Question

Two point charges, A and B, are separated by a distance of 22.0 cm . The magnitude of the charge on A is twice that of the charge on B. If each charge exerts a force of magnitude 47.0 N on the other, find the magnitudes of the charges. g

Answer 1

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.