Recitation 5

Question 1 Solution (35 points) (Average Score: 29.6/35)

Five identical electrons with charge $-e$ and mass $m_{e}$ are arranged as shown in the figure. Give all of your answers below in terms of the variables $e,d, m_{e}$ and $k$ or $ \varepsilon_{0}$. Take the zero of electrical potential to be at $\infty$.

(a) (5 points) For the arrangement shown, what is the magnitude of the electric field at the point labeled $x$?

$E=-\frac{ke}{d^{2}}$

(b) (5 points) For the arrangement shown, what is the electrical potential at the point labeled $x$?

$V=-\frac{ke}{d}-4\frac{ke}{\sqrt{2}d}=-3.828\frac{ke}{d}$

© (5 points) For the arrangement shown, what is the magnitude of the electric field at the point labeled $y$?

$E=-\frac{ke}{4d^{2}}-2\frac{ke}{5d^{2}}\frac{2}{\sqrt{5}}=-0.608\frac{ke}{d^{2}}$

(d) (5 points) For the arrangement shown, what is the electrical potential at the point labeled $y$?

$V=-2\frac{ke}{d}-\frac{ke}{2d}-2\frac{ke}{\sqrt{5}d}=-3.394\frac{ke}{d}$

(e) (5 points) How much electrostatic potential energy is stored in the arrangement shown?

$U=\frac{ke^2}{d}([1]+[\frac{1}{2}+1]+[\frac{1}{2}+\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{8}}]+[\frac{1}{2}+\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{5}}+\frac{1}{2}])=5.602\frac{ke^2}{d}$

(f) (5 points) If a sixth electron is placed at $x$ and released (while all the other electrons are held in place) how fast is the electron moving when it reaches $y$?

$\Delta U = -e (V_{y}-V_{x})=-e(-\frac{3}{2}\frac{ke}{d}-2\frac{ke}{\sqrt{5}d}+4\frac{ke}{\sqrt{2}d})$

$\frac{1}{2}m_{e}v^{2}=-\Delta U$

$v=\sqrt{\frac{2}{m_{e}}(\frac{4}{\sqrt{2}}-\frac{3}{2}-\frac{2}{\sqrt{5}})\frac{ke^2}{d}}=\sqrt{\frac{2}{m_{e}}\times0.434\times\frac{ke^2}{d}}=\sqrt{0.868m_{e}\frac{ke^2}{d}}$

(g) (5 points) What is the maximum speed achieved by the sixth electron which was placed at $x$ and then released?

$\Delta U = -e (V_{\infty}-V_{x})=-e(\frac{ke}{d}+4\frac{ke}{\sqrt{2}d})$

$\frac{1}{2}m_{e}v^{2}=-\Delta U$

$v=\sqrt{\frac{2}{m_{e}}(1+\frac{4}{\sqrt{2}})\frac{ke^2}{d}}=\sqrt{\frac{2}{m_{e}}\times3.828\times\frac{ke^2}{d}}=\sqrt{7.567m_{e}\frac{ke^2}{d}}$

Question 2 Solution (35 points) (Average Score: 22.9/35)

A insulating sphere of radius $r_{0}$ has a total positive charge +Q which is evenly distributed throughout. It is surrounded by an insulating spherical shell with total negative charge -Q, which starts at radius $r_{1}$ and ends at a radius $r_{2}$. The charge in the shell is also evenly distributed. Let $r$ be the distance of a point from the center of the sphere. Write your answers in terms of $Q$, $r$, $r_{2}$, $r_{1}$, $r_{0}$ and $k$ or $\varepsilon_{0}$.

(a) (5 points) What is the magnitude of the electric field $E(r)$ for points $r<r_{0}$?

$E4\pi r^{2}=\frac{1}{\varepsilon_{0}}\rho \frac{4}{3}\pi r^{3}$

$E=\frac{1}{\varepsilon_{0}}\rho\frac{r}{3}$

$\rho=\frac{Q}{4/3\pi r_{0}^3}$

$E=\frac{1}{4\pi \varepsilon_{0}} \frac{Q}{r_{0}^{3}}r$

(b) (5 points) What is the magnitude of the electric field $E(r)$ for points $r_{0}\leq r<r_{1}$?

$E4\pi r^{2}=\frac{1}{\varepsilon_{0}}Q$

$E=\frac{1}{4\pi \varepsilon_{0}} \frac{Q}{r^{2}}$

© (5 points) What is the magnitude of the electric field $E(r)$ for points $r_{1}\leq r<r_{2}$?

$E4\pi r^{2}=\frac{1}{\varepsilon_{0}}[Q-Q\frac{4/3\pi(r^3-r_{1}^{3})}{4/3\pi(r_{2}^3-r_{1}^{3})}]$

$E=\frac{Q}{4\pi \varepsilon_{0}}\frac{1}{r^{2}}[1-\frac{(r^3-r_{1}^{3})}{(r_{2}^3-r_{1}^{3})}]$

(d) (5 points) What is the magnitude of the electric field $E(r)$ for points $r\geq r_{2}$?

$E=0\mathrm{N/C}$

(e) (5 points ) Plot the magnitude of the electric field $E(r)$ against $r$. Make sure to label $r_{0}$,$r_{1}$ and $r_{2}$ on your sketch.

(f) (5 points) If the zero of electric potential is taken to be $r=\infty$ find an expression for the value of the electric potential at r=0. (Hint: You need to evaluate the integral $-\int_{r=\infty}^{r=0}\vec{E}\cdot d\vec{l}$.)

$V=V_{\infty}+\Delta V=V_{\infty}-\int_{r=\infty}^{r=0}\vec{E}\cdot d\vec{l}=\int_{r=0}^{r=\infty}E(r)dr=\int_{r=0}^{r=r_{0}}E(r)dr+\int_{r=r_{0}}^{r=r_{1}}E(r)dr+\int_{r=r_{1}}^{r=r_{2}}E(r)dr$

$=\frac{Q}{4\pi \varepsilon_{0}}\left[ [ \frac{1}{2r_{0}}]+[ \frac{1}{r_{0}}- \frac{1}{r_{1}}]+[(1-\frac{r_{1}^{3}}{r_{2}^{3}-r_{1}^{3}})(\frac{1}{r_{1}}- \frac{1}{r_{2}})+\frac{1}{r_{2}^{3}-r_{1}^{3}}\frac{1}{2}(r_{1}^{2}-r_{2}^{2})]\right]$

(g) (5 points) Plot the potential $V(r)$ against $r$, taking the zero of potential to be $r=\infty$. Make sure to label $r_{0}$, $r_{1}$ and $r_{2}$ on your sketch.

Question 3 Solution (30 points) (Average Score: 24.3/30)

Two square metallic plates with side length 5 cm, separated by a distance of 0.8mm are connected to the terminals of a 9V battery. The permittivity of free space $\varepsilon_{0}=8.85\times10^{-12}\,\mathrm{F\,m^{-1}}$ may be useful in this problem.

(a) (5 points) What is the magnitude of the electric field halfway between the two plates?

$E=\frac{9}{0.8\times10^{-3}}=11250\mathrm{V/m}$

(b) (5 points) What is the magnitude of the charge density on the inside surface of each of the plates?

$\frac{Q}{A}=\varepsilon_{0}E=9.956\times10^{-8}\mathrm{C/m}$

or

$C=\frac{\varepsilon_{0}A}{d}=2.766\times 10^{-11}\mathrm{F}$

$Q=CV=2.489\times10^{-10}\mathrm{C}$

$\frac{Q}{A}=9.956\times10^{-8}\mathrm{C/m}$

© (5 points) How much electrical potential energy is stored in this arrangement?

$U=\frac{1}{2}CV^{2}=\frac{1}{2}\frac{Q^2}{C}=1.12\times10^{-9}\mathrm{J}$

A third plate of the same area with thickness 0.4\,mm is inserted exactly halfway between the two plates.

(d) (5 points) What is the magnitude of the electric field halfway between the two plates now?

$E=0\mathrm{V/m}$

(e) (5 points) What is the magnitude of the charge density on the inside surface of each of the two original plates now?

Field in new capacitors is

$E=\frac{9}{0.4\times10^{-3}}=22500\mathrm{V/m}$

$\frac{Q}{A}=\varepsilon_{0}E=1.991\times10^{-7}\mathrm{C/m}$

or

$C=\frac{\varepsilon_{0}A}{d}=1.11\times 10^{-10}\mathrm{F}$

$Q=CV=4.978\times10^{-10}\mathrm{C}$

where V=4.5V

$\frac{Q}{A}=1.991\times10^{-7}\mathrm{C/m}$

(f) (5 points) How much electrical potential energy is stored in this arrangement?

Each capacitor stores

$U=\frac{1}{2}CV^{2}=\frac{1}{2}\frac{Q^2}{C}=1.12\times10^{-9}\mathrm{J}$

where V=4.5V

but the total energy is $2U=2.24\times10^{-9}\mathrm{J}$. The same result can be found by considering the equivalent capacitance, which is twice the original.

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