I am new to the field of electrodynamics and I recently came across the definition of capacitance as $$C = \frac{Q}{V}$$ It is said that capacitance is constant, which implies that total charge of the one conductor is proportional to the voltage difference between the two conductors. But we know that $$Q = \int ρ(\vec r')dV$$ where the integral is evaluated in one conductor, and $$V(\vec r) = \int \dfrac{ρ(\vec r')}{|\vec r - \vec r'|}dV$$ where the integral is evaluated along the two conductors. How can we prove from these that $Q$ and $V$ are proportional?
My textbook states that:  ...The capacitance $C$ depends only on the geometrical configuration (shape, size, separation)of the system of two conductors. [As we shall see, later, it also depends on the nature of the insulator (dielectric) separating the two conductors.]...  Now I wanted to know if these were the only two factors on which the capacitance of a conductor depends. So I tried thinking about it from the basic definition, which is:  Capacitance is the ratio of the change in electric charge of a system to the corresponding change in its electric potential. (Source: )  i.e., $$C= \frac {dQ}{dV}$$ where, $V$ is the potential of the conductor with respect to zero potential at infinity. Now, I can not think of why $C = f(V)$ isn't a possible scenario, assuming that shape size, etc remains constant. So:  Why cannot $C$ be a function of $V$?
Simple way of creating jQuery Slider How can I get help on terminal commands? I often see things like     read man xyz   For more information read xyz's man page.   see info xyz    so I wonder:   How can I get help on terminal commands generally? What exactly are these man and info pages?   How can I find and view them? Is there anyway to find and remove ligatures from tex file? Sometimes when I copy/paste a text to Winedt, it shows perfect, but when I compile it to Pdf, some issues appear, e.g. finding becomes nding.  I am wondering if there is any way to search and find these characters?  Here is an example file (needs utf8 encoding).  \documentclass{article} \begin{document} ﬁ \end{document} Show that $\mu (E \setminus A)>0$ if $A$ is a nonmeasurable subset of $X$ and $A \subseteq E$, $E$ is measurable. This question was asked , but the question wasn't asking for an answer, just a clarification on the proof by contradiction they already had, so I'm reposting it to ask for help arriving at the answer.  Because the asker above said they proved it by contradiction, that is what I'm attempting. But I'm having issues wrapping my brain around the concept of set measurability. So let's give it a good college try with what I can understand.  For clarity, $\mu$ is an outer measure.  A is nonmeasurable, so $\exists B \subseteq X$ such that $\mu (B) \ne \mu (B \bigcap A) + (B \bigcap A^c)$.  E is measurable, so $\mu (B) = \mu (B \bigcap E) + (B \bigcap E^c)$.  This essentially means $\mu (B) \ge \mu (B \bigcap E) + (B \bigcap E^c)$ (Since monotonicity is alredy implied by the definition of outer measure).  So we can attempt a proof (or the start of one) by contradiction.  Assume $\mu (E \setminus A)=0$. So $E \setminus A$ is a null set.  So $\mu (B) \ge \mu (B \bigcap E) + (B \bigcap E^c)$. 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However, some sources say that to learn something by heart implies that one knows something so well that it has been thoroughly internalised and imbibed.    So is it wrong to conflate the two? Is Java "pass-by-reference" or "pass-by-value"? Showing that $f_0 (x_1, \ldots, x_m) \mathrm tr A =\displaystyle{ \sum_{i=1}^n} f_0(x_1, \ldots, Ax_i,\ldots, x_m)$ Question:  Consider $f: (-\epsilon, \epsilon) \to \mathbb R^{m^2}$ a differentiable path of matrices $m \times m$ such that $f(0) = I_m$ and the function $g: I \to \mathbb R$ is defined by $$g(t) = \det f(t)$$  Show that $g'(0) = \mathrm{tr} \,a $ (trace of the matrix $a$), where $a = f'(0)$.   Attempt:  First we use that $\det f(t) = \det (f_1(t), \ldots, f_m(t)) $, where $f_i$ is the $i$-th row of the matrix $f(t)$ $m \times m$. Considering the operator $': \mathbb R^m \to \mathbb R^m$, $f(t) \mapsto f'(t)$  With this in mind $$\begin{align}g'(t) = \det'f(t) &amp;= \sum_{i=1}^{m} \det (f_1(t),\ldots,f'_i(t),\ldots, f_m(t)) \\&amp;\underbrace{=}_{(*)}\det(f_1(t), \ldots, f_m(t))\, \mathrm tr f'(t) \end{align}$$  For $t = 0$ we have   $$\begin{align}g'(0) &amp;=\det(e_1, \ldots, e_m) \,\mathrm tr f'(0) \\&amp;= 1 \dot\, \mathrm tr \,a \\&amp; = \mathrm tr \, a\end{align}$$  I'm trying to come up with a proof using the fact that $\det$ is a $m$-linear form.   In order to prove $(*)$ I need the following equality, which according to this  holds.   If $A: V \to V$ is a linear operator and $f_0$ is a alternated $m$-linear form, such that $f_0 (e_1, \ldots, e_m) = 1$ then   $$f_0 (x_1, \ldots, x_m) \mathrm tr A = \sum_{i=1}^m f_0(x_1, \ldots, Ax_i,\ldots, x_m) $$  Could someone give a proof to this? How do I uninstall xcode 5 command line tools?