Given: Initial charges: ‌q1=+6µC=6×10−6C ‌q2=+10µC=10×10−6C Force of repulsion: F=30N Using Coulomb's Law: F=‌
1
4πε0
⋅‌
q1q2
r2
Solve for r2 : ‌r2=‌
9×109×6×10×10−12
30
‌r2=18×10−3 Adding Additional Charges: Each charge receives an additional −8µC : ‌q1′=+6−8=−2µC ‌q2′=+10−8=+2µC Recalculate the Force Using Coulomb's Law: F′=‌