science
posted by Brittney on .
If the magnitude of the electric field in air exceeds roughly 3 X10^6 N/C, the air breaks down and a spark forms. For a twodisk capacitor of radius 52 cm with a gap of 1 mm, what is the maximum charge (plus and minus) that can be placed on the disks without a spark forming (which would permit charge to flow from one disk to the other)? The constant å0 = 8.85 X10^12 C2/(N·m2).

Calculate the capacitance using the diameter and gap. Assume the gap is filled with air with the breakdown field strength you were given.
Plate area A = 0.849 m^2
Gap d = 0.001 m
C = epsilon0*A/d = 8.85*10^12*0.849/0.001 = 7.51*10^13 farads
Maximum allowed E = 3*10^6 V/m
Maximum allowed voltage:
Vmax = E*d = 3000 V
Max charge Q = C*Vmax = 2.25*10^9 C 
C = epsilon0*A/d = 8.85*10^12*0.849/0.001 = 7.51*10^13 farads
Your steps are correct but your answer is miscalculated. 
E is approximately (Q/A)/e(o)
so rearranging the formula we get
AEe(o)=Q
hence Q= 2.26e5