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`

Step-by-step 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 = (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 = (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 = (F)/(M)`

=>
`a = ( 1.278 *10^(-13))/(9.11*10^(-31))`

=>
`a = 1.40 *10^(17) \ m/s^2`