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  1. kanwal32
    • one year ago
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    can anyone explain why kq^2/2a and kq^22b is taken

  2. kanwal32
    • one year ago
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    @Michele_Laino @IrishBoy123 hlp

  3. kanwal32
    • one year ago
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    @UnkleRhaukus @amistre64

  4. kanwal32
    • one year ago
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    @ParthKohli

  5. Michele_Laino
    • one year ago
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    here is why: the initial energy is the energy of the electric field generated by our charge q, inside the shell. Now we have the subsequent situation: |dw:1433153535483:dw| Using the theorem of Gauss, we can say that into zone 2 the electric field is zero, whereas into zones 1 and 3 the electric field is given by the subsequent expressions: \[\Large \begin{gathered} {E_1} = \frac{{{q^2}}}{{{r^2}}},\quad 0 < r < a \hfill \\ \hfill \\ {E_3} = \frac{{{q^2}}}{{{r^2}}},\quad b < r < + \infty \hfill \\ \end{gathered} \]

  6. Michele_Laino
    • one year ago
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    now, the electric energy of that electric field, which is the initial energy, is given by the subsequent formula: \[\Large {U_i} = \int_0^a {\frac{{E_1^2}}{{8\pi }}} \;dV + \int_b^{ + \infty } {\frac{{E_3^2}}{{8\pi }}} \;dV\] please keep in mind that the subsequent quantity: \[\Large \frac{{{E^2}}}{{8\pi }}\] is the density of electrostatic energy, with respect to the volume of the space. Here I'm using the CGS system

  7. Michele_Laino
    • one year ago
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    Now, we have to evaluate those two integrals. For the second one there are no problems, since we have, using polar coordinates in the space: \[\large \int_b^{ + \infty } {\frac{{E_3^2}}{{8\pi }}} \;dV = \int_b^{ + \infty } {\frac{1}{{8\pi }}} \frac{{{q^2}}}{{{r^4}}}4\pi {r^2}\;dr = \frac{{{q^2}}}{{2b}}\] where, theright expression of E_3, is: \[\large {E_3} = \frac{q}{{{r^2}}},\quad b < r < + \infty \] Sorry for my typo

  8. Michele_Laino
    • one year ago
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    for the first one, we have a problem since at r=0, we have a divergence, so we are led to introduce the so called self-energy of the charge, namely we have to consider the spatial dimension of the charge. So, if I call with r_0 the classical radius of our charge q, then we have: \[\large \int_0^a {\frac{{E_1^2}}{{8\pi }}} \;dV = \int_0^{{r_0}} {\frac{{E_1^2}}{{8\pi }}} \;dV + \int_{{r_0}}^a {\frac{{E_1^2}}{{8\pi }}} \;dV = {S_e} + \int_{{r_0}}^a {\frac{{E_1^2}}{{8\pi }}} \;dV\] where S_e is the self energy of our charge q. Then evaluating that integral, using the polar coordinate in the space, we have: \[\large {S_e} + \int_{{r_0}}^a {\frac{{E_1^2}}{{8\pi }}} \;dV = \frac{{{q^2}}}{{2{r_0}}} - \frac{{{q^2}}}{{2a}}\] here theright expression of the electric field E_1, is: \[\large {E_1} = \frac{q}{{{r^2}}},\quad 0 < r < a\] again sorry for typo

  9. Michele_Laino
    • one year ago
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    Now, when the charge q is at ininite distance, namely it is located at its final position, then the energy of the generated electrostatic field, or final energy U_f, is: \[\large \begin{gathered} {U_f} = \int_0^{ + \infty } {\frac{{{E^2}}}{{8\pi }}} \;dV = \int_0^{{r_0}} {\frac{{{E^2}}}{{8\pi }}} \;dV + \int_{{r_0}}^{ + \infty } {\frac{{{E^2}}}{{8\pi }}} \;dV = \hfill \\ \hfill \\ = {S_e} + \int_{{r_0}}^{ + \infty } {\frac{1}{{8\pi }}} \;\frac{{{q^2}}}{{{r^4}}}4\pi {r^2}\;dr = {S_e} + \frac{{{q^2}}}{{2{r_0}}} \hfill \\ \end{gathered} \] wher, again I have introduced the self-energy of the charge. SO the requested work is equal to the electrostatic energy variation, as below: \[\large \begin{gathered} W = {U_f} - {U_i} = \left( {{S_e} + \frac{{{q^2}}}{{2{r_0}}}} \right) - \left( {{S_e} + \frac{{{q^2}}}{{2{r_0}}} - \frac{{{q^2}}}{{2a}} + \frac{{{q^2}}}{{2b}}} \right) = \hfill \\ \hfill \\ = \frac{{{q^2}}}{{2a}} - \frac{{{q^2}}}{{2b}} \hfill \\ \end{gathered} \]

  10. Michele_Laino
    • one year ago
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    please keep in mind that the initial energy is given by the subsequent sum: \[\large {U_i} = {S_e} + \frac{{{q^2}}}{{2{r_0}}} - \frac{{{q^2}}}{{2a}} + \frac{{{q^2}}}{{2b}}\]

  11. Michele_Laino
    • one year ago
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    here is how I view the charge q: |dw:1433156135772:dw|

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