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11

You are right - potential is a scalar. But a dipole moment is a vector - it has magnitude and direction. When sodium channels open up, charge flows. Lots of charge moving a little bit causes a change in the dipole moment of the heart. This in turn induces charge to move elsewhere in response (the dielectric properties of tissue cause a propagation of the ...


9

What the picture shows is a corona discharge (see also Wikipedia). It isn't a circuit in the usual sense of the word. It happens because the voltage is so high that it raises the electron energy to above the work function and the electrons just leak off. In effect the coil is charging the air around it. The charge will end up on the furniture, walls, floor, ...


4

Yes you are right. You end up having a varying electric field which generates a varying magnetic field which in turn generates an electric field etc... This causes a particular type of radiation called black body radiation.


4

Equation $(2)$ is indeed a general solution, but that doesn't mean that all the $A_l$ and $B_l$ have to be nonzero all the time. For a problem in which $r=0$ is part of the domain the $B_l$ coefficients are zero, else the potential diverges at $r=0$ due to the $r^{-(l+1)}$ functions. In the case of the point charge, the point $r=0$ is not part of the ...


4

Okay, this seems to me correct since the force imparted by $q'$ on $\left(Q- q'\right)$ is cancelled by the field of $q$ since it is a zero equipotential surface of $q-q'$ system. I have no idea why you think a zero potential surface has anything to so with anything. While the field of $q$ and the field of $q'$ together make a zero potential surface on ...


3

The moment of a quantity is a way of expressing the shape of that quantity. If you have some function $f(\mathbf{x})$ then the function can be decomposed into a sum of moments: $$ \mu_n = \int_\infty^\infty \mathbf{x}^n f(\mathbf{x}) d\mathbf{x} $$ The moments tells us about the shape of the function. The zeroth moment is spherically symmetric, the first ...


2

Complex analysis is very useful in potential theory, the study of harmonic functions, which (by definition) satisfy Laplace's equation. One way to see this connection is to note that any harmonic function of two variables can be taken to be the real part of a complex analytic function, to which a conjugate harmonic function representing the imaginary part of ...


2

What you have to understand is that field lines have no physical significance and are merely something we use to conveniently represent Electric Fields. It doesn't matter whether a million field lines or only one passes through your surface. They represents the exact same thing; as long as your field and configuration remain the same. The flux of any vector ...


2

You refer to plates, so I assume we are talking about a parallel plate capacitor. In this case (to a good approximation) the field is the same everywhere inside the capacitor so if the field is bigger than $E_b$ somewhere inside the capacitor then it is bigger than $E_b$ everywhere. But if the capacitor is not a parallel plate capacitor all bets are off. ...


2

There exist two frameworks in which to look at how fields and interactions appear: a) when talking of elementary particles and their interactions it is the quantum mechanical framework; this is the underlying framework from which emerges b) the classical mechanics and electrodynamics framework. The classical emerges smoothly in a computable way. An ...


2

Your reasoning is correct, but you are not accounting that the flux is the multiplication of $\bf E$ and $d \bf S$, and $d\bf S$ $ = r^2 d\bf\Omega$ increases with $r^2$. So this compensates the decrease of $\bf E$ with $r^2$.


2

Considering the charge in an enclosed surface is always 0 I'm not exactly sure what you mean by this but do understand that the Gaussian surface 'encloses' a volume of space within which the enclosed charge resides. For example, consider an isolated point charge $q$. Due the spherical symmetry, the appropriate Gaussian surface is a sphere of radius $r ...


2

Lets to this step by step and take care of the signs! Let $q_1=3$µC, $q_2=5$µC and $q_3=-8$µC. The formula for the force, acting on particle one due two the presence of particle two, is given by $$\vec{F}_{12}=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{{| \vec{r}_{21}|}^2} {\hat{r}}_{21},$$ where $\hat{r}_{21}$ is the unit vector pointing from charge two to ...


2

If you solve for just inside a sphere you might need to throw out your B terms. If you solve for just outside a sphere you might need to throw out your A terms. But if you are solving for the region between two spherical shells you might need both. That's why it is general, because in general you need them.


2

It looks like you're trying to find a vector field $\vec{A}_m$ such that $\vec{E} = \nabla \times \vec{A}_m$. This is only possible in regions of space that are charge-free: the divergence of the curl of a vector field is always zero, so we necessarily have $$ \frac{\rho}{\epsilon_0} = \nabla \cdot \vec{E} = \nabla \cdot (\nabla \times \vec{A}_m) = 0. $$ ...


2

If you were holding some charge there with some force and always had then an equal charge would distribute throughout the surface of the conductor so that an equal but opposite charge could be right where you are holding your charge. So it is just like the charge was always distributed on the surface. If however you inserted some charge somewhere really ...


2

The binomial expansion says that $(1+x)^n=1+{n \choose 1}x^1+{n \choose 2}x^2 + ...$. This should be familiar to you for positive, integer n just by expanding out the parenthesis. For NEGATIVE n, it still holds, provided you interpret ${n \choose k}$ correctly for negative numbers; for our purposes, we just need to know ${n\choose 1}=n$ always. For very ...


2

Ignore the collecting combs and Leiden jars for the moment. As you noted, the Wimhurst disk and 45° shorting bars form a powerful electrostatic pump such that the left side of both disks gets charged one way and the right side of both disks gets charged the other way. This charge feeds back through the shorting bars and builds up until something leaks ...


1

If you have some (static) charge distribution $\rho(\mathbf{x})$ the the electric field is given by: $$ \nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0} $$ or it might be easier to calculate the potential using Poisson's equation: $$ \nabla^2 V = -\frac{\rho}{\varepsilon_0} $$ and then calculate the field using: $$ \mathbf{E} = -\nabla V $$ In your ...


1

Yes, these thermally generated currents (Johnson noise) generate magnetic fields. This means that even non-magnetic materials generate a very-small magnetic noise if they are conductive. This actually places a limit on very-sensitive magnetic field measurements in shielded environments because the shields are usually conductive. The following Review of ...


1

The expression for the total potential energy stored in the fields is given by $$ \frac{\epsilon_0}{2} \int \left| \mathbf{E}_1 + \mathbf{E}_2 \right|^2 d\tau = \frac{\epsilon_0}{2}\left( \int \left| \mathbf{E}_1 \right|^2 d\tau + \int \left| \mathbf{E}_2 \right|^2 d\tau + 2 \int \mathbf{E}_1 \cdot \mathbf{E}_2 d\tau \right) $$ Notice that the first and ...


1

Speculative question... highly speculative "answer". If force is independent of distance, then a neutral object could still be polarized (positive charges are displaced one way, negative charges displaced the opposite way, in accordance with the force they feel), but since the neutral object contains the same amount of charge after polarization, there could ...


1

As the value of $b$ increases the resistance between the outer and the inner shells will converge to $1/4 \pi \sigma a$. If we consider the outer shell to be at the "infinity", the resistance between the "infinity" and the inner shell will be $1/4 \pi\sigma a$. We can think of the situation in which there are two shells in the infinite sea of poorly ...


1

You've run into a classic approximation problem! Usually, you can approximate $X + \epsilon$ as just $X$ if $\epsilon$ is very small compared to $X$. And usually, you can do this to every variable in a problem, and the answer will come out fine. It only doesn't work when the answer itself is also very small; for example, let's say here the field from the ...


1

I think an insulator does not completely stop charge transfer, if viewing it to act in the same way as maybe ie. a thermos cup, which does significantly increase the cooling time of a hot coffee inside, but does not fully prevent heat from escaping, hence allowing the hot coffee/material inside to cool.


1

Static electricity is not like regular electricity in that it does not involve closing a complete circuit; it just needs a large difference in voltage potential between one object and another. When you shuffle your feed on a nylon carpet and touch your finger to a doorknob, you are not closing a circuit. Instead, you are building up a large negative charge ...


1

After all, you can't say:" since you are studying electrostatics, there must be equipotential region on the surface no matter what happens; that's it" Actually, it is almost that simple - otherwise there is a contradiction (1) Assume the electrostatic case (the electric field is constant with time and there is no electric current, i.e., no electric ...


1

Think of potential as like potential energy. If the mobile charges are electrons and they are mostly at rest they will mostly move towards lower potential energy which is higher potential. So think of it as like a hill with electrons free to roll down hill (higher potential) until they get to a surface of the conductor at which point they are not free to ...


1

Your description is not very complete, but I guess what happens is exactly what you expected: to get significant repulsion (to counteract the atmospheric pressure), you need very high charge, which will be necessarily limited due to air discharge (maybe that is why you observed sparks). I don't see how replacing an aluminum shell with a plastic bag ...


1

Imagine two situations. Situation one, you have a point charge of charge $q'$ at the image charge location. Situation two, you have some charge distributed on the surface of the conductor. In both cases you have the exact same electric field on the surface right outside the conductor. So you have the same electric flux there. But electric flux is ...



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