Michele_Laino
  • Michele_Laino
Tutorial of Electrostatic: On the Method of Images
Physics
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Michele_Laino
  • Michele_Laino
here is the PDF file:
Michele_Laino
  • Michele_Laino
\[\Large \begin{gathered} {\mathbf{Electrostatic:}} \hfill \, {\mathbf{The\; Method\; of\;\,Images}} \hfill \\ \end{gathered} \] \[\qquad \qquad \qquad \qquad \Large \qquad {\mathbf{abstract}}\] In this brief tutorial I will show a very useful method, in order to solve some particular exercises of electrostatic. Such method is called the \(Method\; of\;Images\), and it is used when we have to compute charges, and potentials of conductors, in vicinity of some charged objects, like point-like charges, for example. I will show such Method of Images, solving a particular exercise; namely a point-like charge in front of an infinite conductor plane. Finally, I have written all the equations, using the \((CGS)\) system of electric units. \[\Large {\mathbf{Exercise}}\] Let's suppose to have a point-like charge, with positive charge \(Q\) in vicinity of an horizontal infinite conductor plane, namely a metallic plane. We want to compute the electric charge density on the surface of that conductor plane, and furthermore, the interaction force between the point-like charge and the same conductor plane (see figure below). |dw:1447145829350:dw| \[\Large {\mathbf{Solution}}\] As first step, we note, that, due to electrostatic induction phenomena, on the infinite plane, will be induced an amount of electricity equal to \(-Q\), which is negative, since, by hypothesis it is \(Q>0\). Subsequently we try to draw the Faraday's line, or the field line of the so generated electric field, in the space of the system point-like charge and infinite plane. In order to do that, we note that, in vicinity of the point-like charge, the lines of the electric field are radially outward directed, like below: |dw:1447146139490:dw| Since the conductor plane is at electrostatic equilibrium, namely there isn't any motion of induced electric charge on it, then the electric field in vicinity of such plane is perpendicularly downward directed with respect to such plane, like below: |dw:1447146294340:dw| Now we are able to draw completely those lines, and at the end our drawing will be as follows: |dw:1447146453395:dw| As second step, looking at the so obtained electric field, we have to search for a similar electric field, which we know to belong to a simpler problem: does such electric field exist? Of course, yes; it is exactly the electric field generated by a pair of point-like charges with opposite signs: |dw:1447146631475:dw| So, all what we have to do is to add, to our physical system, a \(fictitious\) second point-like charge \(-Q\), symmetrically placed with respect to the same infinite conductor plane, such that we can consider the conductor plane as an equipotential surface of the system compounded by those two point-like charges. If the distance between the charge \(Q\) from the plane is \(d\), for example, then the distance between the added point-like charge \(-Q\), from that conductor plane, is also \(d\) |dw:1447146855996:dw| Now, it is easy to compute the vertical component of the electric field on the surface of the infinite conductor plane at point \(P\). As we can see, if we consider an appropriate coordinate system, we can write this equations: \[\Large {\mathbf{E}}({\mathbf{P}}) = \frac{Q}{{r_1^2}}{{\mathbf{r}}_1} + \frac{{ - Q}}{{r_2^2}}{{\mathbf{r}}_2}\] Multiplying both sides of (1.1), by the unit vector along the \(z-\)axis: \({\mathbf{\hat z}}\) first, and by the unit vector along the polar \(\rho-\)axis: \({{\mathbf{\rho }}}\), subsequently, then we get the corresponding \(z\) and \(\rho\) component of such electrostatic field, as below: \[\large \begin{gathered} \\{E_z(P) } = \frac{Q}{{{\rho ^2} + {d^2}}}\cos \left( {\pi - \theta } \right) + \frac{Q}{{{\rho ^2} + {d^2}}}\cos \left( {\pi - \theta } \right) = \frac{{ - 2Q\cos \theta }}{{{\rho ^2} + {d^2}}}, \hfill \\ \quad \hfill \\ {E_\rho (P) } = \frac{Q}{{{\rho ^2} + {d^2}}}\cos \left( {\pi + \theta } \right) + \frac{Q}{{{\rho ^2} + {d^2}}}\cos \theta = 0 \hfill \\ \end{gathered} \] Remembering that: \[\Large \cos \theta = \frac{d}{{\sqrt {{\rho ^2} + {d^2}} }}\] we can write: \[\Large {E_z(P) } = \frac{{ - 2Qd}}{{{{\left( {{\rho ^2} + {d^2}} \right)}^{3/2}}}}\] In the above computation, I assumed the cylindrical coordinates, namely polar coordinates on the conductor plane \( \left( {\rho ,\varphi } \right)\) plus a cartesian coordinate \(z\), orthogonal with respect to such infinite conductor plane |dw:1447147684088:dw| The corresponding electric charge density \(\sigma\), at point \( P\), induced on the infinite conductor plane, is given by the subsequent computation: \[\Large \sigma \left( P \right) = \frac{{ - Qd}}{{\left( {2\pi } \right){{\left( {{\rho ^2} + {d^2}} \right)}^{3/2}}}}\] Finally, the magnitude \(F\) of the interaction force between the plane and the charge \(Q\), is the magnitude of interaction force between the same point-like charge \(Q\) and the \(fictitious\) point-like charge \(-Q\), nevertheless I warn you, that the charge \(-Q\) doesn't exist physically! It is an artifice only: \[\Large F = \frac{{{Q^2}}}{{4{d^2}}}\]
anonymous
  • anonymous
Ooh, definitely enjoyed reading this. I wish OS had a "Pinned" section to keep stuff like this for future reference :( I was told that they would be adding it at some point, though!

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Michele_Laino
  • Michele_Laino
thanks, for your appreciation to my tutorial!! :) @CShrix
rvc
  • rvc
awesome tutorial perfect for the beginners :) @Michele_Laino you are always awesome..
Michele_Laino
  • Michele_Laino
Thanks!! :D @rvc
AlexandervonHumboldt2
  • AlexandervonHumboldt2
how could you have not tagged me? nice tutorial :)
IrishBoy123
  • IrishBoy123
yes, this is really interesting idea. never seen it before, will be added to my reading list. as usual, thank you very much, michele. it's very well written too, if i may say so.
anonymous
  • anonymous
nice! I think I'm gonna need this sooner. Thanks for the wonderful notes :D (I'm commenting so that I can easily find this in the future)
Michele_Laino
  • Michele_Laino
thanks!! :) :) @AlexandervonHumboldt2, @IrishBoy123, and @Data_LG2 @AlexandervonHumboldt2 I'm sorry, I will tag all of you at next tutorial! :)

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