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The integrand $\vec E \cdot d\vec r$ is $E\,dr$, not $-E\,dr$. The evaluation of the dot product is sort of done for you when you specify the curve on which you are integrating (i.e., your limits of integration in this case). You've double-accounted for the relative directions of $\vec E$ and $d\vec r$. I suspect the underlying confusion is that you are ...


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The voltage across the battery when there is zero current (no load connected) is called the open circuit voltage. The emf of the battery is equal in magnitude to the open circuit voltage. I'm not certain that there is a standard term for the battery terminal voltage when a load is connected since, in general, this voltage varies with the load. One might ...


2

It's a matter of choice. You can set the potential energy to be any value at any angle. You don't even have to have a zero-value at all; you could make $U$ purely positive or purely negative if you're feeling adventurous. But the advantage for $U(\pi/2)=0$ is, as you said, the simple expression $U(\theta)=-pE\cos\theta = -\vec p \cdot \vec E$ instead of ...


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UPDATED: See below. Your NDSolve inputs seem to be doing what I would expect for a mass around a gravitational center. Using: a = 0; b = 0; traj = Table[ s = NDSolve[{x''[t] == -x[t]/((x[t] - a)^2 + (y[t] - b)^2)^(3/2), y''[t] == -y[t]/((x[t] - a)^2 + (y[t] - b)^2)^(3/2), x[0] == 1, y[0] == 0, x'[0] == 0, y'[0] == v}, {x, y}, {t, -20, ...


1

If your potential is $\propto 1/r$, you're effectively simulating gravity. If it gives you better intuition, imagine it as the earth around the sun. As long as your numerical solver is doing a decent job, you shouldn't expect the ball to spiral in, the correct solution would be a conic section, that is it would orbit the origin in an elliptical path if the ...


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For problem solving, the method of images come in handy in the following cases: conducting sphere, conducting cylinder, conducting ellipsoid, and conducting plane. Another example is to regions of dielectrics, with different $\epsilon$ (permittivity). And that is it. This general rule should save you some time. So, the first problem you mentioned can be ...


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You must first realize that a priori there doesn't need to be a relationship between the 2 things you're comparing here: A. The former, $|\phi(x)|^2= |\langle x|\phi \rangle|^2$ is the square of the position space amplitude of a state $|\phi\rangle$ living in Hilbert space B. The latter has two possible meanings depending on context: If we view $V(\phi)$ ...


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There actually is a current that moves through your body, albeit it not enough to hurt you. You don't need to actually touch the powerline, just standing near it will cause a current to flow in your body. If we model a single powercable hanging a height H above the ground as an electric charged cable of infinite length hanging a height $H$ above a perfect ...


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I think you are a little confused about what problem does the method solve. If you have a known distribution of charges $\rho$, then the potential $\phi$ is simply given by an integral: $$\phi = \dfrac{1}{4\pi \varepsilon _0} \int \frac{\rho \text d V}{r}.$$ This is the integral you have to solve to determine the potential in example 2.7.. Now, if you have ...


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The method of images is useful when you have a point charge near a $V=0$, a.k.a. grounded, surface; usually a plane or a sphere. In this configuration you can substitute the surface by an additional point charge (of opposite charge) and the problem becomes finding the potential due to two point charges. In these other problems, you don't have this ...



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