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.

Answer 1

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 mv<sup>2</sup>`

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