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A very long insulating cylindrical shell of radius
6.00 cm carries charge of linear density 8.50$\mu \mathrm{C} / \mathrm{m}$ spread uni-
formly over its outer surface. What would a voltmeter read if it
were connected between (a) the surface of the cylinder and a point
4.00 cm above the surface, and (b) the surface and a point 1.00 cm
from the central axis of the cylinder?

$\Delta v=\frac{\lambda}{2 \pi \varepsilon_{0}} \ln \left(\frac{r_{b}}{r_{a}}\right)$

$=\frac{\lambda}{2 \pi \varepsilon_0} \ln \left(\frac{r_{b}}{r_{a}}\right)=8.5 * 10^{-6} * 2\left(9 * 10^{9}\right) \ln \left(\frac{10}{6}\right)$

$=7 \cdot 82 * 10^{4} \mathrm{V}=78,200 \mathrm{V}=78.2 \mathrm{kV}$

Charge $Q=5.00 \mu \mathrm{C}$ is distributed uniformly over the
volume of an insulating sphere that has radius $R=12.0 \mathrm{cm} . \mathrm{A}$
small sphere with charge $q=+3.00 \mu \mathrm{C}$ and mass $6.00 \times 10^{-5} \mathrm{kg}$
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?

$K_{1}+U_{1}=K_{2}+U_{2} \rightarrow K_{1}=U_{2}$

$K=\frac{1}{2} m v^{2}$        ,         $U=\frac{k q Q}{r}$

$U_{2}=\frac{k{q}Q}{r_{2}} =\frac{k q Q}{12+8=20 \mathrm{cm}} =\frac{1}{2} m v^2$

$=(0.2 \mathrm{m})$

$\rightarrow V_{1}=\sqrt{\frac{2 k{q} Q}{m r_{2}}} =\sqrt{\frac{2 * 9*10^{4}+3 * 10^{-6} * 5 * 10^{-6}}{6 * 10^{-5} * 0.2}}$

$=150 \mathrm{m/s}$

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