Explanation:
Given that,
Electrostatic force, 
Distance, 
(a)
, q is the charge on the ion


(b) Let n is the number of electrons are missing from each ion. It can be calculated as :


n = 2
Hence, this is the required solution.
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The area of the circle with radius r is
A = πr²
The rate of change of area with respect to time is

The rate of change of the radius is given as

Therefore

When r = 10 ft, obtain

Answer: - 40π ft²/s (or - 127.5 ft²/s)