Answer :
Solution :
a). Using Gauss's law :
[tex]$E=\frac{Q}{4 \pi \epsilon_0r^2}$[/tex] , [tex]$b<r<a$[/tex] .........(1)
Let [tex]$E=E_0,\ r=b$[/tex] in equation (1)
Therefore, [tex]$Q=4 \pi \epsilon_0b^2E_0$[/tex] .............(2)
[tex]$V_b-V_a = \int^a_b \vec E. d\vec l$[/tex]
[tex]$=\int^a_b E \ dx$[/tex]
[tex]$=\frac{Q}{4 \pi \epsilon_0} \int^a_b \frac{1}{x^2} \ dx$[/tex]
[tex]$=\frac{Q(a-b)}{4 \pi \epsilon_0 a b}$[/tex] ....................(3)
Therefore, [tex]$U=\frac{1}{2}Q \Delta V$[/tex]
[tex]$=\frac{1}{2}(4 \pi \epsilon_0 b^2 E_0)\left(\frac{Q(a-b)}{4\pi \epsilon_0 a b}\right)$[/tex]
[tex]$=\frac{4 \pi \epsilon_0}{2a} \ E^2_0 b^3(a-b)$[/tex] .............(4)
Now differentiating the equation (4) w.r.t. 'b', we get
[tex]$b=\frac{3}{4}a$[/tex]
Thus the radius for the inner cylinder conductor is [tex]$b=\frac{3}{4}a$[/tex]
b). For the energy storage, substitute the radius in (4), we get
[tex]$U = 4 \pi\epsilon_0 \frac{27a^3E^2_0}{512}$[/tex]
This is the amount of energy stored in the conductor.