The gap between the plates of a parallel-plate capacitor is filled with istropic dielectrc whose permittlvity `epsilon` varies linearly from `espilon_(1)` to `epsilon_(2) (epsilon_(2) gt epsilon_(1))` in the direction perpendicular to the plates. The area of each plate equals `S`, the separation between the plates is equal to `d`. Find :
(a) the capacitance of the capacitor,
(b) the space density of the bound chagres as a function of a if the charge of the capacitor is `q` and `E` in it is directed toward the growing `epsilon` values.

Explanation

(a) We point the x-axis lowards right and plane the origin on the left hand side plate. The left plate is assumed to be positively chagred.
Since `epsilon` varies linearly, we can write,
`epsilon(x) = a + bx`
where `a` and `b` determined from the boundary condition. We have
`epsilon = epsilon_(1)` at `x = 0` and `epsilon = epsilon_(2)` at `x = d`,
Thus, `epsilon(x) = epsilon_(1) + ((epsilon_(2) - epsilon_(1))/(d)) x`
Now potential difference between the plates
`varphi_(+) - varphia_(-) = int_(0)^(d) vec(E). vec(E). vec(dr) = int_(0)^(d) (sigma)/(epsilon_(0) epsilon (x)) dx`
`= int_(0)^(d) (sigma)/(epsilon_(0) (epsilon_(1) + (epsilon_(2) - epsilon_(1))/(d))x) dx = (sigma d)/((epsilon_(2) - epsilon_(1)) epsilon_(0)) IN (epsilon_(2))/(epsilon_(1))`
Hence, the sought capacitance, `C = (sigma S)/( varphi_(+) - varphi_(-)) = ((epsilon_(2) - epsilon_(1)) epsilon_(0) S)/((IN epsilon_(2)//epsilon_(1)) d)`
(b) `D = (q)/(S)` and `P = (q)/(S) - (q)/(Se (x))`
and the space density of bound charge is
`rho' = div P = - (q (epsilon_(2) - epsilon_(1)))/(Sde^(2) (x))`