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By symmetry, you need only consider the force on one of the charges on the corner, say the top right, due to the other four. There will be a repulsion from the other corners pushing it away from the centre, so you must pick Q to balance this with an attractive force of just the right size. One way to work this out is to calculate the vector forces ...


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The "traditional" form of Coulomb's law, explicitly the force between two point charges. To establish a similar relationship, you can use the integral form for a continuous charge distribution and calculate the field strength at a given point. In the case of moving charges, we are in presence of a current, which generates magnetic effects that in turn exert ...


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I found the answer to be .756N by multiplying .63N and sin(36.87).


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Here is a big hint... I have to believe you can finish it from here. The key is "similar triangles".


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Coulomb's law does not apply to two charged bodies of finite sizes, say two charged spheres. It is because, the distribution of charge does not remain uniform, when the two bodies are bought together.$_1$ Credits: $_1$ Modern's abc of Physics-Satish K Gupta, 23rd edition, pg.14.


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Two dimensional case Let's look at Gauss's law. The law says that if you have a surface $S$ which encloses a charge $Q_\textrm{enc}$, and there is an electric field $\mathbf{E}(\mathbf{r})$ on the surface $S$, then the electric field and the charge enclosed are related by $Q_\textrm{enc} / \epsilon_0 = \iint_S \mathbf{E}(\mathbf{r}) \cdot d\mathbf{A}$. Now ...


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It is really valid for distributions too, but you need to use the integral form. To see why it is only "valid" for point charges, take a look at the equation (for a point particle, the one you have probably been looking at). It refers to the distance from the source, which is only defined for a point, not a distribution. It is however not valid for moving ...


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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 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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Well, I have a way of remembering if you live in the UK - you always drive your motor on the left! The diagram is correct - it shows what force will be exerted by the external magnetic field on the current carrying wire. The direction of the force is found from the left hand rule: splay your thumb and first two fingers out so they are mutually at right ...


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The image you've linked shows us how to find the magnetic field associated with a long current-carrying wire. But we're interested not in the field that the wire creates, but rather the field that the wire experiences. The fields that a charged particle generates don't influence itself (see, perhaps, this question). And so all we need to worry about is: what ...


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Are you talking about the forces two parallel current-carrying wires exert on one another? Given two current-carrying wires, $a$, and $b$, we can determine the force exerted on wire $b$ by wire $a$ with $$F=(µ_oI_a/2πr) I_bL$$ Where F is the force exerted, $µ_o$ is the magnetic permeability of a vacuum, $I_a$ is the current flowing through wire $a$, $I_b$ ...


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For a given amount of resistance (combined resistance of all the circuits in your computer, or home, or city), the amount of current which flows is proportional to the voltage. (I=V/R) When lightning strikes a line, it induces a voltage spike. Traditional circuit breakers are current-sensing devices (whether solid state or electromechanical). So, a ...


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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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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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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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When the current stops, the magnetic field inside a conducting loop diminishes, this produces an Electromotive force ($\nabla \times E=-\frac{\partial B}{\partial t}$). This force can be looked as "resisting" the changing magnetic flux field and producing a current to counter it.


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Question 1: You're running into a problem because you're applying the uniqueness theorem to too large of a volume. The bottom of an appropriate volume lies on the $z=0$ plane, where $\Phi=0$ because there's a grounded conducting plate there. The rest of the volume's boundary is an arbitrary surface that completes the enclosure of the volume and consists ...


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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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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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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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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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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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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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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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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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Without actually providing the mathematical details (which is left for the reader) the basic outline is this: 1.Select a differential segment (a segment of infinitesimal lateral dimensions) on the second cylinder. 2.Write the electric field expression for an infinitesimal charge on the segment. 3.Write the force equation and integrate over the entire ...


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The total amount of charge on the two plates is 0, i.e. $q_1=-q_2$, in the case of a plate capacitor, the field is twice that of a field produced by an infinite charged plate, so $E=\frac{q}{A\epsilon}$ where q is the charge on the plate, A is the area of the plate, and $\epsilon$ the relative dielectric constant of the material between the plates.


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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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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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@Karozo essentially gave you the right answer - but just to help you convince yourself, I encourage you to write down the equations for the "difference" - the analysis that takes account of $\Delta \phi$ being finite. You will find that some additional terms appear in the answer - just as you were suspecting. However, you will see that as $\Delta \phi$ tends ...


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Well the problem is your concept of infinitesimal. You see the picture with a macroscopic $dA$, it isn't a real situation, it's to make you understand. You have to do $dA \rightarrow 0$, in this limit : 1-You can't take another radial line, if you do that the different angle $\Delta\phi \rightarrow 0$ in the limit $dA \rightarrow 0$ 2-The field $\vec{E}$ ...


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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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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 ...


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In practice clean plexiglass aka acrylic is an extremely good insulator for your purposes, assuming, that you want to do precision measurements, rather than deal with high voltage power circuits, in which case you need to go with industrial ceramics for that purpose and maybe even immerse your circuit in oil. In any case, electrostatic shielding is as much ...


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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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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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Why not try (as Nathan hints) assuming a charge density of $\lambda$ $(= q/l)$ per unit length. By symmetry you know that the electric field must be directed along the axis of the cylinder, because in the radial direction, the fields due to a charge on one side of the cylinder have an equal and opposite component generated by a charge on the opposite side of ...


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Depends on what you want to use the capacitance for. If plan to use it to store energy: the high losses due to the conductivity may not prove adequate for this use. The conductivity of water increases with the addition of ions. For example, Sodium or Chloride. Others are presented in this article. Condcutivity changes a a function of temperature too. ...


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The following is taken from the book "The evolution of physics", by A. Einstein,L. Infeld. It describes exactly this type of static electricity example during the course of how physical ideas about electricity evolved. This is an older popular exposition book, but i think is very good for this question without sacrifising scientific methods. Initial views ...


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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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The field would, strangely enough, be equal to that of a uniformly charged sphere of charge $q$. The placement and shape of the cavity doesn't change the field outside the sphere. By using Gauss's law, we see that $$\oint \mathbf E \cdot d \mathbf a = \frac{Q_{enc}}{\epsilon _0}$$ Where $Q_{enc} = q -q +q = q$ The point charge in the cavity will induce ...


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The induced charge on the inner surface of the cavities will be $-q_{a}$ and $-q_{b}$. Also there will be a distributed induced charge on the outer surface of the sphere which is $q_{a}+q_{b}$. Now why these charges do not contribute in the field inside the cavity. The reason is to find out the electric field you will choose a Gaussian surface inside the ...


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We can do some calculations inserting numbers to actually see how is like the force thanks to the water dipole. Let's do the calculations classically. Obviously we need a non-homogeneous field. Therefore, I'll pick a point charge centered in the origin of the coordinates. It can be a uniformely charged sphere as well. Therefore, the eletric field is: $$ ...


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Two small facts to add to Lubos's answer: Rubbing is useful because once the electron has been transferred from one material to the other, you need to "get away" - and there is a chance that the electron comes back with you as you do so. By rubbing, you actually create a large number of "stick-unstick" events, thus improving the chance that charge transfer ...


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Charged objects have an electric field in their vicinity. The air will always contain a small number of ionized particles - this can be a result of cosmic radiation, local electrical activity, or just the chemistry of molecules. Now if your comb is positively charged, negatively charged ions will be attracted to it and positive ones will be repelled. Over ...


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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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There are two ways this can happen. One, the water becomes slightly polarized but net neutral charge. You end up with a dipole and a weak attraction. Two, the polarized water "breaks" so some charge is left behind. This charge can flow back through the main water pipe to ground and the container with the water becomes slightly charged. You could prove ...


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I would like to introduce an approach by complementing the hemisphere. Generally I want to make the left hemisphere bigger. Let the radius of the left hemisphere is $R_1$ and the density $\sigma_1$. The right hemisphere has $R_2$ and $\sigma_2$. ($R_1>R_2$) The force 2 hemisphere acting on each other is $F$, First we complement the left half to make it ...


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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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