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GUJCET · Physics · Electric Charges and Fields
An infinite line charge produces a field of \(9 \times 10^4 \frac{N}{ C }\) at a distance of 2 cm, Then the linear charge density will be ________. \(\left(k=9 \times 10^9 \frac{ Nm ^2}{ C ^2}\right)\)
- A \(1 \frac{\mu C }{m}\)
- B \(10 \frac{\mu C }{ m }\)
- C \(0.01 \frac{\mu C }{m}\)
- D \(0.1 \frac{\mu C }{m}\)
Answer & Solution
Correct Answer
(D) \(0.1 \frac{\mu C }{m}\)
Step-by-step Solution
Detailed explanation
(D)
\(
\begin{array}{l}
E=9 \times 10^4 \frac{N}{C} \\
r=2 cm
\end{array}
\)
Electric field due to infinite line charge is given by,
\(
\begin{array}{l}
E=\frac{\lambda}{2 \pi \varepsilon_0} \cdot \frac{1}{r} \\
E=\frac{2 \lambda}{4 \pi \varepsilon_0} \cdot \frac{1}{r}
\end{array}
\)
\(\begin{array}{ll}\therefore & E=\frac{2 k \lambda}{r} \\ \therefore & \lambda=\frac{E r}{2 k} \\ \therefore & \lambda=\frac{9 \times 10^4 \times 2 \times 10^{-2}}{2 \times 9 \times 10^9} \\ \therefore & \lambda=10^{-7} \\ \therefore & \lambda=0.1 \times 10^{-6} \\ \therefore & \lambda=0.1 \frac{\mu C }{m}\end{array}\)
\(
\begin{array}{l}
E=9 \times 10^4 \frac{N}{C} \\
r=2 cm
\end{array}
\)
Electric field due to infinite line charge is given by,
\(
\begin{array}{l}
E=\frac{\lambda}{2 \pi \varepsilon_0} \cdot \frac{1}{r} \\
E=\frac{2 \lambda}{4 \pi \varepsilon_0} \cdot \frac{1}{r}
\end{array}
\)
\(\begin{array}{ll}\therefore & E=\frac{2 k \lambda}{r} \\ \therefore & \lambda=\frac{E r}{2 k} \\ \therefore & \lambda=\frac{9 \times 10^4 \times 2 \times 10^{-2}}{2 \times 9 \times 10^9} \\ \therefore & \lambda=10^{-7} \\ \therefore & \lambda=0.1 \times 10^{-6} \\ \therefore & \lambda=0.1 \frac{\mu C }{m}\end{array}\)
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