Answer:
1.12×10⁻⁵ C and 2.24×10⁻⁵ C.
Explanation:
From coulomb's law,
F = kAB/r².............................. Equation 1
Where F = Force exerted by each charge, A = charge at point A, B = charge at point B, r = distance of separation between the points, k = constant of proportionality.
Given: F = 47 N, r = 22 cm = 0.22 m.
Constant: k = 9.0×10⁹ Nm²/C²
Let: B = q, the A = 2q.
Substituting these values into equation 1,
47 = 9.0×10⁹(q×2q)/0.22²
47 = 18×10⁹(q²)/0.0484
q² = (47×0.0484)/(18×10⁹)
q² = 0.126×10⁻⁹
q² = 1.26×10⁻¹⁰
q = √( 1.26×10⁻¹⁰)
q = 1.12×10⁻⁵ C
The charge at point A = 2q = 2× 1.12×10⁻⁵ = 2.24×10⁻⁵ C.
Hence the charges are 1.12×10⁻⁵ C and 2.24×10⁻⁵ C.
