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A possible sensible approach to the direct integration involves the very useful expansion $$\frac{1}{|\mathbf{x}-\mathbf{x'}|} = 4 \pi \sum_{l=0}^\infty \sum_{m=-l}^l \frac{1}{2l+1} \frac{r^l_<}{r_>^{l+1}} Y^*_{lm}(\theta',\phi') Y_{lm}(\theta,\phi) \, ,$$ where $r_<$ is the smaller of $|\mathbf{x}|$ and $|\mathbf{x'}|$, ...

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There are two questions here: can the electric field of a static collection of charges have non-zero curl, and do any electric fields of the form $\mathbf{E}(r,\theta,\phi)$ describe the electric field of a static collection of charges? As for the first question, for a static collection of charges, the magnetic field does not change with time. Therefore, one ...

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If you had, say, an extended object with many point charges and wanted to know the force on a point charge, Q, you would use Couldomb's Law to add up the forces on Q due to each individual charge. In the case of a continuous distribution of charge, you do a similar thing but consider the object to be made up of an infinite number of infinitesimal charges, dq,...

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(a) Over here, as we have considered a point charge, we are dealing with spherical/radial symmetry right? thats why we are able to apply Gauss's Law and get $E= -\nabla V= \frac{q}{4π \epsilon_0 r^2}\hat r$ in the radial direction That is correct. (b) What happens to my expressions when i do consider absence of radial symmetry as the statement of the ...

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This is basically a compilation of the discussion I had with secavara and Anonjohn in the comment section so all credits go to them. Thanks, everyone! So, starting from the differential equation $(4)$ which we get on substituting the expanded form of $G(\textbf{x},\textbf{x}')$ into the Green function equation ${\nabla_{r}}^2 G(\textbf{x},\textbf{x}')=-\... 1 The idea is that because the electrostatic field has zero curl, it can be written as the gradient of a potential$\phi$via $$\mathbf{E} = - \nabla \phi.$$ Now, using this starting point, the energy of the electrostatic field of a system of charged conductors can then be expressed in terms of the potential$\phi$as $$U = \frac{1}{8 \pi} \int \mathbf{E}^2 dV ... 1 One of the reasons is the problems with the concept of action at a distance. We could say that moving charges in spacecraft at the asteroid Bennu caused forces that move charges at the NASA control in earth. And call it eletronic communication. But the delay of that action suggests that something (EM waves composed by E and B fields) travelled between the 2 ... 1 The accumulation is due to difference in electric field at two points .HINT try relating flux and electric field for charge accumulation {E} = \frac{J}{\sigma} where j is the current density and {\sigma} is conductivity 1 When electrons zip around in a wire, they occasionally collide with the atoms of the wire. As a result, the electron can impart some (or all) of its kinetic energy to the atoms of the wire. This causes them to vibrate. And it’s this vibration that is macroscopically seen as the heating up. Consider the atoms in the dielectric. They can be thought of as ... 1 Current trough a wire gets the electrons moving, ( I would not talk about traveling energy) they are slowed by bumping in atoms and get them to move or vibrate faster, which is heat. The stuff between the plates of the capacitor ist partly polarized, so the electric field has shorter ways, the same thing as getting the plates closer zigether. 1 The reason that the integral looks complicated is that you haven't made use of your freedom to choose a convenient coordinate system to simplify it. If you choose to make the polar z-axis of a spherical coordinate system pass through the point \vec r where you want to calculate the field, then you have$$\vec r = r \hat z$$and$$\vec r' = r'\sin\theta'\... 1 (a) Over here, as we have considered a point charge, we are dealing with spherical/radial symmetry right? that's why we can apply Gauss's Law and get$E= -∇V=\frac{q}{ 4πϵ_0r^2}$in the radial direction even without starting from the potential expression. Depending on the coordinate system, the operations you do for gradients are different but the important ... 1 Yeah, you are right Purcell didn't consider the possibility of charges other than the spherical shell. So you can do this it in a more general way without taking a spherical shell. Suppose you have Electric field$\mathbf{E}$in space due to some charge distribution. The electric field always undergoes a discontinuity when you cross a surface charge$\sigma\$....

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There is no electric field between the plates. Also the charges are accumulated on the outer surfaces, so anything between the plates does not suffer any consequence of the charges. It is only when the field appears at the edge of the system of two plates that effects occur.

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Because the surface of the plastic wrap is much larger than its thickness, it behaves like a capacitor, accumulating positive charge on one side and negative charge on the other. It creates a potential difference and thus an electric field.

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