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GUJCET · Physics · Electric Charges and Fields
Electric field produced due to an infinitely long straight uniformly charged wire at a perpendicular distance of 2 cm is \(3 \times 10^8 NC ^{-1}\). Then, linear charge density on wire is _________. ( \(k=9 \times 10^9\) SI unit)
- A \(333 \frac{\mu C }{ m }\)
- B \(3.33 \frac{\mu C }{ m }\)
- C \(666 \frac{\mu C }{ m }\)
- D \(6.66 \frac{\mu C }{ m }\)
Answer & Solution
Correct Answer
(A) \(333 \frac{\mu C }{ m }\)
Step-by-step Solution
Detailed explanation
(A)
Electric field at a distance \(r\) from a very long charged wire
\(\begin{aligned} E & =\frac{\lambda}{2 \pi \varepsilon_2 r}=\frac{2 \lambda}{4 \pi \varepsilon_0 r} \\ E & =\frac{2 k \lambda}{r} \\ \therefore \quad \lambda & =\frac{ E r}{2 k} \\ \lambda & =\frac{3 \times 10^8 \times 2 \times 10^{-2}}{2 \times 9 \times 10^9} \\ \lambda & =0.3333 \times 10^{-3} \\ \lambda & =333 \times 10^{-6} \\ \lambda & =333 \frac{\mu C }{m}\end{aligned}\)
Electric field at a distance \(r\) from a very long charged wire
\(\begin{aligned} E & =\frac{\lambda}{2 \pi \varepsilon_2 r}=\frac{2 \lambda}{4 \pi \varepsilon_0 r} \\ E & =\frac{2 k \lambda}{r} \\ \therefore \quad \lambda & =\frac{ E r}{2 k} \\ \lambda & =\frac{3 \times 10^8 \times 2 \times 10^{-2}}{2 \times 9 \times 10^9} \\ \lambda & =0.3333 \times 10^{-3} \\ \lambda & =333 \times 10^{-6} \\ \lambda & =333 \frac{\mu C }{m}\end{aligned}\)
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