Question

Two point charges, A and B, are separated by a distance of 19.0 cm . The magnitude of the charge on A is twice that of the charge on B. If each charge exerts a force of magnitude 45.0 N on the other, find the magnitudes of the charges.

Answer 1

QA = 19μC

QB = 9.5 μC

Step-by-step explanation:

• The force that each charge exerts on the other must obey Coulomb's Law, as follows:

`F_(AB) = (k*Q_(A) * Q_(B))/(r_(AB)^(2)) (1)`

• We know that the value of the magnitude of FAB is 45.0 N, the distance between QA and QB is 0.19 m, and that QA = 2*QB.
• Replacing in (1), we can solve for QB, as follows:

`Q_(B) = \sqrt{(F_(AB)*r_(AB) ^(2))/(2*k) } = \sqrt{(45.0N*(0.19m) ^(2))/(2*9e9N*m2/C2) } = 9.5e-6 C (2)`

• Since QA = 2*QB
• ⇒ QA = 2* 9.5μC = 19.0 μC
• ⇒ QB = 9.5μC