# Tag Info

4

There is indeed a connection. The holomorphy is easily seen in the electrostatic potential. In a charge free (two-dimensional) region, the electrostatic potential solves Laplace's equation and hence is a harmonic function. The real and imaginary parts of a holomorphic function are harmonic functions and thus the electrostatic potential can be identified ...

4

You can use $q/\epsilon_0$ to calculate flux for both cases because that's what Gauss' law says: Just look at the enclosed charge. It's amazing. Flux for any closed surface is $\Phi = \oint \vec{E} \cdot d\vec{A}$. There are two ways to calculate this quantity: The hard way, which means evaluating $\vec{E}$ on every part of the surface, and integrating. ...

3

The gravitational field $\mathbf g$ equals, by definition, negative of the gradient of a correspondonding potential $\Phi$; \begin{align} \mathbf g = -\nabla\Phi. \end{align} Therefore, it suffices to produce a gravitational potential $\Phi$ whose value is zero at a point but whose gradient is non-zero at that point. This is straightforward to do. Let a ...

3

The analogy follows with the right definitions. The "flux" of the "vector" $E(z)$ through a contour $\Gamma$ is $\mathrm{Re}\left(\int_\Gamma E(z)^*\,\mathrm{d} z\right)$. I think you may have forgotten the conjugate in the relationship between the "Electric field" and the complex potential $\Omega$: $E(z) = -(\mathrm{d}_z \Omega(z))^*$. So it is the ...

2

I presume you are referring to process of nuclear fission of uranium-235, which has the equation: $$^1_0\text{n}+^{235}_{\ \ 92}\text{U}\longrightarrow ^{236}_{\ \ 92}\text{U}$$ However, a subsequent reaction is: $$^{236}_{\ \ 92}\text{U}\longrightarrow^{144}_{\ \ 56}\text{Ba}+^{89}_{36}\text{Kr}+3^{1}_{0}\text{n}$$ The production of neutrons is a feature of ...

2

You didn't really simplify the expressions in the same manner, so they're hard to compare. Even if they're not the same, it's often useful to have a clear idea of what differs in the final result. Simplifying the first expression to share the denominator $(b^2-c^2)^2$:  \begin{align} W&=\frac{1}{2}\int\rho V\,d\tau\\ &=\frac{\pi ...

1

But why when I connect a volt-meter across the whole diode will I not see this built in potential? Because at equilibrium (V=0) the Fermi-level throughout the whole device is flat. A voltage will only exist when there is a gradient in Fermi-level over the component (i.e. a voltage drop). Edit Maybe a more visual explanation would help? Think of the ...

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The whole (pedagogical) point of the slide wire generator is to illustrate that not only do changes in the magnetic field generate current in the loop, changes in the area of the loop - in a constant magnetic field - also generate a current. It's the change in magnetic flux that matters. As long as the wire is moving with some velocity, the magnetic flux ...

1

No, your reasoning is incorrect, because there's no reason for the forces to cancel in general. Actually, the charge in general will be attracted by the field of the induced opposite charge on the inside surface of the conductor. This is easy to see by use of the fact that $\nabla^2 V=0$ in the region devoid of charges implies that $V$ is a harmonic ...

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