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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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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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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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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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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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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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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 flux is zero through the cylinder. For charges which are not point charges but are very small The only ...


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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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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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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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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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