(a) According to Gauss law, when 0 < r < R, the electric field is [tex]E(r) = 0[/tex].
When R < r < 2R, the electric field is [tex]E = \frac{1}{4 \pi r^2} \frac{Q}{\epsilon _0} =\frac{1}{4 \pi \epsilon _0} \frac{Q}{r^2}[/tex].
When R < r < 2R, the electric field is [tex]E = \frac{1}{4 \pi r^2} \frac{Q}{\epsilon _0} =\frac{1}{4 \pi \epsilon _0} \frac{Q}{r^2}[/tex].
(b) Electric field inside the conducting sphere is zero, maximum at 'R' then undergoes an exponential decrease. At '2R' there is again a maximum and then afterwards it has an exponential decrease.
Gauss' Law
(a) According to Gauss' law, the electric flux through a close surface with an enclosed charge 'Q' is given by;
[tex]\Phi = \frac{Q}{\epsilon _0}[/tex]
ie; [tex]E. A = \frac{Q}{\epsilon _0}[/tex]
Case 1: When 0 < r < R, the charge enclosed inside is zero.
Therefore, from Gauss law electric field in that area is also zero.
[tex]E(r) = 0[/tex]
Case 2: When R < r < 2R, here the charge enclosed is Q.
Therefore applying Gauss law;
[tex]E \times (4 \pi r^2) = \frac{Q}{\epsilon _0}[/tex]
[tex]E = \frac{1}{4 \pi r^2} \frac{Q}{\epsilon _0} =\frac{1}{4 \pi \epsilon _0} \frac{Q}{r^2}[/tex]
Case 3: When r > 2R, here the charge enclosed is 2Q.
Therefore applying Gauss law;
[tex]E \times (4 \pi r^2) = \frac{2Q}{\epsilon _0}[/tex]
[tex]E = \frac{1}{4 \pi r^2} \frac{2Q}{\epsilon _0} =\frac{1}{4 \pi \epsilon _0} \frac{2Q}{r^2}[/tex]
(b) There is no electric field inside the conducting sphere. Then at the surface, the electric field is maximum and undergoes an exponential decrease. At 2R there is again a maximum and then afterwards it has an exponential decrease.
(See the graph attached below)
Learn more about Gauss'law here:
https://brainly.com/question/13872115
