KCET · Physics · Electrostatics
Two identical charges repel each other with a force equal to \(10 \mathrm{mg}\) wt when they are \(0.6 \mathrm{~m}\) a part in air \(\left(g=10 \mathrm{~ms}^{-2}\right)\). The value of each charge is
- A \(2 \mathrm{mC}\)
- B \(2 \times 10^{-7} \mathrm{C}\)
- C \(2 \mathrm{nC}\)
- D \(2 \mu \mathrm{C}\)
Answer & Solution
Correct Answer
(D) \(2 \mu \mathrm{C}\)
Step-by-step Solution
Detailed explanation
Coulomb force is given by
\(F =\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} \)
\(\therefore\left(10 \times 10^{-3}\right) \times 10 =\frac{\left(9 \times 10^{9}\right) \times q^{2}}{(0.6)^{2}} \)
\(\text {or } q^{2} =\frac{10^{-1} \times 0.36}{9 \times 10^{9}} \left[\because q_{1}=q_{2}=q\right] \)
\(=4 \times 10^{-12} \)
\(\therefore q =2 \times 10^{-6} \mathrm{C}=2 \mu \mathrm{C}\)
\(F =\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} \)
\(\therefore\left(10 \times 10^{-3}\right) \times 10 =\frac{\left(9 \times 10^{9}\right) \times q^{2}}{(0.6)^{2}} \)
\(\text {or } q^{2} =\frac{10^{-1} \times 0.36}{9 \times 10^{9}} \left[\because q_{1}=q_{2}=q\right] \)
\(=4 \times 10^{-12} \)
\(\therefore q =2 \times 10^{-6} \mathrm{C}=2 \mu \mathrm{C}\)
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