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I am glad you are interested in physics. As you know, the leaves of an electroscope are attached together by their bases, which generates a kind of force that we call tension. The horizontal component of this tension force balances the electric force of repulsion. In addition, it is worth mentioning that the vertical component of the tension force ...


4

The charge is distributed uniformly on a spherical surface, but that is a function of the high degree of symmetry on the situation. In general the charge tends to accumulate most strongly near pointy bits and most weakly in depressions in the surface. There are several way to understand this. My favorite is not necessarily the most helpful for a beginner, ...


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Take a look at the conventional form of Maxwell equations. They tell us that Gauss's law actually applies every time. However, to get the field $\vec{E}$ from the charge distribution by the usual methods, we also need to know that $$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} = 0$$ Because otherwise the field could not be generated by the ...


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I'll try to make a simple derivation. Suppose you have a unit point charge located at position $\vec{r}'$. Then the associated charge density is $\delta(\vec{r}-\vec{r'})$, which is a Dirac distribution. The electrostatic potential produced by this charge is given by the Coloumb's law: $$G(|\vec{r}-\vec{r}'|) = ...


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The integrand $\vec E \cdot d\vec r$ is $E\,dr$, not $-E\,dr$. The evaluation of the dot product is sort of done for you when you specify the curve on which you are integrating (i.e., your limits of integration in this case). You've double-accounted for the relative directions of $\vec E$ and $d\vec r$. I suspect the underlying confusion is that you are ...


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These quantities are all infinitesimals. $dx$ and $dl$ are often used to denote infinitesimal line elements, while $dS$ and $dA$ are conventionally used for infinitesimal surface (area) elements. The list is completed with $dV$, the infinitesimal volume element. On the application of these infinitesimals: It is often simple and intuitive to think about ...


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The only way to have the charges on a charged conductor evenly spread is to make sure that there is no horizontal component of electric field in the case of "evenly spread". The only two solutions for that are things with infinite dimensions: planes, cylinders spheres For any other shape of conductor, there comes a point where the evenly distributed ...


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If we say we have two objects, with in general charge $Q_1$ and $Q_2$, which experience a certain electrostatic force $F$ between them, then we know that if we double the charge on one, we will double the force - that's just how electrostatics works. Force is proportional to the charge on each of the two objects. With that insight, we can say that the ...


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If the two plates are made of conducting material, there is nothing preventing charges from flowing as close as possible to each other, which, in this case, means toward the edge of each plate closest to the other, right next to the insulating layer. If we now suppose the layer to be thin (dimension $d$) with respect to the plates' sides $L = 10\; cm$), ...


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Electric field is a vector quantity. So, treat them as vectors and find the vector sum of the electric fields. $$\vec{E}_{net}=\vec{E}_{1}+\vec{E}_{2}+\vec{E}_{3}$$


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It's a matter of choice. You can set the potential energy to be any value at any angle. You don't even have to have a zero-value at all; you could make $U$ purely positive or purely negative if you're feeling adventurous. But the advantage for $U(\pi/2)=0$ is, as you said, the simple expression $U(\theta)=-pE\cos\theta = -\vec p \cdot \vec E$ instead of ...


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There is no such thing as perfectly non-conducting. We just simplify things into conductors and insulators. A current will flow from the plastic comb through you to ground, just not as quickly as it would from a metal object connected by a copper wire. The comb can pick up paper because the paper isn't a good conductor. The electric field from the charged ...


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Do you understand that the electric field within a conductor is zero? The charge is mobile, so the internal charge rearranges itself until there is no longer any force to move them: there is no field in the interior. If you understand that, then you will realize that a test particle within the conductor will feel no force, so no work will be done in ...


1

Voltage has absolutely nothing to do with charge. I can "move" an infinite amount of charge trough a superconductor with zero voltage. Are you asking about the relationship of charge to voltage on a capacitor? That's a linear relationship: Q=C*U. The charges, in that case, are not "created" but merely separated. If you want more charge for the same amount of ...


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It is very hard to use a voltage source to induce charge on an insulator. The reason is that by definition, an insulator does not conduct electricity - so if you apply an electrode at one place, you will not move electrons elsewhere, and so you cannot induce a net charge (the best you can hope for is to create polarization, and maybe pull off a handful of ...


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The image charge only serves for the construction of the field in the half space $z> 0$. The problem with the two point charges is only a model problem to calculate the field for $z>0$. If the field in the real problem is static then it is zero in the conductor half space $z < 0$. Therefore, the construction with the pillbox really delivers a ...


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This can be done in five steps (four integrals). Start with the force of two point charges: you know this equation $$F=\frac{Q_1Q_2}{4\pi\epsilon_0 r^2}$$ Integrate this force over an infinitesimally thin ring of charge: now you have the force of a ring on an off-axis point (hint: you only need the axial component - the radial components will cancel due ...


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you don't need two charges, you can use only one charge and the answer will be the same, you only should note to add a second image charge in the centre of the sphere to make the sphere neutral. in both cases, when you take the limit, you get same electric dipole in the centre of sphere. but because the symmetry of two charges, calculations are somehow ...


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The Triboelectric effect is the process through which materials can become electrically charged through friction when they come in contact with other different materials. These materials do not have to be insulators for this effect to take place however if they are good conductors the charge will usually flow away. There is a series of materials ranging ...


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When you put a charged insulator in air, the reason you lose charge is mostly due to humidity in the air. I gave details of this mechanism in a recent answer to a related question: http://physics.stackexchange.com/a/130988/26969 The curves in the referenced paper (some of which I reproduce in that answer) show you how the leakage current is a function of ...


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Your textbook is talking about a perfect conductor, a model of a real conductor. In a perfect conductor charge is considered to be chopped up into infinitely tiny portions. Charge in a perfect conductor is a continuum, not quantized as we know charge in real conductors to be. There are no discrete charges ... electrons or protons ... to get close to.


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In a conductor (perfect conductor or imperfect conductor) in steady-state there is no macroscopic electric field, i.e. if you average the electric field over a volume that is big enough to include many atoms, the result is zero. That's what the textbook meant. You are certainly right that there are nonzero microscopic electric fields, e.g. the large field ...


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A charged conducting material in the form of a sphere or an infinite plane can only be uniformly charged in the absence of external charges. Any other shape of charged conducting material can be induced to be uniformly charged by the placing the right external charge density around it. A conducting infinite cylinder is also uniformly charged (in the absence ...


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The answers already in here are good; unfortunately the integrals that arise are quite nasty, and don't have solutions in terms of elementary functions. Here is some more detail, in the special case when the tubes have zero length (so they are just charged circular loops), and further they have the same radius $b$, with separation $d$. You'll see that this ...


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According to wikipedia, average duration of a lightning is $30\,\mu s$. If we take a gaussian current splash with $\sigma=30\,\mu s$, its spectrum will be a gaussian with $\sigma_k\approx33\,\text{kHz}$. This doesn't look like DC.


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Of course lightning "has a frequency component". If it didn't have non-zero frequencies, it could never start or would last indefinitely. Lightning is a huge current pulse over a short time, usually several pulses over a few milliseconds. But more importantly, the current is started and stopped very abruptly, which by necessity means it has a broad ...


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For a capacitor, the voltage across must be continuous since the current through since $$i_C = C \frac{dv_C}{dt}$$ Since the current through is proportional to the time derivative of the voltage across, the $v_C(t)$ must be differentiable, i.e., there can be no discontinuous change. There is no such limitation on the capacitor current, the direction ...


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Distance. Electric fields weaken as the distance between them increases, so the force applied on the other leaf shrinks as the leaves separate. Eventually, the force from the electric field balances the force of gravity trying to pull the leaves back to their normal rest position, so the leaves cease to accelerate upwards.


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There are several problems with the classical electron radius of: $$ r_e = 3 \times 10^{-15} \text{ m} $$ first and foremost, and perhaps the only real reason that is important in physics, is that it is wrong. Wrong in the sense that we can do detailed experiments to try to measure this radius, and to date, we have failed to ever observe any structure to the ...


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If I wanted to calculate this current, should I use the same formulae No, since, for one, the capacitance $C$ of the apparatus changes as the capacitor is assembled. In the equation $$i = C \frac{dv}{dt}$$ $C$ is a constant. There are other considerations too but the bottom line is, no, you cannot use the above formula to determine the current in ...



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