Maximum electric field at the surface of a Van de Graaff generator

[archaic]

Homework Statement

We have a Van de Graaff generator consisting of a spherical dome on which charge is continuously deposited by a moving belt until the electric field at the surface of the dome becomes equal to the dielectric strength of air.
If the diameter of the dome $R=38.0$ cm, and the "breakdown" electric field surrounding it is $3.00\times10^6$ V/m
a) What is the maximum value of the potential of the dome in kV?
b) What is the maximum charge on the dome in micro coulombs?

Relevant Equations

N/A

I know that the potential of the sphere at its surface is $V(a)=kQ/a$, and the electric field generated by it is $E(a)=kQ/a^2$, which gives me $V(a)=aE(a)$.
When the electric field at the surface is as in the question, we have $V(a)=38.0\times10^{-2}\times3.00\times10^6=1140000\,V=1140\,kV=1.14\times10^3kV$.
And the charge is $Q=aV(a)/k=\frac{38.0\times10^{-2}\times1.140000}{8.99\times10^9}\times10^6=48.2\,\mu C$.
Do you guys see something missing?
I am confident that the calculations are fine, and that I have the right units..

 

[TSny]

Looks good to me. [EDIT: Ah! They give you the diameter not the radius! Yet they denote the diameter with R.]

 

[archaic]

Looks good to me. [EDIT: Ah! They give you the diameter not the radius! Yet they denote the diameter with R.]

OH! That's actually my bad.. I paraphrased parts of the question.
Thank you very much!