Q#23.34
A ring of diameter 8.00 cm is fixed in place and carries a charge of +5.00 $\mu$C uniformly spread over its circumference.
(a) How much work does it take to move a tiny +3.00 $\mu$C charged ball of mass 1.50 g from very far away to the center of the ring?
(b) Is it necessary to take a path along the axis of the ring? Why?
(c) If the ball is slightly displaced from the center of the ring, what will it do and what is the maximum speed it will reach?
Answer:
The work required is equal to the change in the electrical potential energy of the charge-ring system. We need only look at the beginning and ending points, since the potential difference is independent of path for a conservative field.
(a) $W=\Delta U=q\Delta U$
$W=q(V_{center}-V_{\infty})$
$W=q\left(\frac{1}{4\pi \epsilon_0}\frac{Q}{a}-0\right)$
Substituting numbers gives
$\Delta U=(3.00 \times 10^{-6} \ C)(9.00 \times 10^9 \ N.m^2/C^2)(5.00 \times 10^{-6} \ C)/(0.0400 \ m)$
$\Delta U$ = 3.38 J
(b) We can take any path since the potential is independent of path.
(c) The net force is away from the ring, so the ball will accelerate away. Energy conservation gives
$U_0=K_{max}=\frac{1}{2}mv^2$
Solving for v gives
$v=\sqrt{\frac{2U_0}{m}}$
$v=\sqrt{\frac{2(3.38 \ J)}{0.00150 \ kg}}$
v = 67.1 m/s
Direct calculation of the work from the electric field would be extremely difficult, and we would need to know the path followed by the charge. But, since the electric field is conservative, we can bypass all this calculation just by looking at the end points (infinity and the center of the ring) using the potential.
Q#23.35
A very small sphere with positive charge q = +8.00 $\mu$C is released from rest at a point 1.50 cm from a very long line of uniform linear charge density $\lambda$ = +3.00 $\mu$C/m. What is the kinetic energy of the sphere when it is 4.50 cm from the line of charge if the only force on it is the force exerted by the line of charge?
Answer:
The electric field of the line of charge does work on the sphere, increasing its kinetic energy.
$K_1+U_1=K_2+U_2$ and
$K_1$ = 0, U = qV so
$qV_1=K_2+qV_2$
$K_2=q(V_1-V_2)$
with $V=\frac{\lambda q}{2\pi \epsilon_0}ln\left(\frac{r_0}{r}\right)$
$V_1=\frac{\lambda q}{2\pi \epsilon_0}ln\left(\frac{r_0}{r_1}\right)$
$V_2=\frac{\lambda q}{2\pi \epsilon_0}ln\left(\frac{r_0}{r_2}\right)$, so
$K_2=\frac{\lambda q}{2\pi \epsilon_0}\left[ln\left(\frac{r_0}{r_1}\right)-ln\left(\frac{r_0}{r_2}\right)\right]$
$K_2=\frac{\lambda q}{2\pi \epsilon_0}ln(r_2-r_1)$
$K_1=\frac{\lambda q}{2\pi \epsilon_0}ln\left(\frac{r_2}{r_1}\right)$
$K_1=\frac{(3.00 \times 10^{-6} \ C/m)(8.00 \times 10^{-6} \ C)}{2\pi (8.85 \times 10^{-12} \ C^2/N.m^2)}ln\left(\frac{4.50 \ cm}{1.50 \ cm}\right)$
$K_1$ = 0.474 J
The potential due to the line of charge does not go to zero at infinity but is defined to be zero at an arbitrary distance $r_0$ from the line.
Q#23.36
Charge Q = 5.00 $\mu$C is distributed uniformly over the volume of an insulating sphere that has radius R = 12.0 cm. A small sphere with charge q = +3.00 $\mu$C and mass 6.00 $\times 10^{-5}$ kg is projected toward the center of the large sphere from an initial large distance. The large sphere is held at a fixed position and the small sphere can be treated as a point charge. What minimum speed must the small sphere have in order to come within 8.00 cm of the surface of the large sphere?
Answer:
If the small sphere is to have its minimum speed, it must just stop at 8.00 cm from the surface of the large sphere.
In that case, the initial kinetic energy of the small sphere is all converted to electrical potential energy at its point of closest approach.
$K_1+U_1=K_2+U_2$ and
$K_2$ = 0, $U_1$ = 0. Therefore,
$K_1=U_2$
Outside a spherical charge distribution the potential is the same as for a point charge at the location of the center of the sphere, so
$U=\frac{kqQ}{r}$ and $K=\frac{1}{2}mv^2$
$\frac{kqQ}{r_2}=\frac{1}{2}mv_1^2$
$v_1=\sqrt{\frac{2kqQ}{mr_2}}$
with $r_2$ = 12.0 cm + 8.00 cm = 20.0 cm = 0.200 m
$v_1=\sqrt{\frac{2(8.99 \times 10^9 \ N.m^2/C^2)(3.00 \times 10^{-6} \ C)(5.00 \times 10^{-6} \ C)}{(6.00 \times 10^{-5} \ kg)(0.200 \ m)}}$
v = 150 m/s
If the small sphere had enough initial speed to actually penetrate the surface of the large sphere, we could no longer treat the large sphere as a point charge once the small sphere was inside.
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