#### 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"`

⇒ mv^{2} = qEr .. (1)

Kinetic energy of the electron,`"K" = 1/2 mv^{2}`

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 × 10^{9})J

K = 2.88 × 10^{-17} J