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?
- A \(1.12 \times 10^{10} \mathrm{NC}^{-1}\)
- B \(4.5 \times 10^{10} \mathrm{NC}^{-1}\)
- C \(2.25 \times 10^{10} \mathrm{NC}^{-1}\)
- D \(9 \times 10^{10} \mathrm{NC}^{-1}\)
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
\(\begin{aligned}
E &=\frac{\lambda}{2 \pi \varepsilon_{0} r}=\frac{2 \lambda}{4 \pi \varepsilon_{0} r} \\
&=\frac{9 \times 10^{9} \times 2 \times \frac{1}{4}}{0.2} \\
&=2.25 \times 10^{10} \mathrm{NC}^{-1}
\end{aligned}\)
\(\begin{aligned}
\lambda &=\frac{1}{4} \times 10^{-2} \mathrm{~cm}^{-1} \\
&=\frac{1}{4} \times 10^{-2} \times 10^{2} \mathrm{~m}^{-1}=\frac{1}{4} \mathrm{~m}^{-1} \\
r &=20 \mathrm{~cm}=0.2 \mathrm{~m}
\end{aligned}\)
Electric field at a distance \(r\) from the axis of wire is given as
\(\begin{aligned}
E &=\frac{\lambda}{2 \pi \varepsilon_{0} r}=\frac{2 \lambda}{4 \pi \varepsilon_{0} r} \\
&=\frac{9 \times 10^{9} \times 2 \times \frac{1}{4}}{0.2} \\
&=2.25 \times 10^{10} \mathrm{NC}^{-1}
\end{aligned}\)
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