What is the escape speed of an electron launched from the surface of a 1.1-cm-diameter glass sphere that has been charged to 8.0 nc?

At surface,

v = kq/r

And potential energy of an electron is given by,

PE = - ev = - ekq/r

At escape velocity,

PE + KE = 0.

Therefore,

1/2mv^2 - ekq/r = 0

1/2mv^2 = ekq/r

v = Sqrt [2ekq/mr], where v = escape velocity, e = 1.6*10^-19 C, k = 8.99*10^9 Nm^2/C^2, m = 9.11*10^-31 kg, r = 1.1*10^-2 m, q = 8*10^-9 C

Substituting;

v = Sqrt [ (2*1.6*19^-19*8.99*10^9*8*10^-9) / (9.11*10^-31*1.1*10^-2) ] = 47949357.23 m/s ≈ 4.795 * 10^7 m/s