Two point charges, $q=5.0\mathrm{nC}$ and $Q=10.0\mathrm{nC}$, are separated by a distance of $r=0.20m$ in vacuum. What is the magnitude of the force $F$ on the $5.0\mathrm{nC}$ charge?

• A. $F=1.1\mathrm{\mu N}$
• B. $F=0.56\mathrm{\mu N}$
• C. $F=2.3\mathrm{\mu N}$
• D. $F=5.0\mathrm{\mu N}$

Answer: (A)

Verified step by step guidance

1. Identify the given quantities: charge q = 5.0 nC, charge Q = 10.0 nC, and separation distance r = 0.20 m. Remember to convert nanocoulombs (nC) to coulombs (C) by multiplying by 10^{-9}.
2. Recall Coulomb's law, which gives the magnitude of the electrostatic force between two point charges: \[F = k \frac{\left|qQ\right|}{r^2}\] where \(k\) is Coulomb's constant, \(k = 8.99 \times 10^9 \ \mathrm{N \cdot m^2 / C^2}\).
3. Substitute the converted values of \(q\), \(Q\), and \(r\) into the formula. Make sure to square the distance \(r\) correctly in the denominator.
4. Calculate the numerator by multiplying the absolute values of the charges, and then divide by the square of the distance.
5. Multiply the result by Coulomb's constant \(k\) to find the magnitude of the force \(F\) acting on the 5.0 nC charge.