AP EAMCET · PHYSICS · Electrostatics
If two particles A and B of charges \(1.6 \times 10^{-19} \mathrm{C}\) and \(3.2 \times 10^{-19} \mathrm{C}\) 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 \(5.12 \times 10^{-22} \mathrm{~N}\)
- B \(5.12 \times 10^{-32} \mathrm{~N}\)
- C \(5.12 \times 10^{-12} \mathrm{~N}\)
- D \(5.12 \times 10^{-25} \mathrm{~N}\)
Answer & Solution
Correct Answer
(D) \(5.12 \times 10^{-25} \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
\(F = k \frac{|q_1 q_2|}{r^2}\) \(F = (9 \times 10^9 \mathrm{~N \cdot m^2/C^2}) \frac{|(1.6 \times 10^{-19} \mathrm{~C})(3.2 \times 10^{-19} \mathrm{~C})|}{(0.03 \mathrm{~m})^2}\) \(F = 5.12 \times 10^{-25} \mathrm{~N}\)
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