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13

updated calculations - based on neutrino energy escaping and vapor inhalation risk Your math is close but not quite right. First - the number of tritium atoms. There are 1000/(16+3+3) = 45 moles (as you said) This means there are 45*2*$N_A$ = $5.5 \cdot 10^{25}$ atoms of Tritium Now the half life is 12.3 years or 4500 days, that is $3.9\cdot 10^8 $s. ...


10

The point of the point is to increase the electric field near the point. Small radius curves will have a higher local electric field, eventually creating a localize area where the field is greater than the dielectric strength of the air. This results in what I refer to as "micro-lightning." This microlightning discharges the air (or cloud) before the ...


9

Suppose that you have an negatively charged cloud. Floating over your conductor. Then making your lightning conductor pointy at the edge, facilitates better discharge. Because the electric field set up would be high. ${\sigma}=\frac{q}{4\pi r^2}$, We will take an spherical approximation of the pointed end. It will have a very small radius thus high surface ...


6

Your lecturer got the eigenvalue using the fact that the operator $\hat{p}$ is Hermitian so you can do this: \begin{align} \langle p| \hat{p} &= \left( \hat{p}^\dagger |p\rangle\right)^\dagger\\ &= \left( \hat{p} |p\rangle\right)^\dagger\\ &= \left( p |p\rangle\right)^\dagger\\ &= \langle p| p \end{align} I think it ...


4

In the case of a grounded conducting shell, it is well known that the method of images can be used to calculate how the total charge $-q$ on the inner surface is distributed. The solution is given in the wikipedia link above. Now, you ask what happens if the the potential of the shell is fixed, but not necessarily zero. The electric field inside the ...


3

Use kirchhoff's first law, so for two resistors in parallel: $$ I_\text{total}=I_1+I_2 $$ Then just use I=V/R $$ \\\frac{V}{R_\text{total}}=\frac{V}{R_1}+\frac{V}{R_2} $$ The voltage across any component, whether it be across resistor 1, 2, or the whole parallel portion of the circuit, is the same. It just cancels out so you can divide both sides by V.


3

Very simply, the field of the positive and negative elements of the dipole "almost" cancel out - but not quite. It is because they are some small distance away that there is a residual (third order) term. You can see this by taking two charges $+q$ and $-q$ at a distance $2d$, and look at the field a distance $r$ from the center of the two (on the same ...


3

Why would you expect the images to be the same? The lens formula (object distance $u$, image distance $v$ and focal length $f$; Cartesian sign convention) is: $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$ If you consider two lenses $1$ and $2$, the latter at a distance $d$ from the former, then the first lens forms the image at $v_1$: $$\frac{1}{v_1} - ...


3

Your expression for the electric field in Cartesian coordinates is incorrect. The field is not radial, it points away from the line of charge. If we put the line of charge along the z-axis, then the E-field is $$\vec{E} = \frac{x\vec{i} + y\vec{j}}{x^2 + y^2}$$ with some multiplicative constant involving the charge per unit length and $\epsilon_0$. The ...


3

Schwartz is simply noting that the $\beta$-function has a generic expansion in QED of the form (29) where $\beta_{0,1,2}$ are some numbers that can be computed by explicitly calculating the various Feynman diagrams. For instance the leading $\epsilon/2$ is the tree-level result in $d=4-\epsilon$ dimensions. This can be easily seen as follows. In ...


2

Assuming that you mean $R_3$ is the initial distance from one object to the other, and given that gravitational force goes as the inverse square of the distance $F \propto \frac{1}{R^2}$, it follows that if $R$ increases by $3R$ (making the final distance $(3+1)R = 4R$), then the force changes by $\frac{1}{4^2} = \frac{1}{16}$ - it does indeed become 16x ...


2

Yes, you can use the method of images because unicity of the solution is guaranteed when you Know the total charge of an equipotential surface without knowing the value of the potential itself I'll summarize a procedure to obtain the correct answer: Application of Gauss law tells us that there must be total charge -q on the inner surface then because of ...


2

The essential mathematical point here is that $[H_1^I(t_1),H_1^I(t_2)]$ is a $c$-number, i.e. just a normal number, not an operator. Now, it is not true that $U^I = e^{iH_0t/\hbar} e^{-iH_1t/\hbar}e^{-iH_0t/\hbar}$, this is not a solution to $i\hbar\partial_tU^I = H_1^I U^I$ unless $[H_1,H_0]=0$. The solution instead is usually written as $$ U^I(t) = T\exp ...


2

In reality you have $a(v) = g - \beta v$ and you want $v(t)$ and $x(t)$. This can be achived with direct integration, instead of coefficient matching and diff. equs. $$ t = \int\limits_0^v \frac{1}{a(v)}\,\mathrm{d}v= \int\limits_0^v \frac{1}{g-\beta v}\,\mathrm{d}v = -\frac{1}{\beta} \ln \left(1- \frac{\beta v}{g} \right)$$ or $$ v(t) = \frac{g}{\beta} ...


2

Your analysis will need to be a combination of theory and experiment. You are not maximizing impulse alone: you need to think about drag as well. So you need to think about the shape of your rocket as well as the volume of water and initial pressure. The drag on the rocket is probably well modelled by the ram pressure equation ...


2

In the case of radial freefall is from rest at some initial $r=R$, the motion will be periodic if you treat the gravitating body as a point-mass and ignore collisions. Since the radius is strictly positive, it makes sense to substitute $$r = R\cos^2\left(\frac{\eta}{2}\right) = \frac{R}{2}\left(1+\cos\eta\right)\text{.}$$ while conservation of specific ...


2

Approach sort of valid. See the dimensions of b would be those of V, but for another reason -- you needn't have put zero at RHS. The reason is that only quantities of the same dimension can be added or subtracted to each other. It doesn't matter if RT can equal zero or not. Similarly dimensions of al(V^2) would be those of p. Check out Rules of Evaluating ...


2

This is more intuitive if you think in terms of conductance, which is the inverse of resistance (1/R). When you put two equal resistors in parallel, you double the overall conductance. Why? You are adding a second path for current to flow, so you double that flow. Unfortunately, we tend to speak mostly in terms of resistance, which makes the math a bit ...


2

Firstly, $m$ does not have to be an integer, it is entirely possible for $m$ to be 1/2 for instance. Your points ,1-3 are fine. There are is a maximal and a minimal value of $m$. Call the maximal value $M$ (we have to call it something). Now we can apply the lower operator any number of times, each time it lowers the value of $m$ by a full integer ...


1

Since the charge Q have to be the same for both capacitors and you need more voltage to to push that charge in the capacitor with less capacitance then you must have more valtage difference in $C_2$ The mechanical analogy is a configuration with 2 springs in parallel that move the same distance from their equilibrium position need more force on the spring ...


1

The working voltage of a capacitor depends on the dielectric strength of the insulator. While electrical breakdown is actually a very complicated process with lots of non-linearities, you can simplify the design of a capacitor by saying "the electric field on the insulator must not exceed X". Once you have said that, and you realize that the electric ...


1

I think your problem is that you're adding the electric fields like scalars, rather than breaking then into their vector components. You have: $ E_1 = \frac{q}{2\pi\varepsilon_0\rho_1} $ and $ E_2 = \frac{-q}{2\pi\varepsilon_0\rho_2} $ What you should have is $ E_1 = \frac{q}{2\pi\varepsilon_0}(\frac{1}{\rho_{1x}}\hat{x} + \frac{1}{\rho_{1y}}\hat{y})$ and ...


1

Since this is a homework problem, I won't provide a full solution, but here's a nudge in the right direction. Take a look at these two plots of the effective potential: k = -1, $\alpha$ = 1, L = 0.25 k = -1, $\alpha$ = 1, L = 1 What's different about these two effective potentials? We only changed $L$ between the two graphs; what does that imply about ...


1

This section might help: http://en.wikipedia.org/wiki/Lightning_rod#Should_a_lightning_rod_have_a_point.3F Which also states that : Finding that moderately rounded or blunt-tipped lightning rods act as marginally better strike receptors.


1

Here we are talking about instantaneous velocity. So,its -20 m/s. And the velocity will be always tangent to the circular track. So,it will be 20 m/s in magnitude every second. But the direction will be different so different values in x & y - axes


1

As the particle travels around the cylinder (having radius r) and the taught string wraps around the cylinder, the path of the particle will trace out a spiral with radius that decreases as a function of the cylinder radius. In other words, the radius of the spiral (let's call S) equals the string length at any given point. All you need to do is find the ...


1

Consider these two arrangements of charges: Suppose we ask what is the flux through the surface $S$. If you look at figure (a) with two positive charges the flux lines from the two charges travel in opposing directions and will cancel each other out at $S$. So the flux through $S$ will be the flux from one charge minus the flux from the other charge. ...


1

You set $\rho$ equal to one for no reason. In detail the expression for $\Gamma$ is: $$\Gamma^1_{01}=\frac{1}{2}\sum_\rho g^{1\rho}\left[\frac{\partial g_{1\rho}}{\partial x^0} + \frac{\partial g_{\rho 0}}{\partial x^1} - \frac{\partial g_{01}}{\partial x^\rho}\right].$$ And since $g_{01}$ equals zero (since your metric is diagonal), all four partial ...


1

Two identical hoses with the same pressure difference between their ends will carry twice as much water as each does individually. So you get twice as much flow (current) for the same pressure difference (voltage), which is another intuitive way of thinking about Dave's Answer that it is more enlightening in this case to think in terms of conductances rather ...


1

Answer if anyone is interested. In the end the areas outside the inner edges of the wire cancelled by symmetry and so the surface i was looking for was the area enclosed.



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