## kanwal32 one year ago https://www.youtube.com/watch?v=G_nrvqKctMk&index=14&list=PLYVDsiuOZP5onjQVfPF07-xEU7wiSywJc

1. kanwal32

can anyone explain why kq^2/2a and kq^22b is taken

2. kanwal32

@Michele_Laino @IrishBoy123 hlp

3. kanwal32

@UnkleRhaukus @amistre64

4. kanwal32

@ParthKohli

5. Michele_Laino

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

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

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

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

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

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

here is how I view the charge q: |dw:1433156135772:dw|