Question

A small glass bead has been charged to 20 nC. What is the magnitude of acceleration in m/s^2 of an electron that is 1.0 cm from the center of the bead? (mass of an electron= 9.1x10^-31 kg)

Answer 1

Answer:

The acceleration is 3.16x10¹⁷ m/s².

Explanation:

First, we need to find the magnitude of the Coulombs force (F):

$|F| = \frac{Kq_{1}q_{2}}{d^{2}}$

Where:

K is the Coulomb constant = 9x10⁹ Nm²/C²

q₁ is the charge = 20x10⁻⁹ C  

q₂ is the electron's charge = -1.6x10⁻¹⁹ C

d is the distance = 1.0 cm = 1.0x10⁻² m

$|F| = \frac{Kq_{1}q_{2}}{d^{2}} = \frac{9\cdot 10^{9}Nm^{2}/C^{2}*20 \cdot 10^{-9} C*(-1.6\cdot 10^{-19} C)}{(0.01 m)^{2}} = 2.88 \cdot 10^{-13} N$                                      

Now, we can find the acceleration:

$a = \frac{F}{m} = \frac{2.88 \cdot 10^{-13} N}{9.1 \cdot 10^{-31} kg} = 3.16 \cdot 10^{17} m/s^{2}$

Therefore, the acceleration is 3.16x10¹⁷ m/s².

I hope it helps you!