2017 charged plates

The effect of the plates numbered $3$ to $2015$ is negligible, and the system can be regarded as one consisting of four plates, with the outside two earthed, and the middle two carrying charges of $2Q$ and $Q$ respectively. The distances between successive plates are then $d$ , $D = 2014d$ and $d$ . There must be charges of $-(2-\alpha)Q$ , $(2-\alpha)Q$ , $\alpha Q$ , $-\alpha Q$ , $(1+\alpha)Q$ and $-(1+\alpha)Q$ on the RHS of plate 1, the LHS of plate 2, the RHS of plate 2, the LHS of plate 2016, the RHS of plate 2016 and the LHS of plate 2017 respectively.

If the potentials at plates 1,2,2016,2017 are $0$ , $V$ , $W$ , and $0$ respectively, then $V \; = \; \frac{(2-\alpha)Qd}{A\varepsilon_0} \hspace{1cm} V-W \; = \; \frac{\alpha QD}{A\varepsilon_0} \hspace{1cm} W \; = \; \frac{(1+\alpha)Qd}{A\varepsilon_0}$ and hence $\alpha = \frac{d}{D+2d}$ . Moreover $V \; = \; \frac{(2D+3d)dQ}{A\varepsilon_0(D+2d)} \hspace{2cm} W \; = \; \frac{(D+3d)dQ}{A\varepsilon_0(D+2d)}$ With or without the plates 3 to 2015, the potential between plates 2 and 2016 varies linearly between the two plates, and so the potential at plate $j$ ( $2 \le j \le 2016$ ) is $V_j \; = \; V + \frac{(j-2)d}{D}(W - V) \; = \; \frac{Qd(4033 - j)}{2016A\varepsilon_0}$ and so the potential difference between successive plates (such as plates 1728 and 1729) is $\frac{Qd^2}{A\varepsilon_0(D+2d)} = \frac{Qd}{2016A\varepsilon_0}$ , making the answer $1 + 2016 = \boxed{2017}$ .