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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 ...


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

$q$ and $q'$ together produce a potential that is zero on the surface of the sphere. If you find the surface charge density determined by the radial change in that potential you get a charge distribution whose total over the whole surface is $q'.$ If you take that surface charge and $q$ then they together make a combined field that is exactly zero ...


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 ...



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