Q#23.50
A point charge $q_1$ = +5.00 $\mu C$ is held fixed in space. From a horizontal distance of 6.00 cm, a small sphere with mass 4.00 $\times 10^{-3}$ kg and charge $q_2$ = +2.00 $\mu C$ is fired toward the fixed charge with an initial speed of 40.0 m/s. Gravity can be neglected. What is the acceleration of the sphere at the instant when its speed is 25.0 m/s?
Answer:
As the sphere approaches the point charge, the speed of the sphere decreases because it loses kinetic energy, but its acceleration increases because the electric force on it increases. Its mechanical energy is conserved during the motion, and Newton’s second law and Coulomb’s law both apply.
$K_a+U_a=K_b+U_b$
with $K= 1/2 mv^2$, $U=\frac{kq_1q_2}{r}$, $F=\frac{kq_1q_2}{r^2}$ and F = ma.
Find the distance between the two charges when $v_2$ = 25 0 m/s.
$K_a=\frac{1}{2}mv_a^2=\frac{1}{2}(4.00 \times 10^{-3} \ kg)(40.0 \ m/s)^2=3.20 \ J$
$K_b=\frac{1}{2}mv_b^2=\frac{1}{2}(4.00 \times 10^{-3} \ kg)(25.0 \ m/s)^2=1.25 \ J$
$U_a=\frac{kq_1q_2}{r_a}=\frac{(8.988 \times 10^9 \ N.m^2/C^2)(5.00 \times 10^{-6} \ C)^2}{0.0600 \ m}=1.498 \ J$
$U_b=K_a+U_a-K_b=3.20 \ J+1.498 \ J-1.25 \ J=3.448 \ J$
$U_b=\frac{kq_1q_2}{r_b}$ ⇒ $r_b=\frac{kq_1q_2}{U_b}$
$r_b=\frac{kq_1q_2}{r_a}=\frac{(8.988 \times 10^9 \ N.m^2/C^2)(5.00 \times 10^{-6} \ C)(2.00 \times 10^{-6} \ C)}{3.448 \ J}=0.02607 \ m$
$F_b=\frac{kq_1q_2}{r_b^2}$
$F_b=\frac{(8.988 \times 10^9 \ N.m^2/C^2)(5.00 \times 10^{-6} \ C)(2.00 \times 10^{-6} \ C)}{(0.02607 \ m)^2}=132.3 \ N$, so
$a=\frac{F}{m}=132.3 \ N/4.00 \times 10^{-3} \ m=3.31 \times 10^4 \ m/s^2$
As the sphere approaches the point charge, its speed decreases but its acceleration keeps increasing because the electric force on it keeps increasing.
Q#23.51
A point charge $q_1$ = 4.00 nC is placed at the origin, and a second point charge $q_2$ = -3.00 nC is placed on the -axis at x = +20.0 cm. A third point charge $q_3$ = +5.00 $\mu C$ is to be placed on the -axis between and (Take as zero the potential energy of the three charges when they are infinitely far apart.)
(a) What is the potential energy of the system of the three charges if is placed at
(b) Where should be placed to make the potential energy of the system equal to zero?
Answer:
We have U = $k\left(\frac{q_1q_2}{r_{12}}+\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}}\right)$
(a) given: $r_{12}$ = 0.200 m, $r_{23}$ = 0.100 m and $r_{13}$ = 0.100 m
U = $k\left(\frac{(4.00 \ nC)(-3.00 \ nC)}{0.200 \ m}+\frac{(4.00 \ nC)(2.00 \ nC)}{(0.100 \ m)}+\frac{(-3.00 \ nC)(2.00 \ nC)}{0.100 \ m}\right)$
U = $-3.60 \times 10^{-7} \ J$
(b) let particle 3 have coordinate x, so $r_{12}$ = 0.200 m, $r_{13}$= x and $r_{23}$ = 0.200 − x.
If U = 0, then
0 = $k\left(\frac{q_1q_2}{r_{12}}+\frac{q_1q_3}{x}+\frac{q_2q_3}{r_{12}-x}\right)$
Solving for x we find:
0 = $k\left(\frac{(4.00 \ nC)(-3.00 \ nC)}{0.200 \ m}+\frac{(4.00 \ nC)(2.00 \ nC)}{x}+\frac{(-3.00 \ nC)(2.00 \ nC)}{0.2 -x}\right)$
0 = $-60+\frac{8}{x}-\frac{6}{0.2-x}$
$60x^2-26x+1.6=0$
x = 0.074 m, 0.360 m
Therefore, x = 0.074 m since it is the only value between the two charges.
Q#23.52
A small sphere with mass and charge is released from rest a distance of 0.400 m above a large horizontal insulating sheet of charge that has uniform surface charge density . Using energy methods, calculate the speed of the sphere when it is 0.100 m above the sheet of charge?
Answer:
Two forces do work on the sphere as it falls: gravity and the electrical force due to the sheet. The energy of the sphere is conserved.
The gravity force is mg, downward. The electric field of the sheet is $E=\frac{\sigma}{2\epsilon_0}$ upward, and the force it exerts on the sphere is F = qE. The sphere gains kinetic energy
$K=\frac{1}{2}mv^2$ as it falls.
mg = $4.90 \times 10^{-6}$ N
$E=\frac{8.00 \times 10^{-12} \ C/m^2}{2(8.854 \times 10^{-12} \ C^2/(N.m^2))}=$0.4518 N/C.
The electric force is
qE = $(3.00 \times 10^{-6} \ C)(0.4518 \ N/C)=1.355 \times 10^{-6}$ N, upward.
The net force is downward, so the sphere moves downward when released. Let y = 0 at the sheet. $U_{grav}=mgy$. For the electric force,
$\frac{W_{a→b}}{q}=V_a-V_b$
Let point a be at the sheet and let point b be a distance y above the sheet. Take $V_a=0$. The force on q is qE, upward, so
$\frac{W_{a→b}}{q}=Ey$ and $V_b=-Ey$. $U_b=-Eyq$
$K_1+U_1=K_2+U_2$, $K_1=0$, $y_1=0.400 \ m$, $y_2=0.100 \ m$
$K_2=U_1-U_2=mg(y_1-y_2)-E(y_1-y_2)q$
$K_2=(5.00 \times 10^{-7} \ kg)(9.8 \ m/s^2)(0.300 \ m)-(0.4518 \ N/C)(0.300 \ m)(3.00 \times 10^{-6} \ C)$
$K_2=1.063 \times 10^{-6}$ J. $K_2=\frac{1}{2}mv_2^2$. so
$v_2=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1.063 \times 10^{-6} \ J)}{5.00 \times 10^{-7} \ kg}}=2.06 \ m/s$
Because the weight is greater than the electric force, the sphere will accelerate downward, but if it were light enough the electric force would exceed the weight. In that case it would never get closer to the sheet after being released.
Q#23.53
Determining the Size of the Nucleus. When radium-226 decays radioactively, it emits an alpha particle (the nucleus of helium), and the end product is radon-222. We can model this decay by thinking of the radium-226 as consisting of an alpha particle emitted from the surface of the spherically symmetric radon-222 nucleus, and we can treat the alpha particle as a point charge. The energy of the alpha particle has been measured in the laboratory and has been found to be 4.79 MeV when the alpha particle is essentially infinitely far from the nucleus. Since radon is much heavier than the alpha particle, we can assume that there is no appreciable recoil of the radon after the decay. The radon nucleus contains 86 protons, while the alpha particle has 2 protons and the radium nucleus has 88 protons.
(a) What was the electric potential energy of the alpha–radon combination just before the decay, in MeV and in joules?
(b) Use your result from part (a) to calculate the radius of the radon nucleus.
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