
Andrew R. answered 05/18/18
Tutor
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PhD in Mathematical Physics
a=1cm, b=2cm, d=1m, ε=3000V. The situation describes a capacitor; assume 'plate' a has positive Q and b has -Q.
Approximate the potential at the surface of a to be: Va=kQ/a,
and at the surface of b: Vb=-kQ/b
Then ΔV= Va-Vb=kQ/a+kQ/b=kQ[(a+b)/ab] (ΔV=ε=3000V) then
Q=(ΔV/k)ab/(a+b) (calculate Q numerically) and use
Coulumb's law so that F≈kQ2/d2