Charge distribution on switched capacitors
Although a very "academic" exercise, assuming "1 μΩ" SPDT switches of "0 pF" switching in zero time...

Assuming a clock at 5e8 rate driving both switches at 50% duty cycle;   

from 0 to 1ns  \$dV=\dfrac{Ic*dt}{C_1} = \dfrac{1mA*1ns}{1pF}=10V\$ and  

at 1ns  C1=1pF,10V switches in series with C2=1pF,0V    

after t=1ns switch {C1+C2=0.5pF,10V}//{C2=0.5pF,0V} » {1pF,5V)
from 1 to 2ns since the load is still 1pF dV/dt is the same 10V/ns rate start from 

The Voltage on C3 acquires 50% the charge of C1 thus C3=5V while C1 drops from 10V to 7.5V and C2 drops from 0 to -2.5V totalling the same C3=5V.

Since the 2 switches add up to 2 μΩ the 10V conducts at 5A-pk but oscillates at +/- 5A  then the current doubles the voltage increases after 2ns.