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rock_mit182

  • one year ago

(a) How much charge is on the surface of an isolated spherical conductor that has a radius 10cm and is charged to 2KV (b) What is the electrostatic potential energy of this conductor? (Assume the potential is zero far from the sphere.)

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  1. rock_mit182
    • one year ago
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    @Michele_Laino

  2. rock_mit182
    • one year ago
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    I guess if the radius is infinite the potential is 0.. but im stuck from there

  3. Michele_Laino
    • one year ago
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    the relationship between the potential \(V\) on the surface of our sphere, and its charge is: \[V = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{a}\] where \(a\) is the radius of our sphere

  4. rock_mit182
    • one year ago
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    inside the sphere there will be constant potential, right?

  5. Michele_Laino
    • one year ago
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    yes! since inside the sphere there is no electric field

  6. Michele_Laino
    • one year ago
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    It is supposed that the electric charge is only on the surface of the sphere

  7. rock_mit182
    • one year ago
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    ok i got the first part 2* 10^3 V (4*pi*epsilon_o)(.1) m =Q

  8. rock_mit182
    • one year ago
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    what about b) ?

  9. Michele_Laino
    • one year ago
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    for part (b), we have to consider the energy trapped inside the electrostatic field outside the sphere

  10. rock_mit182
    • one year ago
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    that's why faraday's cage is a success ?

  11. Michele_Laino
    • one year ago
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    yes! inside a metallic box there is no electric field when, of course, outside that metallic box there is one or more than one electric field

  12. rock_mit182
    • one year ago
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    that's really awesome

  13. Michele_Laino
    • one year ago
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    yes! that behaviour it is due to the fact that the electrons of the metallic box will redistribute themselves such that they create an electric field which cancel the external electric field

  14. Michele_Laino
    • one year ago
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    for part (b), we have to compute this integral: \[\Large W = \int_a^{ + \infty } {\frac{1}{{8\pi }}} \frac{{{Q^2}}}{{{r^4}}}4\pi {r^2}dr\] where I have used the \(CGS\) system

  15. Michele_Laino
    • one year ago
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    since I assign a density energy of \[\frac{{{E^2}}}{{8\pi }}\] in the outer space

  16. rock_mit182
    • one year ago
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    ok now i get what is inside the integral

  17. rock_mit182
    • one year ago
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    However idk where did you get that density

  18. Michele_Laino
    • one year ago
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    of course we can show why we have that energy density( \(energy/Volume\) ), nevertheless, when I make computations similar to your exercise I simply remember the formula of that density

  19. Michele_Laino
    • one year ago
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    I remark that I have used \(CGS\) system, since when I was at university I have studied electromagnetism using that system

  20. rock_mit182
    • one year ago
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    mm I see, did you study physics i guess

  21. Michele_Laino
    • one year ago
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    yes! I have a degree in Physics the electric field, written in \(CGS\) system is: \[E = \frac{Q}{{{r^2}}}\]

  22. rock_mit182
    • one year ago
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    well I have another question, maybe is unrealated to this topic, Is there such thing as magnetic potential or electromagnetic potential ?

  23. Michele_Laino
    • one year ago
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    more precisely there is a vector potential \(A\) for magnetic field \(B\), such that the subsequent vector equation holds: \[{\mathbf{B}} = {\text{rot }}{\mathbf{{\rm A}}}\]

  24. Michele_Laino
    • one year ago
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    it is a partial differential equation, and \(rot\) stands for the \(rotor\) operator applied to the vector \(A\)

  25. rock_mit182
    • one year ago
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    i guess I have to learn multivariable calculus to understand such concepts...

  26. rock_mit182
    • one year ago
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    and of course even more differential equations ..

  27. Michele_Laino
    • one year ago
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    yes! correct! Since the vector \(A\) is a function of the spatial variable \(x,y,z\) and, for time-dependent fields, it is also a function of time \(t\)

  28. Michele_Laino
    • one year ago
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    variables*

  29. rock_mit182
    • one year ago
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    well dude you have been very helpful, now i have to study capacitors if a qeustion comes along I'll looking for you. thanks for helping me out

  30. Michele_Laino
    • one year ago
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    thanks! :) Finally, we can write this analogy for both electric and magnetic field: Electric field: \[{\mathbf{E}} = grad\varphi \] and, magnetic field: \[{\mathbf{B}} = {\text{rot }}{\mathbf{{\rm A}}}\] the functions \(\phi\) and \(A\) play analogue role

  31. Michele_Laino
    • one year ago
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    \(grad\) stands for \(gradient\) operator applied to function \(\phi\)

  32. rock_mit182
    • one year ago
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    gradient is a directional derivative, right ?

  33. Michele_Laino
    • one year ago
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    yes! In cartesian coordinates we have this formula: \[{\text{grad}}\varphi = \left( {\frac{{\partial \varphi }}{{\partial x}},\frac{{\partial \varphi }}{{\partial y}},\frac{{\partial \varphi }}{{\partial z}}} \right)\] it is a vector

  34. rock_mit182
    • one year ago
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    damn This is really nice sometimes math could be outstanding

  35. Michele_Laino
    • one year ago
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    of course! :)

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