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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, ...


5

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 ...


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.


3

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.


3

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 $$ \frac{\rho}{\epsilon_0} = \nabla \cdot \vec{E} = \nabla \cdot (\nabla \times \vec{A}_m) = 0. $$ ...


3

You are right! The trick to remember here is that vector fields permeate all of space (literally all of it) and field lines are only a convenient representation of this. When a new field line is added due to the increased magnitude of the field at that point in space, the field vector 'arrow' that is introduced always existed there but was just small ...


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

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

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

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

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

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

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

@Daniel Grissom gives a great answer, but I wanted to drill into the exact question you asked about the direction of flow. There are several points of note here: The comb is not in direct contact with the plates, but is outside it. This means that when the plates have the same charge they are trying to push that charge outwards, if it's energetically ...


1

Steve B is right. I just want to include a diagram. You can see that the surface $\rm S$ consisting of the conducting surfaces $\rm S_1, S_2, S_3,...., S_i$ enclosing the volume $\rm V$. Suppose $\rm S_1$ is at unit potential and all the others are at zero potential. If $\rm S_1$ has $\rm Q$ amount of charge then an equal amount of negative charge is ...


1

A conductor divides the whole space into two parts. (i) outside the conductor and (ii) inside the conductor. We know that the field at a point in region due to charges outside the conductor is cancelled by the charges induced at the surface of the conductor. Suppose that we move the charges outside the conductor. The induced charge pattern will quickly ...


1

A cavity inside a conductor is shielded from outside electric influences. Shielded from electrostatic effects, yes. if you put charges inside the cavity, the exterior of the conductor is not shielded from the fields by the inside charges. The exterior region know how much charge in is on and inside the conductor. But the exterior region does not ...


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

Let's look at the simplest case: two like charges a distance d away. If free to move, they move away from each other and away from the center. Now imagine three like charges on the corners of an equilateral triangle, if free to move they also move away from the center. Now imagine four like charges on the corners of a square, if free to move they also move ...


1

Don't worry, I think I understand now. One simply applies a Uniqueness theorem to the region below the conducting sheet (which I should have mentioned in the question, is grounded) to conclude that $\vec{\mathbf{E}}_{below} = 0$.


1

My question is how the three sides don't contribute any flux whereas we can see that a small fraction amount of flux can pass through the three sides. Not really. You see, the electrostatic field $\vec{E}$ of the charge is always radially outwards. If the charge is situated at the exact corner of the cube, then the field is exactly coplanar with the ...


1

Suppose you have a charge in a plane. The electric field from the charge is spherically symmetric, and that means every field line from the charge that intersects the plane has an equal and opposite field line intersecting the plane. So when we integrate $\mathbf{E}\cdot\text{d}\mathbf{A}$ the two field lines will cancel out and the net flux will be zero. ...


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

Of course, The number of lines passing through any surface is infinite. But still we stick to our old concept of numb of field lines because it gives us intutive idea about how strong the field is there. It depends on you how much lines you draw around any charge but remember id 2q charge have 8 lines around then 4q charge deserve to be 16 lines. Electric ...



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