Respuesta :
Answer:
[tex]E=0[/tex] at r < R;
[tex]E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}[/tex] at 2R > r > R;
[tex]E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}[/tex] at r >= 2R
Explanation:
Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,
[tex]\int {E} \, dA=\frac{Q_{enclosed}}{\epsilon}[/tex] (integral over a closed surface)
where,
[tex]E[/tex] = Electric field
[tex]Q_{enclosed}[/tex] = charged enclosed within the closed surface
[tex]\epsilon[/tex] = permittivity of free space
Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .
at r < R,
[tex]Q_{enclosed}[/tex] = 0 and hence [tex]E[/tex] = 0 (since the sphere is conducting, all the charges get repelled towards the surface)
at 2R > r > R,
[tex]Q_{enclosed}[/tex] = Q,
therefore,
[tex]E\times4\pi r^{2}=\frac{Q_{enclosed}}{\epsilon}[/tex] Â Â Â
(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to [tex]4\pi r^{2}[/tex])
or, [tex]E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}[/tex]
at r >= 2R
[tex]Q_{enclosed}[/tex] = 2Q
Hence, by similar calculations, we get,
[tex]E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}[/tex]
