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8

If you have an excess of electron in your body, your hair might stand on end and you might feel a bit negative (I couldn't help that pun), and you should probably avoid touching people or metal object if you don't want a static shock, but other than that, it's mostly harmless. The real danger comes from flowing electrons. Because the body basically runs on ...


6

Typically this is explained by the saying, "current kills." It's not the charge (or potential above ground) that a body attains that hurts biological systems, it's the current that flows through them and either 1) heats them or 2) disrupts important electrical signals in the body. Heating damage occurs and can "cook" (cause 1st, 2nd, or 3rd degree burns ...


5

So in the first case, when talk about a plain circular ring, I assume you mean an annular ring, with a well defined inner radius and a different well defined out radius. With a positive charge at the center of the annular ring, positive charges will be repelled outward and negative charges attracted inward. Incidentally, not all the positive charge will go ...


4

Remember that we keep leading order terms. So for the second part of the expression in parentheses, as $R \rightarrow 0$, we don't just get 1. Using the taylor expansion, we get $$ \frac{1}{\sqrt{1+\frac{R^2}{x^2}}}\Rightarrow 1 - \frac{1}{2}\frac{R^2}{x^2}+....$$ Plug this into the original equation while remembering $\sigma= \frac{Q}{\pi R^2}$ gives $$ ...


4

Potentials are defined up to an arbitrary constant, so there is no particular meaning associated to a negative potential. It is the difference of potential that really matters, since the arbitrary constant is then washed away.


4

The product of the permittivity and permeability is encoded into the geometry of spacetime because the product $\varepsilon_0\mu_0 = 1/c^2$ and the speed of light is special. So the value of the product is telling us about the geometry of spacetime. The relative values of $\varepsilon_0$ and $\mu_0$ tell us about the relative strengths of the electric and ...


3

Take a capacitor and put it across a battery. There will be a transient current as the electrons go towards the anode . This happens very fast and the current is small. If you short the capacitor with a wire, the battery will empty all its charge on the short, which, depending on the battery can really be damaging. Your body accumulates some charge which ...


3

For $H\gg R,L$ And for $L\gg R,H$ you get pretty much the same thing. First off, $(H+L)^2\sim H^2$ and the same goes for $(H-L)^2$. That means that $(H+L)^2+R^2\approx (H+L)^2$. However, $(H+L)\not\approx H$, which means that $\sqrt{R^2+(H+L)^2}\approx H+L$. This makes the first approximation have $\frac{1}{H+L}-\frac{1}{H-L}$ in it. The second ...


2

What's wrong about my process? You're answer cannot be correct since it does not satisfy KVL. For your final charge solution, the final voltages across the capacitors are given by $$V_{C1} = \frac{16.08}{4.08}\mathrm V = 3.94\mathrm V $$ $$V_{C2} = \frac{24.39}{6.19}\mathrm V = 3.94\mathrm V $$ $$V_{C3} = \frac{13}{3.3}\mathrm V = 3.94\mathrm V ...


2

Variations of this problem show up all the time. If you start with the spring "locked" and the spheres charged, then release the spring, it will expand to the new length and when it gets there the spheres will have a velocity - essentially you have a simple harmonic oscillator and the point of (new) equilibrium is the point where the oscillator moved ...


2

The whole electrical power grid is connected to ground. I don't know the details of other regions, but if you are in North America, the two current carrying conductors in a residential electrical outlet are called "hot" and "neutral". The "neutral" conductor is connected to the Earth at many places. If your bare feet touch wet Earth, and your hand touches ...


2

Without equations: The ideal dipole is made up of two oppositely charged particles infinitely close to each other. So we can immediately deduce that if the electric field does not change along the direction of the dipole, it exerts no force (because the force it exerts at the positive particle will identically cancel that at the negative one). The only way ...


2

My guess; you are mixing up quadripoles and quadrupoles. Quadripoles are two-port networks used in electric circuit analysis. The original German word is "Vierpol Theorie", which means Four-pol because of 4 Poles. https://en.wikipedia.org/wiki/Two-port_network Quadrupoles are related to multipole expansion used in electromagnetic, atomic orbital,.. theory. ...


2

If you had a spherical piece of paper, any point on the paper would be surrounded by paper on two dimensions. You could cut out a little circle with that point in the center. If you had a normal sheet of paper, most of the paper would be like that, but there'd be a boundary where the points only have paper on one side and you could only cut out a semicircle. ...


2

The total energy can be expressed as the sum of the pairwise energys excluding the self-interaction energy. $$W=\frac{1}{2}\sum_{i\neq j}\frac{q_iq_j}{r_{ij}}=\frac{1}{2}\sum_iq_i\sum_{j\neq i}\frac{q_j}{r_{ij}}=\frac{1}{2}\sum_iq_iV_i,$$ where $r_{ij}$ is the distance between two point charge. The factor $\frac{1}{2}$ comes from the fact that the summation ...


2

In a parallel plate capacitor there's accumulation of electrons on one side and lack of them on the other. Since one plate is in front of another, the fields on each are equal in magnitude and opposite and therefore the field lines are straight (away from the boundaries) and cancel. In a battery the field is chemically produced inside the structure of the ...


1

The solid angle approach works, it's just when using that formula you need the full charge. The distance from the charge to the edge of the disc formed by your hemisphere is just $R\sqrt{2}$. So draw a sphere of radius $R\sqrt{2}$ around your point charge. The total flux through this sphere will be the full $\frac{Q}{\epsilon_0}$. Because the electric field ...


1

Since the sphere has no electrical properties, I am taking it to just be a mathematical surface. In this case, we are just computing a special case of the electric flux through a disk of radius $R$ from a charge centered on its axis a distance $\ell$ away which is $$ \Phi = \int_0^R E(r) \cos\theta\,dS = \int_0^R dr \frac{Q}{\ell^2 + r^2} (2\pi ...


1

Start from the assumption that space is isotropic (independent of direction). Under that assumption, if you rotate your imagined empty-universe-with-two-charges by any angle around the line joining the two charges, then you will wind up with exactly the same physical system you started with. In particular, the charge configuration is unchanged. Now, if the ...


1

If you are not in a complete electrical circuit, any electric shock caused by touching a charged object or wire is brief. These "static shocks" are slightly painful, but they are (rarely) dangerous or fatal. I'm sure you've experienced a minor static shock. By wearing insulating footwear, you break a complete circuit and forbid a flow of electricity from ...


1

The best rubber material to use is pure gum rubber. Rubber from an inner tube (tire) contains carbon and balloons may have other colorants which are conductive and will not work well. A wider belt will carry more charge. The 'combs' or charge pick-offs at the base and collecting sphere should not touch the belt but come very close - about 1/16". The tube ...


1

Your brain/mind might be processing the information in the wrong way. As field is uniform, force remains constant and acceleration remains constant. You see, the acceleration remains constant, but velocity doesn't remain constant, as the time to which your proton is accelerated increases, your proton's velocity also increases. So, your proton will be ...


1

We are forbidden to solve home-works, but we are allowed to help. So, I can remind you that you know the formula for the total capacitance of $C_2$ and $C_3$ together. Also, since after the charges are re-distributed, the potential difference $V$ between the upper plate of $C_2$ and lower plate of $C_3$, is the same as between the plates of $C_1$. As ...


1

Generically the potential is given by the integral, $$ \Phi(x) = \int \rho(x') \frac{1}{\vert x - x' \vert} d\tau' ,$$ which is based on Coulomb's law of Electrostatics. It is possible to use the spherical symmetry of $\rho$ to show that $\Phi(\Omega x ) = \Phi(x) $ for any proper rotation $\Omega$. This means we can set $x$ to be on the $z$-axis without ...


1

My question is - where does the energy go? Someone has to insert or remove the dielectric and that will require work (either positive or negative depending on the situation). Work is where the energy goes.


1

Your final expression is off by a minus sign, and is not what you want. For #1, $H$ is large, so I'd factor $H$ out. For #2, $L$ is large, so I'd factor $L$ out. When you factor out a large thing, the remaining things are either numbers (which are what they are) or small things. And when something is small you can approximate it by comparing it to other ...


1

When considering these things, at least as a warm up to a more rigorous answer, it is worth thinking about what it means to say that $H>>R$. I take this to mean that, roughly, if I add $H$ to $R$ I'm going to get something close to $H$, as it is much larger. For example $1000000>>1$ so $1000000+1\simeq1000000$. When the variables are squared, as ...


1

I don't see any problems up until maybe the very very last step (where you lost some constants at the very least and I can't tell what you were trying to do or why you think there is a problem). So you were at: $$\int_{-\infty}^{\infty}\frac{\sigma}{2\pi\epsilon_0 (x^2+z_0^2)} x dx.$$ I'm not sure if you then tried a u-substitution or just found an ...


1

For a Gaussian surface between the two plates, the total flux through the surface is zero. For the particular surface you give, all of the electric field lines crossing one of the circular faces cross the other face but in an opposite sense. This is because the outward normal vector for one face of the cylinder is opposite in direction to the outward ...



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