ExamBro
ExamBro
KCET · Physics · Electrostatics

An infinitely long thin straight wire has uniform charge density of \(\frac{1}{4} \times 10^{-2} \mathrm{~cm}^{-1}\).
What is the magnitude of electric field at a distance \(20 \mathrm{~cm}\) from the axis of the wire?

  1. A \(1.12 \times 10^{10} \mathrm{NC}^{-1}\)
  2. B \(4.5 \times 10^{10} \mathrm{NC}^{-1}\)
  3. C \(2.25 \times 10^{10} \mathrm{NC}^{-1}\)
  4. D \(9 \times 10^{10} \mathrm{NC}^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2.25 \times 10^{10} \mathrm{NC}^{-1}\)

Step-by-step Solution

Detailed explanation

Given, charge density of uniformly long thin wire,
\(\begin{aligned}
\lambda &=\frac{1}{4} \times 10^{-2} \mathrm{~cm}^{-1} \\
&=\frac{1}{4} \times 10^{-2} \times 10^{2} \mathrm{~m}^{-1}=\frac{1}{4} \mathrm{~m}^{-1} \\
r &=20 \mathrm{~cm}=0.2 \mathrm{~m}
\end{aligned}\)
Electric field at a distance \(r\) from the axis of wire is given as
\(\begin{aligned}
E &=\frac{\lambda}{2 \pi \varepsilon_{0} r}=\frac{2 \lambda}{4 \pi \varepsilon_{0} r} \\
&=\frac{9 \times 10^{9} \times 2 \times \frac{1}{4}}{0.2} \\
&=2.25 \times 10^{10} \mathrm{NC}^{-1}
\end{aligned}\)