Answer:
The acceleration is 3.16x10¹⁷ m/s².
Explanation:
First, we need to find the magnitude of the Coulombs force (F):
[tex] |F| = \frac{Kq_{1}q_{2}}{d^{2}} [/tex]
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
[tex]|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[/tex]
Now, we can find the acceleration:
[tex] 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} [/tex]
Therefore, the acceleration is 3.16x10¹⁷ m/s².
I hope it helps you!
