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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Gauss's Law

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Gauss's Law

A long cylindrical wire possesses a positive charge with a linear...

A long cylindrical wire possesses a positive charge with a linear density of 2.0×10−8 C m−12.0 \times 10^{-8} \text{ C m}^{-1}. An electron orbits the wire in a circular path, driven by the attractive electrostatic force. Determine the kinetic energy of the electron. (Note that this energy is independent of the orbital radius.)

Options

  1. A2.88×10−17 J2.88 \times 10^{-17} \text{ J}
  2. B2.88×10−16 J2.88 \times 10^{-16} \text{ J}
  3. C5.76×10−17 J5.76 \times 10^{-17} \text{ J}
  4. D1.44×10−17 J1.44 \times 10^{-17} \text{ J}

Correct answer

A. 2.88×10−17 J2.88 \times 10^{-17} \text{ J}

Step-by-step solution

The electric field at a distance rr from a long cylindrical wire is given by:

E=λ2πϵ0rE = \dfrac{\lambda}{2 \pi \epsilon_0 r}

The electrostatic force on the electron provides the necessary centripetal force for its circular motion:

mv2r=eE=eλ2πϵ0r\dfrac{m v^2}{r} = eE = \dfrac{e \lambda}{2 \pi \epsilon_0 r}

mv2=eλ2πϵ0m v^2 = \dfrac{e \lambda}{2 \pi \epsilon_0}

The kinetic energy of the electron is:

K=12mv2=eλ4πϵ0K = \dfrac{1}{2} m v^2 = \dfrac{e \lambda}{4 \pi \epsilon_0}

Substituting the given values λ=2.0×10−8 C m−1\lambda = 2.0 \times 10^{-8} \text{ C m}^{-1}, e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C}, and 14πϵ0=9×109 N m2 C−2\dfrac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}:

K=(9×109)×(1.6×10−19)×(2.0×10−8)K = (9 \times 10^9) \times (1.6 \times 10^{-19}) \times (2.0 \times 10^{-8})

K=28.8×10−18 J=2.88×10−17 JK = 28.8 \times 10^{-18} \text{ J} = 2.88 \times 10^{-17} \text{ J}
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