We start by considering two parallel plate conductors. Suppose that one of them has a charge density of \(\sigma\), and that the other has a charge density of \(-\sigma\). Gauss' law implies an electric field of \(\frac{\sigma}{\varepsilon_0}\).
Now Gauss' law was perfectly satisfying, but I couldn't help but wonder where I went wrong in reasoning that each parallel plate, acting as a conductor, contributes an electric field of \(\frac{\sigma}{\varepsilon_0}\) over the gap between the two plates, resulting in a net electric field of \(\frac{2\sigma}{\varepsilon_0}\). I had already seen many interpretations of the correct result coming from considering each plate to be a sheet of charge (which would cause each plate to contribute \(\frac{\sigma}{2\varepsilon_0}\) by Gauss' law, yielding the correct answer). Why was it incorrect to consider each plate conductor to produce electric fields in accordance with conductors instead of sheets of charges? I took the issue to Stack Exchange. My question was marked as a duplicate, and I found that someone else had asked the same question and received a beautifully explained response from user David Z. Here is the key run down. We can consider infinitely thin plates. In this case, the conductors behave as sheets of charge, since there is not thickness to the plate where there could possibly be an internal electric field to violate electrostatic equilibrium. If we were adamant on the condition that our plates must have thickness, we can note that charges migrate to the inner surface (where inner denotes the side of the plate that faces the gap between the plates), and to achieve electrostatic equilibrium in the system, an identical charge distribution (though with opposite polarity of course) forms on the inner surface of the other plate. The key insight is that the symmetric charge distribution formation on the inner surface of the second plate is a direct consequence of the electric field induced by the first plate, hence it does not contribute its own independent electric field. The electric field in the gap is hence simply that obtained by considering one of the plates to be a conductor, which is \(\frac{\sigma}{\varepsilon_0}\). Anyway, this system of charged parallel plate conductors is a capacitor. It functions as a device to store electrostatic potential energy. To find exactly how much energy is stored in a capacitor, we consider the work it takes to separate both plates from each other to a separation of \(h\). We mustn't apply a force of \(EA\sigma\). This is because we must consider the average electric field of one of the conductors. Since it is zero on the outer side and \(E\) on the inner side, the correct force is \(\frac{1}{2}EA\sigma\). Therefore, the work done is: \[U=\frac{1}{2}EA\sigma h\] But notice that \(Ah\) is the volume enclosed, hence by rearranging the equation, we can obtain the energy density of a capacitor: \[u=\frac{1}{2}E\sigma=\frac{1}{2}\varepsilon_0 E^2\] By using the definition of capacitance, \(C=\frac{Q}{V}\), and the fact that \(V=Eh\) in this scenario, we obtain the common formulas for the energy stored in a capacitor: \[U=\frac{1}{2}CV^2=\frac{1}{2}QV=\frac{Q^2}{2C}\] University Physics was kind enough to point out to me that the last expression above demonstrates that a capacitor behaves analogously to a spring in storing potential energy, where the reciprocal of capacitance is analogous to the spring constant and charge is analogous to the stretch from equilibrium of the spring. Simple calculations show that in a circuit, capacitance enjoys additive properties that are opposite to those of resistance. I'll try to wrap up this post soon. In our discussion so far, we assumed that the gap between the plates of a capacitor was a vacuum. In fact, it is possible to put some other material in this gap. Doing so will form a substantial dielectric for sufficiently strong electric fields (resulting from sufficiently high charge densities). The dielectric actually increases the capacitance of the capacitor by a factor of \(\kappa\). This is why in the real world, capacitor plates (which are rolled up in cylinders to maximize surface area for a given volume) are separated by some insulating material. Not only does this prevent electrical breakdown and discharge through contact, but it also increases the capacitance of the capacitor! Also, the permittivity \(\varepsilon\) is defined by \(\varepsilon=\kappa\varepsilon_0\). This explains why \(\varepsilon_0\) is called the permittivity of free space. Because the dielectric constant is \(1\) in a vacuum! There are many, many more results that are direct corollaries of what I've discussed here. But these are the essential pieces that I have learned thus far. Good night!
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This is from David Morin's Introductory Classical Mechanics.
The problem is to consider a rope of linear mass density \(\rho\) and length \(\ell\) lying on an inclined plane with an initial angle of \(0\) which is then raised to some angle \(\theta\). Suppose a coefficient of friction, \(\mu\) exists between the plane and the rope, and that the topmost end of the rope is nailed to the plane. What is the tension at the top of the rope? Morin approaches this by first parameterizing the position along the rope as \(z=x\sec{\theta}=y\csc{\theta}\), where the plane is oriented with the bottommost point of the plane (also the bottom end of the rope) at the origin, and the plane extends into the first quadrant. Next, he considers an infinitesimal segment of the rope, from \(z\) to \(z+\textrm{d}z\). The upward forces on this segment are \(T(z+\textrm{d}z)\), the tension on the upper end of the rope, and \(F_f(z)\textrm{d}z\), the force of friction on the segment, where the function \(F_f(z)\) gives the force of friction per unit length for a given \(z\) coordinate. The downward forces on the segment are \(T(z)\), the tension on the lower end of the rope, and \(\rho g\sin{\theta}\textrm{ d}z\), the weight down the plane of the segment. Since the rope is at rest everywhere: \[T(z+\textrm{d}z)+F_f(z)\textrm{d}z=T(z)+\rho g\sin{\theta}\textrm{ d}z\] Morin then states: "Expanding this to first order in \(\textrm{d}z\) gives" \[\frac{\textrm{d}T}{\textrm{d}z}=\rho g\sin{\theta}-F_f(z)\] Now I don't know what it means to expand something to first order in some differential unit, but I can arrive where Morin did by rearranging the first equation as: \[\frac{T(z+\textrm{d}z)-T(z)}{\textrm{d}z}=\rho g\sin{\theta}-F_f(z)\] And then observing that the LHS is equivalent to the limit definition of the derivative: \[\frac{T(z+\textrm{d}z)-T(z)}{\textrm{d}z}=\lim_{h\rightarrow 0}{\frac{T(z+h)-T(z)}{h}}=\frac{\textrm{d}T}{\textrm{d}z}\] Is that what it means to "expand to the first order"? Anyway, Morin continues by asserting that if there is to be any tension at all in the top portion of the rope (or any portion of the rope for that matter), we must have \(\tan{\theta}>\mu\) (which is a well-known, easily obtainable result). In that case, frictional force is maximized: \[F_f(z)=\rho g\mu\cos{\theta}\] The result follows upon integration: \[\int{\textrm{d}T}=\int{\rho g\sin{\theta}-\rho g\mu\cos{\theta}\textrm{ d}z}\] Using the initial condition \(T(0)=0\), we conclude: \[T(\ell)=\rho g(y_0-\mu x_0)\] Where \((x_0,y_0)\) is the location of the top end of the rope. I went about this problem in an almost correct way. I forgot to account for the fact that the tension in the bit of rope above each infinitesimal segment pulled upward on that segment. :(. Hey, now I know better. Yes, solutions to problems from Introductory Classical Mechanics will be up soon. I will present solutions to problems that I have solved myself correctly (so not this one). Otherwise, I'm effectively regurgitating the solutions manual. I'm working through Walter Lewin's lectures on E&M. Boy it's hard. However, it's just a matter of seeing things from different perspectives. For example, an electric field is always perpendicular to equipotential surfaces. Professor Lewin proves this by contradiction. Suppose to the contrary, \(\mathbf{E}\) was not perpendicular to an equipotential surface. Then, \(\mathbf{E}\) has some components along the surface. Thence, it requires work to move a charge from point \(A\) to point \(B\) on an equipotential surface. However, this means that the potential at \(A\) is not equal to the potential at \(B\), violating our assumption that \(A\) and \(B\) lie on an equipotential surface. As a side note, along with the fact that the surfaces of conductors attain electrostatic equilibrium, this argument implies that the surface of a conductor is an equipotential surface. Lewin then provided another (beautiful) analogy. Consider a topographical contour map. Each contour line represents a gravitational equipotential surface, as each contour lines corresponds with a particular altitude which in turn corresponds to a particular gravitational potential energy. Obviously, the gravitational field everywhere points straight down at the ground. But on our contour map, this is perpendicular to each contour line! In particular, the field lines point from each contour lines to one that surrounds it, where the distance between the contour lines represents a differential unit of altitude, \(\textrm{d}h\). A more mathematical way of thinking about it is by considering a contour map of some multivariable function, say, \(f(x,y)\). Then, the gradient of the function, \(\nabla f\), is a vector field that maps each contour line to one that surrounds it from some infinitesimally small distance away. Locally, these two infinitesimally close contour lines are parallel. Furthermore, the gradient represents the greatest direction and rate of change (the vector sum of the rates of change in the \(y\) and \(x\) directions). Therefore, the length of the vectors of \(\nabla f\) are minimal, and must be perpendicular to the contour lines. How does this connect with electric fields and equipotential surfaces? Electric potential is defined as the negative of the work done by the electric field on a test charge (or the work you have to do to move a test charge in an electric field at constant velocity) divided by the charge of the test charge when the test charge is moved from infinity to some position \(\mathbf{r}\). This is the line integral: \[V=-\int_{C}{\mathbf{E}\cdot\textrm{d}\mathbf{l}}\] Where \(C\) is any path from infinity to \(\mathbf{r}\). If instead \(C\) is a path between two arbitrary points, the potential difference is found. Anyway, this implies that: \[\mathbf{E}=-\nabla V\] And the result follows when we consider a contour map of \(V\) over space (with each contour line representing an equipotential surface). Another cool result I learned helped explain why plasma discharge always seemed to occur at "spiky" or "sharp" points. For instance, see this image: The plasma discharge occurs at the sharp tips of this contraption. To explain this, we consider a configuration of a conductor with a sphere (radius \(R_1\)) connected to another sphere (radius \(R_2>R_1\)) by a wire (length \(\ell\gg R_2\)). Since \(\ell\) is large, we can assume that the potential on the surface of either sphere is independent of the existence of the other sphere. Hence, the potential on the surface of either sphere is given by:
\[V_n=\frac{Q_n}{4\pi\varepsilon_0 R_n}\] We have shown earlier that the surface of a conductor must be an equipotential surface, hence we set the potentials on the surfaces of both spheres equal to each other. Cancelling constants yields: \[\frac{Q_1}{R_1}=\frac{Q_2}{R_2}\] Now, when the radius of a sphere is scaled by a factor \(k\), the charge of the sphere is also scaled by \(k\). However, the surface area of the sphere is scaled by a factor \(k^2\). This means that the surface charge density, \(\sigma\), is scaled by a factor of \(\frac{1}{k}\). We consider a Gaussian surface just above an infinitesimally small section of the surface of the conductor, and one just below. Since the conductor is in electrostatic equilibrium, the flux through the surface within the conductor is 0. Hence, by Gauss' Law: \[E\cdot\textrm{d}A=\frac{\textrm{d}q}{\varepsilon_0}\] Rearranging this gives us an electric field of \(\frac{\sigma}{\varepsilon_0}\). This result was one that I was familiar with and is already on my website. But the point of this is that the larger sphere has a smaller surface charge density, and since \(\sigma\propto E\), it must have a smaller electric field as well. With a smaller electric field, electrons are less compelled to accelerate to the extent required to ionize atoms in the air causing electrical breakdown and plasma discharge! Hence, smaller spheres with higher curvatures and smaller radii would be needed to produce electric fields strong enough to lead to plasma discharge! This argument extends to all conductors as any conductor's surface can be represented by an infinite number of osculating circles. Tough stuff. There's still lots more to learn. But it's pretty darn rewarding. Remember that little lemma from last time? The one that showed that \(q\leq p\)? Let us define a new sequence:
\[p_n=\frac{a_n}{n}\] According to our lemma, \(p_n\) constitutes a decreasing sequence. Now, let us construct another sequence that bounds \(p_n\). In particular, I want the \(n^{\textrm{th}}\) term of this sequence to be less than or equal to \(p_n\). I noted that I could accomplish this by modifying \(a_n\) such that I rounded down each time, instead of merely to the nearest multiple of \(n\). Hence, I defined my new sequence as: \[b_1=p_1\] And then, each new \(b_n\) would be obtained by rounding down to the nearest \(n\) and then dividing by \(n\): \[b_{n+1}=\left\lfloor \frac{n}{n+1}b_n \right\rfloor\] I rewrote this recursive definition by splitting the floor into the difference of the actual ratio and the fractional part: \[b_{n+1}=\frac{n}{n+1}b_n-\left\{\frac{n}{n+1}b_n\right\}\] This rearranged to: \[\frac{n}{n+1}b_n-b_{n+1}=\left\{\frac{n}{n+1}b_n\right\}\] Now observe that the fractional part of any number is between \(0\) (inclusive) and \(1\) (exclusive). Hence: \[0\leq\frac{n}{n+1}b_n-b_{n+1}<1\] At this point, I observed that the first inequality in this chain could be rearranged to: \[b_{n+1}\leq\frac{n}{n+1}b_n\] So the sequence was decreasing! Next, I observed that since \(b_1>0\), then all \(b_{n+1}=\left\lfloor \frac{n}{n+1}b_n \right\rfloor\) must be at least \(0\). This implied that the sequence was bounded as well! Hence, \(\lim_{n\rightarrow\infty}{b_n}\) existed. Since \(p_n\) is bounded below by \(b_n\) and is also decreasing, \(\lim_{n\rightarrow\infty}{p_n}\) existed as well, and all sequences of \(a_n\) must therefore converge to arithmetic sequences! This was such a delightful problem! I do hope that my reasoning is correct. I have never quite proven something in this way before, but it makes perfect sense to me! Hooray! Here is my current progress on the problem I described in my previous post.
First, I observed that \(|a_{n+1}-a_n|\) is maximized when \(a_n\) is equidistant (or as close as possible to being equidistant) to consecutive multiples of \(n+1\). Hence: \[|a_{n+1}-a_n|\leq\left\lfloor \frac{n+1}{2} \right\rfloor\] Next, I considered a general term \(a_n=np\) and \(a_{n+1}=(n+1)q\) (where of course \(p,q\in\mathbb{Z}\). WLOG, I let \(a_{n+1}>a_n\). This was the case of interest anyway. Under these assumptions, I had: \[np+R=(n+1)q\Rightarrow q=\frac{np+R}{n+1}\] I set this to be less than or equal to \(p\). At the time, it just assumed this via heuristics. It wasn't until later until I was able to actually prove it. Lemma: Let \((n+1)q>np\) be the closest multiple of \(n+1\) to \(np\). Then \(p\geq q\) Proof: Suppose that to the contrary, \((n+1)r>np\) was the closest multiple of \(n+1\) to \(np\), where \(r>p\). Then, the difference between the two numbers is: \[nr+r-np=n(r-p)+r\] Observe that \(r-p\geq1\Rightarrow n(r-p)\geq n\) and \(r>1\) (by implicit assumption). Adding these two inequalities yields: \[n(r-p)+r>n+1\] The maximum possible distance between a multiple of \(n\) and the closest larger multiple of \(n+1\) is \(n+1\), and this occurs when the aforementioned multiple of \(n\) is also a multiple of \(n+1\). However, we have shown that the distance between \((n+1)r\) and \(np\) is greater than \(n+1\), contradicting our initial assumption that \((n+1)r\) was the closest multiple of \(n+1\) that was larger than \(np\). Hence, \(\exists q\leq p\) such that \((n+1)q\) is the closest larger multiple of \(n+1\) to \(np\). \(\square\) Ok that's cool. So that means that we can safely set our expression for \(q\) to be lesser than or equal to \(p\): \[\frac{np+R}{n+1}\leq p\] We can rearrange this inequality to obtain: \[p\geq R\] The equality condition is more interesting. When \(p=R\), we must also have \(p=q\), and the sequence becomes arithmetic (\(n=k\)). This can more easily be obtained by noting that \(R=(n+1)q-np\). But when does \(p=R\)? Here is what I did know: \[0\leq R\leq\left\lfloor \frac{n+1}{2} \right\rfloor\] \(0\) wasn't particularly interesting... but what about the upper bound? Letting \(R\) equal its upper bound seemed to match experimental results of my program for \(a_k\)! So I strongly suspect that: \[a_k=k\left\lfloor \frac{k+1}{2} \right\rfloor\] The question is, does this \(a_k\) always occur? How do I go about showing that it does? I have shown that \(q\leq p\), but how do I show that \(q=p\) will eventually occur? |
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