Question

an electron is moving around a infinite linear charge in a circular path of diameter0 .30m.if linear charge density is 10^- 6 c/m^2. then calculate the speed of electron​

Answer 1

Question

An electron is moving around a infinite linear charge in a circular path of diameter0 .30m.if linear charge density is 10^- 6 C/m²,then calculate the speed of electron

Solution

Velocity is 5.6 × 10^20 m/s

Given

• Diameter,D = 0.30 m

• Radius,R = 0.15 m

• Linear Charge Density = 10^-6 C/m²

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• An electron is revolving around a charge with some velocity V.

• The charge would exert an electrostatic force on the electron.

• Like every revolving body,the electron would exert a centripetal force to the centre of the charge.

• These forces would be equal in magnitude but opposite in direction

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Assume a uniformly cylindrical guassian surface around the charge of infinite length. The electric flux would be :

$\sf \: \phi \: = \dfrac{Q}{ \epsilon {}_{o} }$

But $\sf \: \lambda \: = \dfrac{Q}{L}$

$\longrightarrow \: \sf \: \phi \: = \dfrac{ \lambda \: L}{ \epsilon {}_{o} } - - - - - - - - (1)$

Electric Flux in a infinitely long charged cylindrical wire would be :

$\sf \: \phi = E \: (2\pi \: r L) - - - - - - - (2)$

From relations (1) and (2),we write :

$\huge{ \boxed{ \boxed{ \sf{E = \frac{ \lambda}{2\pi \: r \: \epsilon {}_{0} } }}}}$

$Here \begin{cases} \sf{E \longrightarrow Electric\ Field } \\ \sf{\phi \longrightarrow Flux } \\ \sf{\epsilon_{o} \longrightarrow Permittivity } \\ \sf{\lambda \longrightarrow Linear \ Charge \ Density } \end{cases}$

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Now,

$\boxed{\boxed{\sf Electrostatic \ Force = Centripetal \ Force}} \\ \\ \longrightarrow \sf E \times Q = \dfrac{mv^2}{r} \\ \\ \longrightarrow \ \sf v^2 = \dfrac{E Q \times r}{m} \\ \\ \longrightarrow \sf v = \sqrt{\dfrac{2 \lambda \times Q_E \times \cancel{r}}{M_E \times \cancel{r }} \times \dfrac{1}{4 \pi \epsilon_o}}$

$\huge{\longrightarrow \boxed{\boxed{\sf v = \sqrt{\dfrac{2 \lambda K Q_E}{M_E} }}}}$

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Substituting the values,we get :

$\sf v = \sqrt{\dfrac{2 \times 9 \times 10^9 \times 10^{-6} \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} \\ \\ \longrightarrow \sf v = \sqrt{3.16 \times 10^{15}} \\ \\ \sf \longrightarrow v = \sqrt{31.6 \times 10^{14} } \\ \\ \longrightarrow \underline{\boxed{\sf v = 5.6 \times 10^{7} m s^{-2}}}$

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