The magnitude of the second charge given that the first is –6×10¯⁶ C and is located 0.05 m away is +3.0×10¯⁶ C
<h3>Coulomb's law equation </h3>
F = Kq₁q₂ / r²
Where
- F is the force of attraction
- K is the electrical constant
- q₁ and q₂ are two point charges
- r is the distance apart
<h3>How to determine the second charge </h3>
- Charge 1 (q₁) = –6×10¯⁶ C
- Electric constant (K) = 9×10⁹ Nm²/C²
- Distance apart (r) = 0.05 m
- Force (F) = 65 N
F = Kq₁q₂ / r²
Cross multiply
Fr² = Kq₁q₂
Divide both side by Kq₁
q₂ = Fr² / Kq₁
q₂ = (65 × 0.05²) / (9×10⁹ × 6×10¯⁶)
q₂ = +3.0×10¯⁶ C (since the force is attractive)
Learn more about Coulomb's law:
brainly.com/question/506926
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