Chapter 17, Problem 21P

To determine

Speed of electron.

Answer to Problem 21P

The answer is $4.8\times {10}^{3}m/s$ .

Explanation of Solution

Given:

Charges $Q=+9.5\mu C=9.5\times {10}^{-6}\text{}C$

Radius, $r=3.5cm=0.035m$

Mass, $m=1.0mg=1.0\times {10}^{-6}kg$

Formula used:

Because of conservation of energy, all of initial potential energy is converted to kinetic energy (at the time when the charges are far apart). No charge is fixed.Hence, both are moving at the same speed away from each other. So,

Where,

  $Q$ is amount of charge

  $r$ is radius

  $k$ is constant, $\left(8.99\times {10}^{9}N\cdot m^2/C^2\right)$

  $m$ is mass, (mass of one electron $9.11\times {10}^{-31}kg$ )

Calculation:

Substitute all the known and given values in expression $v=\sqrt{\frac{k{Q}^2}{mr}}$ to find the velocity:

  $v=\sqrt{\frac{\left(8.99\times {10}^{9}N\cdot m^2/C^2\right){\left(9.5\times {10}^{-6}C\right)}^2}{\left(1.0\times {10}^{-6}kg\right)\left(0.035m\right)}}=4.8\times {10}^{3}m/s$

Conclusion:

The required speed is $4.8\times {10}^{3}m/s$ .