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Physics 2 Exam 2 Test Prep

by Tyler Clarke

Helpfully enough, the physics department published a test outline! There will be,

•A question on GlowScript (for the uninitiated: GlowScript is Python for WebGL-based visualizations).
•A question on electric fields of distributed charges.
•A question on potential difference.
•A question on circuits.

There are a bunch of old practice exams from whence can be pulled sample questions that align with these. Let’s go through a few!

I’m not going to include the GlowScript problem, because GlowScript is literally just Python and because the GlowScript questions are long. If this is a thing you really want to practice, consider reading through a bunch of GlowScript programs to get familiar with the quirks of the language. Main point of consideration: it’s an incredibly stupid language.

Electric Fields of Distributed Charges

First off, very helpfully provided by one of my classmates on their own dungeon trawl:

This “single question” actually has three parts, as is customary in this hellish class.

Part one asks us to break up the slab into an infinite number of infinitely large charged discs with thickness dx and charge density pdx. Then, we’re asked to find the integral representing the net electric field at some point x much larger than L.

Because the electric field of a flat infinite disc is

\frac{pdx}{2 e_{0}}

– a constant, related to the charge density and completely ignorant of distance from the plate (think about it: does distance from the plate actually change anything if the plate is infinitely large?) – we know that the field of each plate

dE = \frac{pdx}{2 e_{0}}

. Integrating dE is absurdly easy:

\int_{0}^{L} dE \to \int_{0}^{L} \frac{p}{2 e_{0}} dx \to {\frac{px}{2 e_{0}} |}_{0}^{L} \to \frac{pL}{2 e_{0}}

It also wants to know the direction, which is just (because we know x >> L) the unit vector of the x-axis

\hat{x}

. That gives a final answer

\frac{pL}{2 e_{0}} \hat{x}

Part 2 of this question wants to know the magnitude and direction of the electric field on the other side. Because the electric field here is constant through all of space, this is also quite simply our answer from part 1 multiplied by -1:

- \frac{pL}{2 e_{0}} \hat{x}

The final part of our “single question” wants to know the magnitude and direction of the electric field inside the slab.

This is an interesting one.

The Obvious Solution™ is to just set up two integrals: one for the infinitesimal discs in front of our x point, and one for behind, and algebra from there. Seems reasonable:

\int_{0}^{x} \frac{p}{2 e_{0}} \hat{x} dx - \int_{x}^{L} \frac{p}{2 e_{0}} \hat{x} dx

This boils down to

\frac{px}{2 e_{0}} \hat{x} - (pL - px) \frac{\hat{x}}{2 e_{0}}

Which reduces further to

\frac{2 px - pL}{2 e_{0}} \hat{x}

. This is reasonable! At

x = \frac{L}{2}

, this is 0 as expected, and it seems to point in the right direction as x approaches each extrema.

I was mildly surprised to check the answers and see all of the above were correct. This was a deceptively simple problem.

Circuits

On one of the tests from Spring 2022, there’s this little gem of a problem:

This one (surprisingly) only has one question, and it is pretty simple: having waited long enough for the capacitor to charge and the circuit to reach steady state, what’s the current through each of the bulbs?

It’s immediately obvious that there are two loops: battery 1 → bulb 1 → bulb 3, and battery 2 → bulb 4 → bulb 2. This means we’re dealing with identical loops, each with the resistance of two bulbs in series, which is just 2R. Ohm’s law is helpful here:

V = IR \to Emf = I 2 R \to I = \frac{Emf}{2 R}

. Because current must be the same at any point in a series circuit, this is the current in all of the bulbs.

Potential Difference

Diligently hunted down by my auspicious peers is this precious pearl:

There are a bunch of steps before we can do any actual potential difference math.

First, we need to know

Q_{i}, Q_{v},

and

Q_{o}

. These are pretty easy to predict: because the net charge inside the volume must be 0, and the charges must otherwise match up to produce a zero field,

Q_{i} = - 24 Q, Q_{v} = 0, Q_{o} = 24 Q

The electric field vectors at each area marked with an “x” are also easy to find, and must be as follows:

•at R: 0 (the electric field inside a charged shell is 0)
•at 3R: pointing away from center
•at 5R: 0
•at 8R: weakly pointing away from center

Now for the meat of the question: finding the electric potential at 8R, 5R, 3R, and R. Because electric potential to infinity due to a charged object is just

\frac{kQ}{r}

, we have:

•8R: $k \frac{24 Q}{8 R} = 3 \frac{kQ}{R}$
•5R: because potential inside a conductor is zero, this is the same as the potential at 6R, which is $4 \frac{kQ}{R}$
•3R: this is the potential at 4R plus the potential between 4R and 3R: $P_{3 R} = P_{4 r} + (P_{3 R} - P_{4 R})$ . 4R is the same as at 6R (because there is no potential over the conductor), so this yields $P_{3 R} = 4 \frac{kQ}{R} + 24 kQ (\frac{1}{3 R} - \frac{1}{4 R}) = 4 \frac{kQ}{R} + 24 kQ (\frac{1}{12 R}) = 4 \frac{kQ}{R} + 2 \frac{kQ}{R} = 6 \frac{kQ}{R}$
•R: Because this is inside the shell, the potential here is the same as the potential at 2R. How do we find the potential at 2R? Same technique as above! The potential is $P_{2 R} = P_{3 R} + (P_{2 R} - P_{3 R})$ . The difference $P_{3 R} - P_{2 R} = 24 kQ * (\frac{1}{2 R} - \frac{1}{3 R}) = 4 \frac{kQ}{R}$ , and the potential at 3R is already known to be $6 \frac{kQ}{R}$ , so the potential at R is $10 \frac{kQ}{R}$ .

Potential difference is a difficult topic, so let’s do another.

The first question asks us to find the potential difference between A and B, knowing that d is much larger than s.

Because the electric field is produced by a dipole, we’re gonna have to use an integral. The dipole moment here is

[- qs, 0, 0]

(if you don’t know how this was calculated, review dipoles!). This means the field at any position r along the x axis is

2 k \frac{[- qs, 0, 0]}{r^{3}}

. Potential difference between A and B is just the integral from A to B of that formula:

[- 2 kqs, 0, 0] \int_{d}^{2 d} \frac{1}{r^{3}} = [- 2 kqs, 0, 0] {\frac{- 1}{2 r^{2}} |}_{d}^{2 d}

Since all we want is the magnitude, we can discard the directional vector, to get

- 2 kqs {\frac{- 1}{2 r^{2}} |}_{d}^{2 d} = 2 kqs (\frac{1}{8 d^{2}} - \frac{1}{2 d^{2}}) = \frac{- 3 kqs}{4 d^{2}}

. This is very close to correct, but is still wrong for a subtle reason: the voltage is the negative integral of the electric field, meaning our final result is

\frac{3 kqs}{4 d^{2}}

The second question asks about the kinetic energy of an electron released at location B, when it is far away from the dipole. Because the kinetic energy is starting at 0, we can assume that our final kinetic energy is equal to the work done by the electric field on the charge – which is potential difference times charge.

To find the potential difference between position B and infinity, we can use the same integral as above, but from B to infinity rather than from A to B. This works out nicely to

\frac{- kqs}{4 d^{2}}

. Multiplying this by the charge of an electron

e^{—}

, we get a total kinetic energy at infinity equal to

\frac{- e^{—} kqs}{4 d^{2}}

Finally, we need to find the potential difference between B and C. We know the electric potential at B to be

\frac{- kqs}{4 d^{2}}

, and the electric potential at C can be found to be

\frac{- kqs}{2 d^{2}}

;

V_{C} - V_{B} = \frac{- kqs}{2 d^{2}} - \frac{- kqs}{4 d^{2}} = \frac{- kqs}{4 d^{2}}