The plate separations do not affect the distribution of charge in this problem.
In the figure
`q_0=q_1-q_a,q_d=q_2-q_c`
`and q_f=q_3-q_e`
electric field at point P is zero because this point is lying inside a conductor.
`E=0`
At `P` charge `q_a` will give an electric field towards right. All other charges `q_0, q_c.`.........etc. will give electric field towars left. So,
`1/(2Aepsilon_0)[q_a-(q_1-q_1)-q_c-(q_2-q_c)-q_e-(q_3-q_e)]=0`
or `2q_a-q_1-q_2-q_3=0`
or `q_a=(q_1+q_2+q_3)/2`
similarly the conditionn
`E_R=0`
will give the result
`q_f=(q_1+q_2+q_3)/2`
from here we may conclude that half of the sum of all charges appears on each of the two outermost surfaces of the system of plates.
Further we have a condition.
`E_Q=0`
`1/(2Aepsilon_0)[q_q+(q_1-q_a)+q_c-(q_2-q_c)-q_e-(q_3-q_e)]=0`
or `q_1+2q_c-q_2-q_3=0`
`:. q_c=(q_2+q_3-q_1)/2`
`q_b=q_1-q_a=(q_1-q_2-q_3)/2=-q_c`
similarly, we can show that
`q_d=-q_e`
from here we can find another important resultant that the pairs of opposite surfaces like `b,c`, and `d,e` carry equal and opposite charges.
