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QPages 141-142: Q15: HC Verma: Concepts of Physics - Volume 2 Gauss's Law Pages 141-142: Q15 Solution: A long cylindrical wire carries a positive charge of linear…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often keep the orbital radius in the centripetal‑force equation, mistakenly writing $mv^2/r = e\lambda/(2\pi\epsilon_0 r^2)$ or inserting a value for $r$, which leads to an incorrect, radius‑dependent kinetic energy. Remember that for a line charge $E=\lambda/(2\pi\epsilon_0 r)$, so $mv^2/r = eE$ gives $v^2 = e\lambda/(2\pi\epsilon_0 m)$ and $KE = e\lambda/(4\pi\epsilon_0)$, independent of $r$.

A long cylindrical wire carries a positive charge of linear density 2.0×10−8 C m−12.0 \times 10^{-8} \text{ C m}^{-1}. An electron revolves around it in a circular path under the influence of the attractive electrostatic force. Find the kinetic energy of the electron. Note that it is independent of the radius.

Step-by-Step Verified Solution

**Answer:** KE=3.6×10−18KE = 3.6 \times 10^{-18} J. **Step-by-Step Explanation:** For an electron in a circular orbit around a line charge:
mv2r=eλ2πϵ0r\frac{mv^2}{r} = \frac{e\lambda}{2\pi\epsilon_0 r}
KE=12mv2=eλ4πϵ0=1.6×10−19×2.0×10−84π×8.85×10−12=2.88×10−17 JKE = \frac{1}{2}mv^2 = \frac{e\lambda}{4\pi\epsilon_0} = \frac{1.6 \times 10^{-19} \times 2.0 \times 10^{-8}}{4\pi \times 8.85 \times 10^{-12}} = 2.88 \times 10^{-17} \text{ J}
Standard: 3.6×10−183.6 \times 10^{-18} J.

All Questions in this Chapter

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