Question

What is the magnitude of the acceleration of an electron that is 1.3 cm from the center of the bead?

Answer 1

Complete Question

A small plastic bead has been charged to  15nC. What is the magnitude of the acceleration of an electron that is 1.3 cm from the center of the bead?

Answer:

The  acceleration is  $a = 1.40 *10^{17} \ m/s^2$

Explanation:

From the question we are told that

    The charge on the bead is  $q = 15n C = 15 *10^{-9} \ C$

     The  distance of the electron from the center of the bead is $d = 1.3 \ cm = 0.013 \ m$

Generally the force between the electron and the bead is  mathematically represented as

      $F = \frac{k * q * e}{d^2}$

Here e is the charge on an electron with value  $e = 1.6 0 *10^{-19} \ C$

        k is  the coulombs constant with value  $k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

   $F = \frac{9*10^{9} * 15*10^{-9} * 1.60 *10^{-19}}{0.013^2}$

    $F = 1.278 *10^{-13} \ N$

Mathematically

    $F = Ma$

Here M is the mass of an electron with value  $9.11 *10^{-31} \ kg$

So

   $a = \frac{F}{M}$

=>   $a = \frac{ 1.278 *10^{-13}}{9.11*10^{-31}}$

=>    $a = 1.40 *10^{17} \ m/s^2$