Physics Final Exam Review

By Tyler Clarke in Physics on 2025-4-26

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.

Electric Field and Force

Let's start with some easy stuff (editors note: I feel deeply stupid for having written the preceding). 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.

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.

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 always 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.

It gets worse. There's no good way to find the force on just 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 }`.

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 much 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}`.

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!

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.

Charged Shapes

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.

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`.

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.

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.

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 linear charge density: 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 }`.

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.

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!

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.

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 hard.

Potential Difference

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.

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)`.

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.

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`.

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}`.

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!

Radiation from Accelerated Charges

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!

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.

Faraday's Law and magnetism intentionally omitted as I already covered those in some detail in a previous post