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Remember that force is the negative gradient of the potential for a single particle in a potential $\vec F(\vec x)=-\nabla V(\vec x)$. Now for an interaction potential $U(\vec x_1,\vec x_2)$, the forces on each of the two particles are given by the gradients with respect to the coordinate of the particle. One usually denotes that $\vec ...


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A constant charge density does not imply a zero magnetic field. Even considering a set of isolated charges, suppose they were (mechanically) moved along a circular path. The charge density could remain the same but there would be a current flow. The curl of the magnetic field produced would be $\mu_0 \vec{J}$, where $\vec{J}$ is the current density. If the ...


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For point charges The charge is not enclosed by the cylinder since it is on the apex. Gauss's law states, $\int \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\epsilon_0}$ Since no charge is enclosed, $Q_{enc}$ is zero, so by Gauss's law the electric field is zero. For charges which are not point charges but are very small The only way for the flux ...


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I'm not clear why you would wan't to try to approach this without Gauss' law - its just a fact about the way integration works. Anyway, taking the wire to be along the $z$ axis, and the plate to lie on the $y=a$ plane, with its centre at $(0,a,0)$, we can write the electric field as $$\vec E(\vec r)=\frac{1}{4\pi\epsilon_{0}}\int dV' \,\rho(\vec ...


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Whether a charged particle is in equilibrium or not depends on its potential energy. A particle is in classical equilibrium when its potential energy is minimized (A consequence of this is that the net force on the particle becomes zero.). If you fiddle around with the position of the proton a bit you'll notice that the minimum potential energy the proton ...


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Can you tell from the image below if Q1 and Q2 are attracted or repelled? No, you do not have enough information. Will Q2 only be attracted to the sphere if Q2 is enough bigger than Q1? For any nonzero values of Q1 and Q2 you can compute the distance at which there is no net force. Will the positive charge inside the shell attract electrons interior to ...


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Ok, it was obviously not apparent to people reading this question - there is a perfectly conducting, infinite cylinder (with a circular cross-section) at some distance from the plane of separation. After some research, I am convinced there is not analytic solution to the problem, and instead used numeric methods to receive an approximate solution.


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The way I would proceed is first to exchange the spherical electrode by a point charge at the centre : $$ Q = \frac{VR}{K} $$ With : $$ K = \frac{1}{4 \pi \epsilon_0} $$ Then that charge would have an image across the dielectric surface, the method has been described quite well in introduction to electrodynamics or here The image charge will itself ...


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I think the electric field is zero on A, B, C, D and E, because otherwise there would be current, which would be odd And you are totally right for an electrostatic system (with no current). Instead of explaining it by, "this would be odd", let's have a look at what happens in the instant you add the wire to the battery pole.: Before the wire touches ...


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The electric field due to the charge is uniformly distributed in all directions in the space.., the flux will also be uniform for the charge on all directions regardless of shape if we consider the surface vector is parallel to the electric field lines.


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The rest of the energy is basically emitted as heat energy. Why? You have two capacitors in the circuit, and the connecting wires offer negligible resistance. Hence, when electrons flow from the charged capacitor to the uncharged one, the electrons basically face no resistance, and they collide with high speed with the uncharged capacitor. This collision ...


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According to Gauss's law, the flux through a closed surface is equal to $q_{enclosed}/\epsilon_o$. This doesn't talk about the type of closed surface that encloses the charge. Hence, the flux remains the same regardless of the shape or size of the surface in question. (as long as the charge remains enclosed in it.)


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You miss the condition that the field should be constant. Your equation should be: $$E*4\pi r^2=(Q+2\pi A(r^2-a^2))/\epsilon$$ $$E=(Q+2\pi A(r^2-a^2))/\epsilon4\pi r^2=(\frac{Q}{r^2}+2\pi A-\frac{2\pi Aa^2}{r^2})/\epsilon4\pi $$ I leave the rest for yourserlf


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The potential, $\phi$, is defined as the the potential energy, $U$, of a system of charges, $Q$ and a unit charge. The potential energy of a system of charges is defined to be the negative of the work done by the electric field of one charge, $Q$, on another charge $q$, as $q$ is moved from infinitely far away to some distance $r$ from $Q$. If $q$ is a unit ...


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It is a convention that field lines point from positive to negative. Now because they point from positive to negative we get $$E = -{dV\over dr}$$ (in spherical symmetry). If you have a positive charge at the origin then the potential at the origin is positive and it decreases as you move away from the origin - thus $dV \over dr$ is negative because ...


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There are multiple ways to interpret Coulomb's law in quantum electrodynamics (QED). Interestingly, they don't lead to quite the same conclusion (but there is no inconsistency because they are not defined in the same way). The most commonly used way (that ACuriousMind refers to in his comment) consists in relating the notion of classical potential with that ...


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If the shell and its charge distribution are spherically symmetric and static, and if electric field lines begin and end on charges, then we know that any electric field that might be present inside the shell must be directed radially (in or out). From there, a simple application of Gauss's law, using a spherical surface centered on the center of the shell ...


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Integrate to get $Q(r)=\int_0^rdq$ next apply Gauss law to get E(r). Use espherical coordenates to integrate dq


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The outward area vector convention we use is for closed surfaces. (For example a cube). For non-closed surfaces, the net flux would be the flux through one side subtracted from the flux through the other. A somewhat weak analogy would be forces on a body. If there is a force directed towards the left, and one towards the right, (on the same body, of ...


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Consider these two arrangements of charges: Suppose we ask what is the flux through the surface $S$. If you look at figure (a) with two positive charges the flux lines from the two charges travel in opposing directions and will cancel each other out at $S$. So the flux through $S$ will be the flux from one charge minus the flux from the other charge. ...


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Two electrons when they move experience these forces $$ F_{electrostatic repulsion } = \frac{ke^2}{r^2}$$ And, $$ F_{magnetic attraction} = \frac{μ_0 . e^2 v^2}{4 \pi . r^2}$$ As you can see from the formulae for attraction there must be a velocity. For the two forces to be the same the speed of the electrons must as fast as light, practically these two ...


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It's not sum because its at different sides of the surface. So flux made by one will be cancelling the other partially. So taking the difference is correct. Had the question been both charges above the surface taking sum is perfect.


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you can check the discussion here. There is a certain case in which a phonon mediates attraction between two electrons. Indeed, acoustic phonons correspond to a slowly varying in-space displacement of atoms which produces a charge. This charge, in turn, results in an electric potential for the electrons. This means that the electron distorts the crystal ...


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In certain scenarios there can be a magnetic attraction, but the electrostatic replusion will greatly overpower it.


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Indeed they can. This can even be seen in a classical set of equations if the interaction between the two polarizable objects is strong enough. Take a simple 1d problem with two polarizable point particles in a line. For low polarizations, we assume a linear relation: $$ p_1 = \alpha E(r_1) = \alpha (E_2 + E_{ext}) = \alpha (k p_2 + E_{ext})$$ $$ p_2 = ...


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While Gauss' law does imply that the electric field terminates only on charges, it certainly does not imply that it cannot form closed paths. Gauss' law may be expressed as follows: $$\nabla\cdot\mathbf{E} = \frac{\rho}{\epsilon_0}$$ From the Helmholtz theorem of vector calculus, as any vector can be expressed in terms of two parts ($\phi$ and $\mathbf{A}$ ...


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Consider a plane of charge. For distances from the plane which are small compared to the size of the plane the electric field will be constant. In the capacitor, with two oppositely-charged plates and relatively small gap, the electric field is a constant, depending only on the overlapping plate area and the amount of charge, $E_{gap}=Q/(k\epsilon_o A)$, ...


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Gauss law states that electric flux through a closed surface equals $\frac{q}{\epsilon_0}$, where $\epsilon_0$ is a constant and $q$ is the charge contained in the surface. So, if we corner a 'point charge' using a spherical closed surface, with the charge at the center of the sphere, all we have to do is to start reducing the radius of the sphere. Because ...


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The working voltage of a capacitor depends on the dielectric strength of the insulator. While electrical breakdown is actually a very complicated process with lots of non-linearities, you can simplify the design of a capacitor by saying "the electric field on the insulator must not exceed X". Once you have said that, and you realize that the electric ...


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Since the charge Q have to be the same for both capacitors and you need more voltage to to push that charge in the capacitor with less capacitance then you must have more valtage difference in $C_2$ The mechanical analogy is a configuration with 2 springs in parallel that move the same distance from their equilibrium position need more force on the spring ...


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To do this you must use the electrostatic image method : The problem with two spheres is that you will have image charges of the image charges Here is a diagram of what it will look like after two iterations : Using the method of images we have the image charges inside the spheres: $Q_1$ has an image $q'_1$ located at $O_2 - ( \frac{R_2^2}{D} )$ with ...


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Answering your three questions: He knows the relationship between radius and terminal velocity, but the drops are too small to measure their radius with any accuracy (1 µm is tiny - looking at such a drop from a distance makes it no more than a speck of light, even with a "micro telescope"). Meauring the terminal velocity, he can deduce the size ...


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Rob Jeffries's answer is correct, but here's another way to see it. Consider two solutions to Laplace's/Poisson's equation: $V_1(\vec{r})$ is the solution (in all space) for a spherical conducting shell at potential $V_0$, with $V \to 0$ as $r \to \infty$. $V_2(\vec{r})$ is the solution (in all space) for a grounded spherical conducting shell with an ...


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There have been attempts to use ionized air to reduce drag on aircraft It's a generator that sends a beam of microwaves upstream into the Mach 6 flow, ripping apart the gas ahead of the model so that it is flying through a plasma--a boiling mix of positive ions and electrons--rather than ordinary gas. The experiment, at NASA's Langley ...


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Since this is not a homework, I want to add some thoughts about the limits for this calculations. First, every body, if accelerated, radiates and this happens on every curved path (if the path is not the geodesic gravitational line). Second, every charged particle has a magnetic dipole moment too and this moments get aligned under the each other influence. ...


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When you ask questions about things "in the limit", the answer is almost always "It depends". In this case, the answer is "it depends". The equation $Q=CV$ assumes linear behavior of the capacitor - in reality the dielectric of most capacitors has hysteresis as well as a nonlinear component, so as you increase the voltage, the capacitance will change. This ...


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First, I'm a bit confused by your statement about two batteries. The capacitance of a two-sided capacitor is defined to be the ratio of the magnitude of separated charge (commmonly, the magnitude of charge on each plate) to the resultant potential difference (aka, voltage) between the plates. It's technically a "what if" formula. The actual capacitance ...


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Batteries are not capacitors. Closing the switch doesn't redistribute some fixed amount of charge in the circuit. Instead, the batteries can create new charge through chemical reactions. The redistribution would slightly lower the potential, allowing the chemical reaction to proceed until it comes back to the resting voltage. While real batteries are ...


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First of all, the substitution is not entirely correct, because $ds$ is not the same as $d\vec{s}$, for $ds$ is the modulus of $d\vec{s}$ and is hence a scalar. How to approach the problem is to convert the problem from one of vectors to one of scalars, and thus be able to manipulate the moduli of vectors. Let $v$ be the modulus of $\vec{v}$ to simplify ...


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Short answer: plasma ball emits RF "noise". The touch screen is sensitive to noise (as it tries to detect the presence of your finger by detecting very tiny currents that appear between transparent electrodes as a result of the presence of your finger (which is dielectrically different than air ). Hoping someone else will elaborate for you.


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Yes, you can use the method of images because uniqueness of the solution is guaranteed when you Know the total charge of an equipotential surface without knowing the value of the potential itself I'll summarize a procedure to obtain the correct answer: Application of Gauss law tells us that there must be total charge -q on the inner surface then because of ...


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The boundary conditions you mention, $$\vec{D} = \rho_s\hat{a}_z$$ are for charges distributed on the surface of a volume conductor, following from the requirement that the electric field is zero inside the conductor. The expression $$\vec{E} = \frac{\rho_s}{2\epsilon}\hat{a}_z$$ is instead valid for a planar sheet of a conductor alone (or for that matter, ...


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If we assume potential at infinity to be zero and Earth to be spherical then the potential at the surface of the earth is given by ${kq}/{r}$, where $k$ is a constant, $q$ is the charge on earth and $r$ its radius. As $q$ is extremely small and $r$ very large, the potential at Earth's surface is almost zero. So for all practical purposes we assume its ...


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Yes, the electric field will be zero inside the cavity. The explanation is given on the following page: http://en.wikipedia.org/wiki/Faraday_cage


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Electric potential (voltage) is the strength of an electric field surrounding a point charge "q". The electric field may be evaluated by the work done to move a charge FROM infinity to a point "r" in the field around "q", against the strength of the field. In this case, the zero of potential is set at infinity, where no work is done. This link explains ...


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Do you mean that the field is created by a stationary charge $C$, and $r$ is the distance from your particle to it? Even then, if you want acceleration you have to multiply your expression by $q/m$, where $q$, $m$ are the mass and the charge of your particle, respectively. More generally $a=\frac{q}{m}\,E$, where $E$ is the electric field strength. See ...


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Any formula used in physics is for calculating exactly what it says. The formula above says : If you want to know the integrated electric field on a surface sum the quantities on the right. So it will yield one real number, in units of charge.


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You have ignored the mobile charges in the conductor. In your plot the field lines are not perpendicular to the surface, particularly near the charges. That will cause the conduction electrons to move. The positive charges will attract electrons until the field inside the conductor is zero. This means that the whole conductor, including the inner ...


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We assume that the the electric field is uniform for a charged solid sphere. It follows that the electric charge of the sphere is equal to $$ Q = \rho V$$ Where $\rho$ is the charge density and $V$ is the volume. Therefore, $$ Q = \rho V = \frac{4}{3}\rho R^3$$ We create a Gaussian surface in the form of a sphere of radius $r <R$. Thus, using Gauss's ...


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You have made two elementary errors. First, you forgot a negative sign and second, you forgot your limits (V at infinity is conventionally taken to be zero) $V = -\int_{\infty}^{r} E dr$ Since I believe you are taking the electric field at the x axis, instead of r you can use x, making that part of your work correct. These two things will fix your ...



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