exam review

site/posts/physics-final-exam.html

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+[!]
+[=post-]
+<p>
+    Hello, everyone! I've only a bit more than two days until my physics final (officially at the point of counting hours...) and haven't studied at all. Fortunately,
+    the physics department are the absolute goats, and have kindly posted a really detailed overview of the test, so I know exactly what's necessary to study.
+    Contained herein are my worked solutions to a bunch of different problems adapted from homework and old exams.
+</p>
+<h2>Electric Field and Force</h2>
+<p>
+    Let's start with some easy stuff <i>(editors note: I feel deeply stupid for having written the preceding)</i>. The critical point for all of these is Coulomb's law - `E = frac 1 {4 pi e_0} frac q {r^2} hat r`, `F = qE`. In simple English:
+    electric field due to a point charge is the magnitude of the charge, times a constant `k = frac 1 {4 pi e_0}`, divided by distance cubed, in the direction
+    of the distance vector `r`. Force on a particle is charge times electric field. This can be turned into an integral expression that finds the net electric field
+    due to almost any shape by representing it as a bunch of point charges.
+</p>
+<p>
+    Most point charge questions are far too easy to merit inclusion here, even for the sake of completeness; let's start with an interesting one. If you're in 2212 at
+    Tech, you'll recognize this as being a rough adaptation of GPS 2 #2. We're given a dipole with negative end facing an uncharged polarizable atom (with polarizability `alpha`).
+    The center is `d` meters away. We need to find the force on the dipole due to the atom.
+</p>
+<p>
+    The first step is, of course, to find the induced dipole in the neutral atom. Because `p = alpha E`, and `E_"para" = frac {2kp} {r^3}` for a dipole, it's easy enough to find
+    `p_"atom" = frac { 2 k p alpha } { d^3 }` (my mathjax-fu is not strong enough to represent these properly, but rest assured `p_"dipole"` and `p_"atom"` are both
+    vectors). Now we need to find the electric field produced by this induced dipole at the location of the permanent dipole. If you aren't keeping track of the vectors either,
+    that's okay; just remember the rule of thumb that induced dipoles <i>always</i> attract the charge that induced them. We can plug our formula for the
+    induced dipole into our formula for electric field of a dipole on axis: `E = frac { 4 k^2 p alpha } { r^3 d^3 }` (why the separation? `r^3` is the distance from the center
+    of the dipole to the center of the atom; you'll see in a second that `d^3` will have some other added terms). Yikes.
+</p>
+<p>
+    It gets worse. There's no good way to find the force on <i>just</i> a dipole - we have to think about the dipole as two separate point charges separated by a distance `s`.
+    We then sum the force on both. The force on the negative charge is going to be `E = frac { 4 k^2 q p alpha } { d^3 (d - frac s 2)^3 }` attractive, and on the positive charge
+    `E = frac { 4 k^2 q p alpha } { d^3 (d + frac s 2)^3 }` repulsive. Adding these, we get a total attractive force of `frac { 4 k^2 q p alpha } { d^3(d - frac s 2)^3 } - frac { 4 k^2 q p alpha } { d^3 (d + frac s 2)^3 }`.
+</p>
+<p>
+    This is nice (well, it isn't, but ykwim), but it doesn't give us a particularly useful or elegant equation. We need to combine these terms somehow. Time for... binomial expansion theorem!
+    There is no situation that cannot be made worse with binomials. The formula for general binomial expansion can be represented with a surprisingly elegant sum
+    `(a + b)^n = sum_(k=0)^n C_n a^{n - k} b^k`, where `C_n` is the binomial coefficient given by `frac {n!} {k! * (n - k)!}` - for example, this yields the well known FOIL
+    `(a + b)^2 = a^2 + 2ab + b^2` for `n=2`. The easiest way to find these coefficients is Pascal's triangle - I won't explain it here, but I highly recommend having
+    a copy on hand for binomial expansions. Pascal helpfully gives us `\[1, 3, 3, 1]` for our binomial coefficients where `n=3`, so the general expansion will be
+    `a^3 + 3a^2 b + 3b^2 a + b^3`. Eugh. Substituting this in yields the truly monstrous
+    `frac { 4 k^2 q p alpha } { d^3(d^3 - 3 frac {d^2s} 2 + 3 frac {d s^2} 4 - frac {s^3} 8) } - frac { d^3(d^3 + 3 frac {d^2s} 2 + 3 frac {d s^2} 4 + frac {s^3} 8) }`
+    Fortunately, there's a ray of light: those denominators can be simplified massively because we know `s` is <i>much</i> smaller than `d`. If we take `d^3 - 3 frac {d^2s} 2 + 3 frac {d s^2} 4 - frac {s^3} 8`
+    to be `d^3 (1 - 3 frac {s} {2 d} + 3 frac {s^2} {4d^2} - frac {s^3} {8d^3})`, we can actually just assume that `frac {s^2} {d^2}` and `frac {s^3} {d^3}` are both much, much smaller than
+    1, and pretend they're 0: `(d - frac s 2)^3 = d^3 (1 - 3 frac {s} {2 d}) = d^3 - 3 frac {s d^2} {2}`. For the repelling one, this is also true:
+    `(d + frac s 2)^3 = d^3 (1 + 3 frac {s} {2d}) = d^3 + 3 frac {s d^2} {2}`. 
+</p>
+<p>
+    This means that our terrible equation above simplifies down quite a bit to `frac { 4 k^2 q p alpha } {d^6} (frac 1 {1 - 3 frac {s} {2d} } - frac 1 { 1 + 3 frac {s} {2d} })`.
+    Now to combine these! The standard rule for fraction combination gives us `frac {1 + 3 frac {s} {2d}} {(1 - 3 frac {s} {2d})(1 + 3 frac {s} {2d}) } - frac {1 - 3 frac {s} {2d}} { (1 - 3 frac {s} {2d})(1 + 3 frac {s} {2d}) }`,
+    which simplifies to `frac {6 frac {s} {2d}} {(1 - 3 frac {s} {2d})(1 + 3 frac {s} {2d}) }`. FOILing this out gives us `frac {3 frac {s} {d}} { 1 + frac {9 s^2} {4d^2} }`.
+    Because we know `frac {s^2} {d^2}` is much less than 1, the entire denominator can be assumed to be `1`, leaving us with... `3 frac s d`. Anticlimactic! Recombining this with the main formula gives us
+    `4 k^2 q p alpha frac {3s} {d^7}`. That's the correct answer!
+</p>
+<p>
+    Arriving at this conclusion is not terribly easy (this one problem was longer than most of my Thomas reviews in the multivariable section). It requires a pretty good understanding
+    of when you can handwave what: if you're assuming `d` is so much larger than `s` that `frac s d` is effectively 0, the end result would be invariably 0, which would be true for
+    very far distances. Likewise, if you assumed `d` and `s` are close enough for `frac {s^2} {d^2}` to be significant, the end result would be much more complicated, but would also
+    be closer to universally true. The balance we struck here worked out nicely, but was also somewhat arbitrary.
+</p>
+<h2>Charged Shapes</h2>
+<p>
+    After that nightmare above, let's do an easy one.
+    We're given a long conductive pipe with an outer radius `r`. It's charged to produce an electric field with magnitude `E` directed radially inward on the surface.
+    We insert a long, thin charged rod into the pipe, and the electric field is reversed: we need to find the charge density of the rod.
+</p>
+<p>
+    This problem is actually fairly simple, relatively speaking. We know fundamentally that the electric field due to any charged rod with charge density `lambda` is
+    `k frac {2 lambda} {r}`, and in this case our distance `r` is literally the radius. We already know the field is `E`, so for the field to become `-E`, we would need
+    a superposition with a field twice as strong: this means we have an equation `k frac { 2 lambda } {r} = 2E`. This is really very easy to solve to get
+    `lambda = frac {Er} k`.
+</p>
+<p>
+    Next, let's do a superposition problem. We have a sphere of charge with charge `Q_1`, with radius `d`, inside a circular shell with thickness `d` and charge `Q_2`.
+    The inner surface of the shell has a radius of `2d`, so there is a distance `d` between the ball and the shell. We need to find the electric field
+    at some point `1.5d` between the ball and the shell.
+</p>
+<p>
+    This one is difficult (lots of surface integrals) without Gauss' law, but with Gauss' law we know that `int_S E dA = frac {q_enc} {e_0}`. To use this, we visualize
+    a thin shell with radius `1.5d`: the surface integral of the electric field about this shape `int_S E dA` is equal to the enclosed charges divided by `e_0`. The
+    only enclosed charge is `Q_1` - we aren't touching the outer shell at all. One more piece of ancient wisdom is necessary: the surface integral around a shell with
+    constant E is just the area of the shell times the constant value of E; `int_S E dA = 4 pi r^2 E`. This allows us to set up an equation `4 pi r^2 E = frac {q_enc} {e_0}`,
+    which solves to `E = frac {q_enc} {4 pi e_0 r^2}`. Look familiar? This is Coulomb's law! We can pretend the ball is a point charge. 
+</p>
+<p>
+    Let's wrap this up with a somewhat more difficult problem: we need to find the electric field at the origin produced by a rod bent into a circular arc,
+    with radius `R`, spanning from `0` to `theta` radians. It has a total charge `Q`. This is going to require some integration. Step one is to find the <i>linear charge density</i>:
+    this is the charge of a unit length of the rod. In this case, the length of the arc is going to be `2 pi r frac {theta} {2pi}` - quite simply `r theta`. 
+    Thus the charge density is `frac Q { r theta }`.
+</p>
+<p>
+    Next up is to find `dL` in terms of `d theta`, so we can integrate the arc. We already have a formula for length in terms of theta `L = r theta`, so taking the derivative
+    gives us `dL = r d theta`. This is actually consistent with our exploration of radial coordinates over in the multivariable section, if you've been reading that.
+</p>
+<p>
+    Now we need to find the electric field contribution in terms of `theta`. The R-vector from a point on the arc is just `R = \[-r cos(theta), -r sin(theta)]`. Coulomb's law
+    applied here reads `k frac {q hat R} {R^2}`. The magnitude of `R` is trivially `r` (if you aren't sure how I know that, try calculating it yourself). This means
+    `hat R = frac R {abs R} = \[-cos(theta), -sin(theta)]`. Substituting in gives us `E = -k frac {q} {r^2} \[cos(theta), sin(theta)]`. This should be all we need to calculate
+    the integral!
+</p>
+<p>
+    Setting up the integral gives us `-k frac {q} {r^2} int_0^{theta} \[cos(theta), sin(theta)] r d theta`. Integrating this is trivial; the result is
+    `-k frac {q} {r} \[sin(theta), -cos(theta) + 1]`. If `theta = 2pi`, we expect this to be 0 (field at the center of a constant ring charge is always 0);
+    substituting in does indeed give us `\[0, 0]`; this produces the correct results for a number of other easily predictable situations. 
+</p>
+<p>
+    It's still rather shocking to me how three separate charged shape problems took less time and space to write than a single point charge problem. That binomial expansion is
+    <i>hard</i>.
+</p>
+<h2>Potential Difference</h2>
+<p>
+    Let's do some potential difference problems. First up is a classic: we have two spherical shells on the x-axis, one of which has a positive charge `Q` and is positioned at
+    `\[-L, 0]`, and one of which has a negative charge `-Q` and is positioned at `\[L, 0]`. We need to find the potential difference between the center of the positively charged
+    shell and the origin. We'll represent both of these as point charges: the electric potential at a location due to a point charge is just `V = k frac Q r`. It's critical to note
+    that this is entirely meaningless alone: there is no such thing as a standing electric potential; it must be a potential between two points. The canonical way to find
+    this is with the integral `int_{oo}^r E dl`: the integral of electric field between some point at infinity and the target point, in terms of distance from a point charge.
+    The potential between a point `a` and a point `b` is `V_b - V_a` - because this essentially eliminates the `oo` terms.  
+</p>
+<p>
+    The easiest way to do this is to find the potential due to each charge and superpose. The negatively charged shell produces a total potential between `-L` and `0`
+    of `-kQ * (frac 1 {L} - frac 1 {2L}) = -kQ frac 1 { 2L }`. Straightforward enough. The positively charged one is a bit trickier: We can't take the potential at its
+    center with this method, because that would yield infinity. Fortunately, we know the potential between any two points inside a shell is 0, so we can just take the potential
+    at the edge. `kQ * (frac 1 {L} - frac 1 {R})`. The total potential is the sum of these: `kQ * (frac 1 {L} - frac 1 {R} - frac 1 { 2L }) = kQ * (frac 1 {2L} - frac 1 R)`.
+</p>
+<p>
+    This is a pretty quick solution, but difficult to understand. This is because it's really just syntax sugar for a much more understandable but much more verbose integral.
+    To prove that the simpler method does in fact work, let's work out the integral. This is a fairly tricky operation. We'll actually have to do this in two integrals:
+    if we attempt to do it in one, our bounds will end up incompatible; either we'll fail to consider some of the field from the negatively charged shell, or we'll
+    produce an impossible `frac 1 0` situation.
+</p>
+<p>
+    The bounds for the positively charged shell are necessarily `R - L <= x <= 0`. The electric field due to the shell inside the shell is 0, so we don't have to consider it
+    at all; integrating from the edge of the shell ensures that we don't try to find the field due to the charge at `\[-L, 0, 0]`, which is undefined. The bounds for the negative
+    shell have no such problem: it can be safely integrated between `-L <= x <= 0`.
+</p>
+<p>
+    The integrand for each shell is, by Coulomb's law, `frac {kQ} {d^2}`. `d` is the distance between the shell and `x`. We can safely ignore the `y` component of the field, due to symmetry;
+    we also know both shells are producing a field in the `+ hat x` direction, so we don't have to worry about directionality (this is why we aren't using `-Q` for the positive shell -
+    were we considering direction, it would be multiplied by `- hat x`, which would leave it positive). The integrand for the positive shell is, then,
+    `frac {kQ} {(x + L)^2}`, and for the negative shell `frac {kQ} {(x - L)^2}`.
+</p>
+<p>
+    Our full integral expression considering all of these factors works out to `int_{R - L}^0 frac {kQ} {(x + L)^2} dx + int_{-L}^0 frac {kQ} {(x - L)^2} dx`. This is really
+    very easy to evaluate - we know fundamentally that `int frac 1 {(x + l)^2} dx = - frac 1 {x + l}`. We get `|_{R-L}^{0} -frac {kQ} {x + L}` + `|_{-L}^{0} -frac {kQ} {x - L}`.
+    This evaluates to `- frac {kQ} {L} + frac {kQ} {R} + frac {kQ} {L} - frac {kQ} {2L}`. Simplified down, we get... `frac {kQ} {R} - frac {kQ} {2L}`! This is `-1` times the result
+    we got above, which is fine - `V = - int_C E cdot dl`, meaning we have to multiply this by `-1` to get the real value. Yippee!
+</p>
+<h2>Radiation from Accelerated Charges</h2>
+<p>
+    Time for some problems on new material! We're given a system with an electron at a point `d` meters above a detector. Suddenly, the electron is accelerated with magnitude
+    `a`, `theta` East of North. We need to find the electric field component at the detector immediately after the wave propagates to it. We have the handy equation
+    `E = frac {mu_0} {4 pi} frac {-q a_"perp"} {abs r}`: the electric field is the negative of the charge times the perpendicular component of acceleration projected onto `r`,
+    divided by the distance, multiplied by the magnetic constant. The charge is about `-1.6e-19`, as it's an electron. `r` is just `d`. `a_"perp"` is a bit trickier - some trigonometry
+    tells us it's `a sin(theta)`. Substituting these in gives us `E = frac {mu_0} {4 pi} frac {1.6e-19 a sin(theta)} {d}`. Easy enough!
+</p>
+<p>
+    What if we wanted to find the magnetic field? This part is actually very simple. The magnetic field is always perpendicular to `E` and the direction of propagation `v`,
+    so it's pretty easy to use the right hand rule for this. The magnitude is controlled by the equation `E = cB`, so `B = frac E c` - the electric field divided by the speed
+    of light.
+</p>
+<p>
+    <i>Faraday's Law and magnetism intentionally omitted as I already covered those in some detail in a previous post</i>
+</p>
+[/]
+[=title "Physics Final Exam Review"]
+[=author "Tyler Clarke"]
+[=date "2025-4-26"]
+[=subject "Physics"]
+[#post.html]