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Question
A long cylindrical wire carries a positive charge of linear density 2.0 × 10-8 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.
Solution
Let the linear charge density of the wire be λ.
The electric field due to a charge distributed on a wire at a perpendicular distance rfrom the wire,
`"E" = λ/ (2 pi ∈ _0 "r")`
The electrostatic force on the electron will provide the electron the necessary centripetal force required by it to move in a circular orbit. Thus,
`"qE" = ("m""v"^2)/"r"`
⇒ mv2 = qEr .. (1)
Kinetic energy of the electron,`"K" = 1/2 mv2`
From (1),
`"K" = ("qEr")/2`
`"K" = "qr"/2 λ /(2 pi ∈_0 "r")` `[∵ "E" = λ/((2 pi ∈_0 "r")) ]`
K =(1.6 ×10-19) × ( 2 × 10-8) × ( 9 × 109)J
K = 2.88 × 10-17 J
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