What minimum radius must a large conducting sphere of an electrostatic generating machine have if it is to be at 20000 v without discharge into the air? how much charge will it carry?

First of all,

The electric field at the surface of the sphere is given by

 E = kQ/r²

The field strength at which breakdown occurs in the air is 3.0 MV/m

So,  E = 3.0 MV/m

The sphere potential is defined as

 V = kQ/r

If we divide E/V  we get

 E = V/r 

r = V/E  = 20000V / 3.10^6 V/m  =  6.66 10^-3 m = 6.66 mm

2. Charge

V = kQ/r >>

Q = Vr/k = 20000V *( 6.66 10^-3 m)/ (9.10^9 N m2/C2) = 1.481 10^-8 C

Gravity bc it’s gravity