An electron is moving around an infinite linear charge in a circular path of diameter $large{0.30 m}$. If linear charge density is $large{10^{–6} C/m}$, then calculate the speed of an electron. ($large{m_e= 9.0 times 10^{–31} kg, e = 1.6 times 10^{–19} C}$)[latexpage]

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An electron is moving around an infinite linear charge in a circular path of diameter $0.30 m$ . If linear charge density is $10^{-6} C/m$ , then calculate the speed of an electron. ( $m_e= 9.0 \times 10^{-31} kg, e = 1.6 \times 10^{-19} C$ )

The intensity of electric field due to a infinite linear charge is given by the formula:

$E=\dfrac{2k\lambda}{r}$

Where $k=9\times10^9Nm^2C^{-2}$ , $\lambda$ is the linear charge density and $r$ is the perpendicular distance.

Since electron is moving in a circular path, therefore a centripetal force must be acting on the electron. This centripetal force is provided by the electrostatic force of attraction.

$F_{cp} = F_e$

$\dfrac{m_ev^2}{r}=eE$

If $m_e$ is the mass of electron ( $9.0\times10^{-31}kg$ ), $v$ is the speed of electron, $e$ is the charge on electron( $1.6\times 10^{-19}C$ ) , and $r$ is the radius of the circular path.