Q#23.1
A point charge $q_1$ = +2$\mu$C is held stationary at the origin. A second point charge $q_2$ = $-4.30 \mu$C moves from the point x = 0.150 m, y = 0 to the point x = 0.250 m, y = 0,250 m. How much work is done by the electric force on $q_2$?
Answer:
Apply Eq. (23.2) to calculate the work. The electric potential energy of a pair of point charges is given by Eq. (23.9).
Let the initial position of $q_2$ be point a and the final position be point b, as shown in Figure 23.1.
$r_a$ = 0.150 m
$r_b=\sqrt{(0.250 \ m)^2+(0.250 \ m)^2}$ = 0.3536 m, so
$W_{a→b}=U_a-U_b$
$U_a=k\frac{q_1q_2}{r_a}$
$U_a=(8.988 \times 10^9 \ N.m^2/C^2)\frac{(+2.40 \times 10^{-6} \ C)(-4.30 \times 10^{-6} \ C)}{0.150 \ m}=-0.6184$ J
$U_b=(8.988 \times 10^9 \ N.m^2/C^2)\frac{(+2.40 \times 10^{-6} \ C)(-4.30 \times 10^{-6} \ C)}{0.3536 \ m}=-0.2623$ J
$W_{a→b}=-6184 \ J-(-0.2623 \ J)=-0.356$ J
The attractive force on $q_2$ is toward the origin, so it does negative work on $q_2$ when $q_2$ moves to larger r.
Q#23.2
A point charge is held stationary at the origin. A second charge $q_2$ is placed at point and the electric potential energy of the pair of charges is +5.4 $\times 10^{-8}$ J. When the second charge is moved to point b, the electric force on the charge does $-1.9 \times 10^{-8}$ J of work. What is the electric potential energy of the pair of charges when the second charge is at point b?
Answer:
Apply $W_{a→b}=U_a-U_b$
$U_a=+5.4 \times 10^{-8}$ J. Solve for $U_b$.
$W_{a→b}=-1.9 \times 10^{-8} \ J=+5.4 \times 10^{-8} \ J-U_b$
$U_b=5.4 \times 10^{-8} \ J+1.9 \times 10^{-8} \ J=7.3 \times 10^{-8} \ J$
When the electric force does negative work the electrical potential energy increases.
Q#23.3
How much work is needed to assemble an atomic nucleus containing three protons (such as Be) if we model it as an equilateral triangle of side +2.00 $\times 10^{-15}$ m with a proton at each vertex? Assume the protons started from very far away.
Answer:
The work needed to assemble the nucleus is the sum of the electrical potential energies of the protons in the nucleus, relative to infinity.
The total potential energy is the scalar sum of all the individual potential energies, where each potential energy is $U=\frac{1}{4\pi \epsilon_0}\frac{qq_0}{r}$.
Each charge is e and the charges are equidistant from each other, so the total potential energy is
$U=\frac{1}{4\pi \epsilon_0}\left(\frac{e^2}{r}+\frac{e^2}{r}+\frac{e^2}{r}\right)=\frac{3e^2}{4\pi r \epsilon_0}$
Adding the potential energies gives
$U=\frac{3(1.60 \times 10^{-19} \ C)^2(9.00 \times 10^9 N.m^2/C^2)}{2.00 \times 10^{-15} \ m}=3.26 \times 10^{-13} \ J=2.16$ MeV
This is a small amount of energy on a macroscopic scale, but on the scale of atoms, 2 MeV is quite a lot of energy.
Q#23.4
(a) How much work would it take to push two protons very slowly from a separation of +2.00 $\times 10^{-10}$ m (a typical atomic distance) to 3.00 $\times 10^{-15}$ m (a typical nuclear distance)?
(b) If the protons are both released from rest at the closer distance in part (a), how fast are they moving when they reach their original separation?
Answer:
The work required is the change in electrical potential energy. The protons gain speed after being released because their potential energy is converted into kinetic energy.
(a) Using the potential energy of a pair of point charges relative to infinity, $U=\frac{1}{4\pi \epsilon_0}\frac{qq_0}{r}$
we have W=$\Delta U=U_2-U_1=\frac{1}{4\pi \epsilon_0}\left(\frac{e^2}{r_2}-\frac{e^2}{r_1}\right)$
Factoring out the $e^2$ and substituting numbers gives
W$=(9.00 \times 10^9 \ N.m^2/C^2)(1.6 \times 10^{-19} \ C)^2\left(\frac{1}{3.00 \times 10^{-15} \ m}-\frac{1}{2.00 \times 10^{-10} \ m}\right)=7.68 \times 10^{-14}$ J
(b) The protons have equal momentum, and since they have equal masses, they will have equal speeds and hence equal kinetic energy.
$\Delta U = K_1+K_2=2K=mv^2$. Solving for v gives
$v=\sqrt{\frac{\Delta U}{m}}=\sqrt{\frac{7.68 \times 10^{-14} \ J}{1.67 \times 10^{-27} \ kg}}=6.78 \times 10^6$ m/s
The potential energy may seem small (compared to macroscopic energies), but it is enough to give each proton a speed of nearly 7 million m/s.
Q#23.5
A small metal sphere, carrying a net charge of $q_1$ = $-2.80 \mu C$, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of $q_2$ = $-7.80 \mu C$ and mass 1.50 g, is projected toward $q_1$. When the two spheres are 0.800 m apart, $q_2$ is moving toward $q_1$ with speed 22.0 m/s (Fig. E23.5). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.
(a) What is the speed of $q_2$ when the spheres are 0.400 m apart? (b) How close does $q_2$ get to $q_1$?
Answer:
(a) Use conservation of energy:
$K_a+U_a+W_{other}=K_b+U_b$
U for the pair of point charges is given by Eq. (23.9).
Let point a be where $q_2$ is 0.800 m from $q_1$ and point b be where $q_2$ is 0.400 m from 1q , as shown in Figure 23.5a.
Only the electric force does work, so $W_{other}=0$ and $U=\frac{1}{4\pi \epsilon_0}\frac{q_1q_2}{r}$
$K_a=\frac{1}{2}mv_a^2=\frac{1}{2}(1.50 \times 10^{-3} \ kg)(22.0 \ m/s)^2=0.3630$ J
$U_a=\frac{1}{4\pi \epsilon_0}\frac{q_1q_2}{r_a}=(8.988 \times 10^9 \ N.m^2/C^2)\frac{(-2.80 \times 10^{-6} \ C)(-7.80 \times 10^{-6} \ C)}{0.800 \ m}=+0.2454$ J
$K_b=\frac{1}{2}mv_b^2$
$U_b=\frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r_b}=(8.988 \times 10^9 \ N.m^2/C^2)\frac{(-2.80 \times 10^{-6} \ C)(-7.80 \times 10^{-6} \ C)}{0.400 \ m}=+0.4907$ J
The conservation of energy equation then gives
$K_b=K_a+(U_a-U_b)$
$\frac{1}{2}mv_b^2=+0.3630 \ J+(0.2454 \ J-0.4907 \ J)$ = 0.1177 J
$v_b=\sqrt{\frac{2(0.1177 \ J)}{1.50 \times 10^{-3} \ kg}}=12.5 \ m/s$
The potential energy increases when the two positively charged spheres get closer together, so the kinetic energy and speed decrease.
(b) Let point c be where $q_2$ has its speed momentarily reduced to zero. Apply conservation of energy to points a and c:
$K_a+U_a+W_{other}=K_c+U_c$
Points a and c are shown in Figure 23.5b.
$K_a=+0.3630 \ J$ (from part (a))
$K_b=+0.2454 \ J$ (from part (a))
$K_c$ = (at distance of closest approach the speed is zero)
$U_c=\frac{1}{4\pi \epsilon_0}\frac{q_1q_2}{r_c}$
Thus conservation of energy
$K_a+U_a=U_c$ gives
$\frac{1}{4\pi \epsilon_0}\frac{q_1q_2}{r_c}=+0.3630 \ J+0.2454 \ J=0.6080 \ J$
$r_c=(8.988 \times 10^9 \ N.m^2/C^2)\frac{(-2.80 \times 10^{-6} \ C)(-7.80 \times 10^{-6} \ C)}{+0.6084 \ J}$ = 0.323 m
U → ∞ as r → 0 so $q_2$ will stop no matter what its initial speed is.
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