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kanwal32
 one year ago
https://www.youtube.com/watch?v=G_nrvqKctMk&index=14&list=PLYVDsiuOZP5onjQVfPF07xEU7wiSywJc
kanwal32
 one year ago
https://www.youtube.com/watch?v=G_nrvqKctMk&index=14&list=PLYVDsiuOZP5onjQVfPF07xEU7wiSywJc

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kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0can anyone explain why kq^2/2a and kq^22b is taken

kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino @IrishBoy123 hlp

kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus @amistre64

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0here 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: \[\Large \begin{gathered} {E_1} = \frac{{{q^2}}}{{{r^2}}},\quad 0 < r < a \hfill \\ \hfill \\ {E_3} = \frac{{{q^2}}}{{{r^2}}},\quad b < r < + \infty \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0now, 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0Now, 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0for the first one, we have a problem since at r=0, we have a divergence, so we are led to introduce the so called selfenergy 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0Now, 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 selfenergy of the charge. SO the requested work is equal to the electrostatic energy variation, as below: \[\large \begin{gathered} W = {U_f}  {U_i} = \left( {{S_e} + \frac{{{q^2}}}{{2{r_0}}}} \right)  \left( {{S_e} + \frac{{{q^2}}}{{2{r_0}}}  \frac{{{q^2}}}{{2a}} + \frac{{{q^2}}}{{2b}}} \right) = \hfill \\ \hfill \\ = \frac{{{q^2}}}{{2a}}  \frac{{{q^2}}}{{2b}} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0please keep in mind that the initial energy is given by the subsequent sum: \[\large {U_i} = {S_e} + \frac{{{q^2}}}{{2{r_0}}}  \frac{{{q^2}}}{{2a}} + \frac{{{q^2}}}{{2b}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0here is how I view the charge q: dw:1433156135772:dw
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