AP EAMCET · PHYSICS · Electrostatics
Two particles of charges 4 nC and Q are kept in air with a separation of 10 cm between them. If the electrostatic potential energy of the system is \(1.8 \mu \mathrm{~J}\). Then \(\mathrm{Q}=\)
- A 12 nC
- B 9 nC
- C 5 nC
- D 7 nC
Answer & Solution
Correct Answer
(C) 5 nC
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & U=1.8 \mu J, r=10 \mathrm{~cm} \\ & \therefore U=\frac{\mathrm{kQ}_1 Q_2}{r} \\ & \Rightarrow 1.8 \times 10^{-6}=\frac{9 \times 10^9 \times 4 \times 10^{-9} \times Q}{10 \times 10^{-2}} \\ & \therefore \quad Q=5 \mathrm{nC}\end{aligned}\)
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