Question

If two particles A and B of charges \(1.6 \times 10^{-19}\,\text{C}\) and \(3.2 \times 10^{-19}\,\text{C}\) respectively are separated by a distance of 3 cm in air, then the magnitude of electrostatic force on particle A due to particle B is

• A. \(5.12 \times 10^{-22}\,\text{N}\)
• B. \(5.12 \times 10^{-32}\,\text{N}\)
• C. \(5.12 \times 10^{-12}\,\text{N}\)
• D. \(5.12 \times 10^{-25}\,\text{N}\)

Hint:

Always convert distance to meters in SI units and use Coulomb's constant \(9 \times 10^9\,\text{Nm}^2/\text{C}^2\) when calculating electrostatic force.

Answer 1

The Correct Option is D

Step 1: Use Coulomb's Law: \[ F = \frac{1}{4\pi \varepsilon_0} . \frac{q_1 q_2}{r^2} \] Step 2: Substitute values: \[ q_1 = 1.6 \times 10^{-19}\,\text{C},
q_2 = 3.2 \times 10^{-19}\,\text{C},
r = 3\,\text{cm} = 0.03\,\text{m},
\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \] \[ F = 9 \times 10^9 . \frac{(1.6 \times 10^{-19})(3.2 \times 10^{-19})}{(0.03)^2} \] \[ F = 9 \times 10^9 . \frac{5.12 \times 10^{-38}}{9 \times 10^{-4}} = \frac{46.08 \times 10^{-29}}{9} \approx 5.12 \times 10^{-25}\,\text{N} \]