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Menurut saya begini,Maaf tulisannya jelek&Semoga bermanfaatPersamaan gaya Coulumb :
F = k q1q2 / r^2
ket :
F = Gaya Coulomb
k = konstanta
q1 = muatan 1
q2 = muatan 2
r = jari jari
Persamaan 1
F = k q1q2 / r^2
100 = k . q . 2q / r^2
100 = k 2q^2 ...1) Persamaan I
F = k q1q2 / r^2
F = k 2q. 2q / 2r^2
F = k 4q^2 / 2 r^2 ... 2) Persamaan II
Bandingkan kedua keadaan
100 = k 2q^2 / r^2 100 = q^2 / 1
----------------------- = ----------------------
F = k 4q^2 / 2r^2 F = 2q^2 / 4
*k, dan r saya coret* , *q saya sederhanakan*
maka didapati q^2 = 100
F = 2q^2 / 4
F = 2 . 100 / 4
F = 200 / 4
F = 50 N
Maaf kalau salah
Semoga membantu:)