View page as slide show

Midterm Exam 1 Information

The first midterm exam will take place in class at 10AM on Tuesday March 4th. You will have 80 minutes to complete the exam. Make sure to be on time as the exam will finish promptly at 11:20 AM.

You may bring one sheet of letter size paper which has handwritten notes on both sides. You should also bring a calculator.

The exam will consist of 3 questions. 2 of the questions will concern electrostatics, ie. Electric force, field and potential. One of these questions will involve integral calculus, the other will not. The third question will cover capacitors and/or resistors and/or dc circuits and will not involve integral calculus.

The lecture on Monday 3rd March will be a review lecture and in place office hours that day there will be a review session from 2:00-3:30PM in B131 in the Physics building.

Results

The exam was definitely challenging, but overall the class did well and the average was 76.8%, which is very similar to previous years. A histogram of the scores is shown below.

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.

Practice Exams

There are two practice exams posted on Mastering Physics. There are also several extra practice problems, also on Mastering Physics. Below you can find midterms from previous years and solutions to them.

Spring 2011 Midterm 1

Spring 2011 Midterm 1 Solutions

Spring 2012 Midterm 1

Spring 2012 Midterm 1 Solutions

Spring 2013 Midterm 1

Spring 2013 Midterm 1 Solutions

Review Videos

2014 review part 1

2014 Review Part 2

phy142/examprep/m1s2014.txt · Last modified: 2014/04/14 21:28 by mdawber
CC Attribution-Noncommercial-Share Alike 3.0 Unported
Driven by DokuWiki