the positive test charge $q_{0}$ moves downward, in the same direction as $\overset{\to}{\mathbf{\it{E}}}$; the displacement is in the same direction as the force $\overset{\to}{\mathbf{\it{F}}}=q_{0}\overset{\to}{\mathbf{\it{E}}},$ so the field does positive work and $U$ decreases."
Now we introduce the electric potential $V_{a}$ at a point $a$. This is NOT voltage, but electric potential. We define it to be the potential energy per unit charge: $V_{a}=U_{a}/q_{0}$. Here $U_{a}$ is the potential energy at point $a$ relative to some zero point, and $q_{0}$ is the test charge at point $a$. If we wish to write the work as a function of $V$, we divide through by the test charge to obtain
$$ \frac{W_{a \to b}}{q_{0}}=- \frac{ \Delta U}{q_{0}}=
- \left( \frac{U_{b}}{q_{0}}- \frac{U_{a}}{q_{0}}\right)=
-(V_{b}-V_{a})= V_{a}-V_{b}.$$
Finally, we introduce the concept of voltage, which is a potential difference. That is,
$$V_{ab}:=V_{a}-V_{b},$$
and we say that $V_{ab}$ is the potential of $a$ with respect to $b$. Hence,
$$ \frac{W_{a \to b}}{q_{0}}=V_{ab}.$$
Now we change our physical setup. Up until now, we have considered a constant electric field $E$. Suppose, instead, that we have a single point charge $q$ generating an electric field. The force it exerts on a test charge $q_{0}$ is given by the usual Coulomb's Law formula
$$F_{r}=\frac{kqq_{0}}{r^{2}}.$$
Following the procedure before, we must integrate to find the work done in moving a test charge from point $a$ to point $b$:
$$W_{a \to b}= \int_{r_{a}}^{r_{b}}F_{r} \,dr
=kqq_{0} \int_{r_{a}}^{r_{b}}r^{-2} \,dr
=kqq_{0} \left( \frac{1}{r_{a}}- \frac{1}{r_{b}} \right).$$
Recall that we still have $W=- \Delta U$, and hence it must be that
$$-(U_{b} - U_{a}) = U_{a}-U_{b}=kqq_{0} \left( \frac{1}{r_{a}}- \frac{1}{r_{b}} \right).$$
Hence, it is consistent to write that
$$U_{a}= \frac{kqq_{0}}{r_{a}} \quad \text{and} \quad
U_{b}= \frac{kqq_{0}}{r_{b}}.$$
So, from this, we generalize and claim that the potential energy at a test charge $q_{0}$ that is $r$ away from a charge $q$ is $U=kqq_{0}/r$.
Since the potential is $V=U/q_{0}$, we divide this equation by $q_{0}$ to obtain
$$V= \frac{kq}{r},$$
which was the goal of this exercise.
The incorrect thing to do in the point charge case is to say this:
$$W_{a \to b} = Fd = q_{0} E d,$$
but also
$$W_{a \to b}=- \Delta U=-q_{0}(V_{b}-V_{a}) = -q_{0}V_{ba}.$$
Hence,
$$-q_{0}V_{ba}= q_{0} E d \implies V_{ba}=-Ed=-\frac{kq}{r^{2}}\,d=-\frac{kq}{r}.$$
The problem with this, of course, is that the force is not constant in the point charge case, and hence the equation $W = F d$ doesn't work. You must integrate, which yields the minus sign.
The goal of this post is to arrive at the correct sign information for the potential due to a point charge:
$$V=\frac{kq}{r}.$$
Background: suppose a ball rolls down a hill, with a starting height of $y_{i}$ and a stopping height of $y_{f}$. These heights are measured relative to a position of low potential energy. Then the work done on the ball by the gravitational force is given by $$W_{i\to f}=-\Delta U=-(U_{f}-U_{i})=-(mgy_{f}-mgy_{i})=mg(y_{i}-y_{f}).$$
Since $y_{i}>y_{f}$, the work done by the gravitational force on the ball is positive. This makes sense, because, as Young and Freedman say on page 195 of University Physics, 9th Edition, "When the body moves up, $y$ increases, the work done by the gravitational force is negative, and the gravitational potential energy increases ($\Delta U>0$). When the body moves down, $y$ decreases, the gravitational force does positive work, and the gravitational potential energy decreases ($\Delta U<0$)." So far so good.
Moving on to the electrical case. Suppose I have a uniform electric field of field strength $E$ pointed down. This could happen in the middle of a capacitor, e.g. By the definition of the electric field, the lines go away from positive charge, and towards negative charge. Hence, I have positive charges up top (say, on the top plate of a capacitor), and negative charges down below. Now suppose I have a positive test charge $q_{0}$ that is in this field pointed down. Because the test charge is positive, it's going to want to go towards the negative plate, so the electric field will be trying to move the test charge down - with the field. The electric force on the test charge $q_{0}$ is $F=q_{0}E$, a constant. Hence, the work done by the field on the test charge in moving the test charge from $a$ to $b$ is given by
$$W_{a\to b}=\int_{r_{a}}^{r_{b}}F\,dx=Fd=q_{0}Ed,$$
where $d$ is the distance from point $a$ to point $b$, and equals $|r_{a}-r_{b}|$.
If we measure distance $y$ from the negative plate, which is typical, then the potential energy is $U=q_{0}Ey$. The work done by the electric field in moving the test charge from point $a$ to point $b$ is
$$W_{a\to b}=-\Delta U=-(U_{b}-U_{a})=-(q_{0}Ey_{b}-q_{0}Ey_{a})
=q_{0}E(y_{a}-y_{b}).$$
Again, Young and Freedman on page 732:
"When $y_{a}$ is greater than $y_{b}$ ..., the positive test charge $q_{0}$ moves downward, in the same direction as $\overset{\to}{\mathbf{\it{E}}}$; the displacement is in the same direction as the force $\overset{\to}{\mathbf{\it{F}}}=q_{0}\overset{\to}{\mathbf{\it{E}}},$ so the field does positive work and $U$ decreases."
Now we introduce the electric potential $V_{a}$ at a point $a$. This is NOT voltage, but electric potential. We define it to be the potential energy per unit charge: $V_{a}=U_{a}/q_{0}$. Here $U_{a}$ is the potential energy at point $a$ relative to some zero point, and $q_{0}$ is the test charge at point $a$. If we wish to write the work as a function of $V$, we divide through by the test charge to obtain
$$ \frac{W_{a \to b}}{q_{0}}=- \frac{ \Delta U}{q_{0}}=
- \left( \frac{U_{b}}{q_{0}}- \frac{U_{a}}{q_{0}}\right)=
-(V_{b}-V_{a})= V_{a}-V_{b}.$$
Finally, we introduce the concept of voltage, which is a potential difference. That is,
$$V_{ab}:=V_{a}-V_{b},$$
and we say that $V_{ab}$ is the potential of $a$ with respect to $b$. Hence,
$$ \frac{W_{a \to b}}{q_{0}}=V_{ab}.$$
Now we change our physical setup. Up until now, we have considered a constant electric field $E$. Suppose, instead, that we have a single point charge $q$ generating an electric field. The force it exerts on a test charge $q_{0}$ is given by the usual Coulomb's Law formula
$$F_{r}=\frac{kqq_{0}}{r^{2}}.$$
Following the procedure before, we must integrate to find the work done in moving a test charge from point $a$ to point $b$:
$$W_{a \to b}= \int_{r_{a}}^{r_{b}}F_{r} \,dr
=kqq_{0} \int_{r_{a}}^{r_{b}}r^{-2} \,dr
=kqq_{0} \left( \frac{1}{r_{a}}- \frac{1}{r_{b}} \right).$$
Recall that we still have $W=- \Delta U$, and hence it must be that
$$-(U_{b} - U_{a}) = U_{a}-U_{b}=kqq_{0} \left( \frac{1}{r_{a}}- \frac{1}{r_{b}} \right).$$
Hence, it is consistent to write that
$$U_{a}= \frac{kqq_{0}}{r_{a}} \quad \text{and} \quad
U_{b}= \frac{kqq_{0}}{r_{b}}.$$
So, from this, we generalize and claim that the potential energy at a test charge $q_{0}$ that is $r$ away from a charge $q$ is $U=kqq_{0}/r$.
Since the potential is $V=U/q_{0}$, we divide this equation by $q_{0}$ to obtain
$$V= \frac{kq}{r},$$
which was the goal of this exercise.
The incorrect thing to do in the point charge case is to say this:
$$W_{a \to b} = Fd = q_{0} E d,$$
but also
$$W_{a \to b}=- \Delta U=-q_{0}(V_{b}-V_{a}) = -q_{0}V_{ba}.$$
Hence,
$$-q_{0}V_{ba}= q_{0} E d \implies V_{ba}=-Ed=-\frac{kq}{r^{2}}\,d=-\frac{kq}{r}.$$
The problem with this, of course, is that the force is not constant in the point charge case, and hence the equation $W = F d$ doesn't work. You must integrate, which yields the minus sign.
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