Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

We can find magnitude of the charge that was transferred. $Q = (3.0\times 10^9)(1.6\times 10^{-19}~C)$ $Q = 4.8\times 10^{-10}~C$ We can find the area of each disk. $E = \frac{Q}{A~\epsilon_0}$ $A = \frac{Q}{E~\epsilon_0}$ $A = \frac{4.8\times 10^{-10}~C}{(2.0\times 10^5~N/C)(8.85\times 10^{-12}~C^2/N~m^2)}$ $A = 2.71\times 10^{-4}~m^2$ We can find the radius of each disk. $\pi~r^2 = A$ $r = \sqrt{\frac{A}{\pi}}$ $r = \sqrt{\frac{2.71\times 10^{-4}~m^2}{\pi}}$ $r = 0.0093~m = 9.3~mm$ Since the diameter is twice the radius, the diameter of each disk is 18.6 mm