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

Answer:

QA = 19μC

QB = 9.5 μC

Explanation:

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

       $F_{AB} = \frac{k*Q_{A} * Q_{B}}{r_{AB}^{2}} (1)$

• 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:

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

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