The idea of electric potential energy isn’t restricted to the special case of a uniform electric field. Indeed, we can apply this concept to a point charge in any electric field caused by a static charge distribution. Recall from Chapter 21 that we can represent any charge distribution as a collection of point charges. Therefore it’s useful to calculate the work done on a test charge moving in the electric field caused by a single, stationary point charge q.
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| Fig.1:Test charge $q_0$ moves along a straight line extending radially from charge $q$. As it moves from a to b the distance varies from $r_a$ to $r_b$. |
We’ll consider first a displacement along the radial line in Fig. 23.5. The force on $q_0$ is given by Coulomb’s law, and its radial component is
$F_r=\frac{1}{4\pi \epsilon_0}\frac{qq_0}{r^2}$
If q and have the same sign (+ or -) the force is repulsive and is positive; if the two charges have opposite signs, the force is attractive and is negative. The force is not constant during the displacement, and we have to integrate to calculate the work $W_{a→b}$ done on $q_0$ by this force as $q_0$ moves from a to b:
$W_{a→b}=\int_{r_a}^{r_b}F_rdr=\int_{r_a}^{r_b}\frac{1}{4\pi \epsilon_0}\frac{qq_0}{r^2}dr=\frac{qq_0}{4\pi \epsilon_0}\left(\frac{1}{r_1}-\frac{1}{r_b}\right)$
The work done by the electric force for this particular path depends only on the endpoints.
Now let’s consider a more general displacement (Fig. 23.6) in which a and b do not lie on the same radial line. From Eq. (23.1) the work done on $q_0$ during this displacement is given by
$W_{a→b}=\int_{r_a}^{r_b}F \ cos \ \phi dl=\int_{r_a}^{r_b}\frac{1}{4\pi \epsilon_0}\frac{qq_0}{r^2} \ cos \ \phi \ dl$
But Fig. 23.6 shows that $cos \ \phi \ dl=dr$ That is, the work done during a small displacement $d\mathbf{l}$ depends only on the change dr in the distance r between the charges, which is the radial component of the displacement. Thus Eq. (23.8) is valid even for this more general displacement; the work done on $q_0$ by the electric field $\mathbf{E}$ produced by q depends only on $r_a$ and $r_b$ not on the details of the path. Also, if $q_0$ returns to its starting point a by a different path, the total work done in the round-trip displacement is zero (the integral in Eq. (23.8) is from back $r_a$ to $r_b$). These are the needed characteristics for a conservative force, as we defined it in Section 7.3. Thus the force on $q_0$ is a conservative force.
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| Fig.2: The work done on charge $q_0$ by the electric field of charge $q$ does not depend on the path taken, but only on the distances $r_a$ and $r_b$. |
We see that Eqs. (23.2) and (23.8) are consistent if we define the potential energy to be $U_a=qq_0/4\pi \epsilon_0 r_a$ when $q_0$ is a distance $r_a$ from q, and to be $U_b=qq_0/4\pi \epsilon_0 r_b$ when is a distance $r_b$ from q. Thus the potential energy U when the test charge $q_0$ is at any distance r from charge q is
$U=\frac{1}{4\pi \epsilon_0}\frac{qq_0}{r}$ (electric potential energy of two point charges $q$ and $q_0$)
Equation (23.9) is valid no matter what the signs of the charges $q$ and $q_0$. The potential energy is positive if the charges $q$ and $q_0$ have the same sign (Fig. 23.7a) and negative if they have opposite signs (Fig. 23.7b).
Potential energy is always defined relative to some reference point where $U=0$. In Eq. (23.9), U is zero when $q$ and $q_0$ are infinitely far apart and $r=\infty$. Therefore U represents the work that would be done on the test charge $q_0$ by the field of $q$ if $q_0$ moved from an initial distance $r$ to infinity. If $q$ and $q_0$ have the same sign, the interaction is repulsive, this work is positive, and $U$ is positive at any finite separation (Fig. 23.7a). If the charges have opposite signs, the interaction is attractive, the work done is negative, and $U$ is negative (Fig. 23.7b).
We emphasize that the potential energy U given by Eq. (23.9) is a shared property of the two charges. If the distance between $q$ and $q_0$ is changed from $r_a$ to $r_b$, the change in potential energy is the same whether $q$ is held fixed and $q_0$ is moved or $q_0$ is held fixed and $q$ is moved. For this reason, we never use the phrase “the electric potential energy of a point charge.” (Likewise, if a mass $m$ is at a height $h$ above the earth’s surface, the gravitational potential energy is a shared property of the mass $m$ and the earth. We emphasized this in Sections 7.1 and 13.3.)
Equation (23.9) also holds if the charge $q_0$ is outside a spherically symmetric charge distribution with total charge $q$; the distance $r$ is from $q_0$ to the center of the distribution. That’s because Gauss’s law tells us that the electric field outside such a distribution is the same as if all of its charge $q$ were concentrated at its center (see Example 22.9 in Section 22.4).
Example 1
(a) Determine the electrostatic potential energy of a system consisting of two charges 7 μC and –2 μC (and with no external field) placed at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively. (b) How much work is required to separate the two charges infinitely away from each other? (c) Suppose that the same system of charges is now placed in an external electric field $E = A (1/r^2)$; $A = 9 \times 10^5 N.C^{–1}.m^2$. What wouldthe electrostatic energy of the configuration be?
Solution:
(a) $U=\frac{1}{4\pi \epsilon_0}\frac{qq_0}{r}$
$U=(9.0 \times 10^9 \ N.m^2/C^2)\frac{(7 \times 10^{-6} \ C)(-2 \times 10^{-6} \ C)}{0.18 \ m}$
$U=-0.7 \ J$
(b) $W=U_2-U_1= U_{\infty}-U_1$
$W=0-(-0.7 \ J)=0.7 \ J$
(c) The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find,
$q_1V\mathbf{(r_1)}+q_2V\mathbf{(r_2)}=A\frac{7 \times 10^{-6}}{0.09 \ m}+A\frac{-2 \times 10^{-6}}{0.09 \ m}$
and the net electrostatic energy is
$q_1V\mathbf{(r_1)}+q_2V\mathbf{(r_2)}+\frac{q_1q_2}{4\pi \epsilon_0r_{12}}$
$=A\frac{7 \times 10^{-6}}{0.09 \ m}+A\frac{-2 \times 10^{-6}}{0.09 \ m}-0.7 \ J=49.3 \ J$
Example 2
A positron (the electron’s antiparticle) has mass $9.11 \times 10^{-31} \ kg$ and charge $q_0=+e=+1.60 \times 10^{-19} \ C$ Suppose a positron moves in the vicinity of an (alpha) particle, which has charge $q=+2e=3.20 \times 10^{-19} \ C$ and mass $6.64 \times 10^{-27} \ kg$. The $\alpha$ particle’s mass is more than 7000 times that of the positron, so we assume that the $\alpha$ particle remains at rest. When the positron is $1.00 \times 10^{-10} \ m$ from the $\alpha$ particle, it is moving directly away from the $\alpha$ particle at $3.00 \times 10^6 \ m/s$. (a) What is the positron’s speed when the particles are $1.00 \times 10^{-10} \ m$ apart? (b) What is the positron’s speed when it is very far from the $\alpha$ particle? (c) Suppose the initial conditions are the same but the moving particle is an electron (with the same mass as the positron but charge $q_0=e$). Describe the subsequent motion.
Solution:
(a) Both particles have positive charge, so the positron speeds up as it moves away from the particle. From the energy conservation equation, Eq. (23.3), the final kinetic energy is
$K_b=\frac{1}{2}mv_b^2=K_a+U_a-U_b$
In this expression,
$K_a=\frac{1}{2}mv_a^2$
$K_a=\frac{1}{2}(9.11 \times 10^{-31} \ kg)(3.00 \times 10^6 \ m/s)^2$
$K_a=4.10 \times 10^{-18} \ J$
$U_a=\frac{1}{4\pi \epsilon_0}\frac{qq_0}{r_a}$
$U_a=(9.0 \times 10^9 \ N.m^2/C^2)\left(\frac{(3.20 \times 10^{-19} \ C)(1.60 \times 10^{-19} \ C)}{(1.00 \times 10^{-10} \ m)}\right)$
$U_a=4.61 \times 10^{-18} \ J$
$U_b=\frac{1}{4\pi \epsilon_0}\frac{qq_0}{r_a}$
$U_b=(9.0 \times 10^9 \ N.m^2/C^2)\left(\frac{(3.20 \times 10^{-19} \ C)(1.60 \times 10^{-19} \ C)}{(2.00 \times 10^{-10} \ m)}\right)$
$U_b=2.30 \times 10^{-18} \ J$
Hence the positron kinetic energy and speed at $r=r_b=2.00 \times 10^{-10} \ m$ are
$K_b=\frac{1}{2}mv_b^2$
$K_b=4.10 \times 10^{-18} \ J+4.61 \times 10^{-18} \ J-2.30 \times 10^{-18} \ J$
$K_b=6.41 \times 10^{-18} \ J$
$v_b=\sqrt{\frac{2K_b}{m_e}}$
$v_b=\sqrt{\frac{2(6.41 \times 10^{-18} \ J)}{9.11 \times 10^{-31} \ kg}}$
$v_b=3.8 \times 10^6 \ m/s$
(b) When the positron and particle are very far apart so that $r=r_c →\infty$ the final potential energy approaches zero. Again from energy $U_c$ conservation, the final kinetic energy and speed of the positron in this case are
$K_c=K_a+K_b-U_c$
$K_c=4.10 \times 10^{-18} \ J+4.61 \times 10^{-18} \ J-0$
$K_c=8.71 \times 10^{-18} \ J$
$v_c=\sqrt{\frac{2K_c}{m_e}}$
$v_c=\sqrt{\frac{2(8.71 \times 10^{-18} \ J)}{9.11 \times 10^{-31} \ kg}}$
$v_c=4.4 \times 10^6 \ m/s$
(c) The electron and particle have opposite charges, so the force is attractive and the electron slows down as it moves away. Changing the moving particle’s sign from $+e$ to $-e$ means that the initial potential energy is now $U_a=-4.61 \times 10^{-18} \ J$, which makes the total mechanical energy negative:
$K_a+U_a=(4.10 \times 10^{-18} \ J)-(4.61 \times 10^{-18} \ J)$
$=-0.51 \times 10^{-18} \ J$
The total mechanical energy would have to be positive for the electron to move infinitely far away from the $\alpha$ particle. Like a rock thrown upward at low speed from the earth’s surface, it will reach a maximum separation $r=r_d$ from the $\alpha$ particle before reversing direction. At this point its speed and its kinetic energy $K_d$ are zero, so at separation $r_d$ we have
$U_d=K_a+U_a-K_d$
$U_d=\frac{1}{4\pi \epsilon_0} \frac{qq_0}{r_d}=-0.51 \times 10^{-18} \ J$
$r_d=\frac{qq_0}{4U_d \pi \epsilon_0}$
$r_d=\frac{(9.0 \times 10^9 \ N. m^2/C^2)}{(-0.51 \times 10^{-18} \ J)}\times (3.20 \times 10^{-19} \ C)(-1.60 \times 10^{-19} \ C)$
$r_c=9.0 \times 10^{-10} \ m$
For $r_b=2.00 \times 10^{-10} \ m$ we have $U_b=-2.30 \times 10^{-18} \ J$, so the electron kinetic energy and speed at this point are
$K_b=\frac{1}{2}mv_b^2$
$K_b=4.10 \times 10^{-18} \ J+(-4.61 \times 10^{-18} \ J)-(-2.30 \times 10^{-18} \ J)$
$K_b=1.79 \times 10^{-18} \ J$
$v_c=\sqrt{\frac{2K_c}{m_e}}$
$v_c=\sqrt{\frac{2(1.79 \times 10^{-18} \ J)}{9.11 \times 10^{-31} \ kg}}$
$v_c=2.0 \times 10^6 \ m/s$
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