Answer
$a = 3.93\times 10^{16}~m/s^2$
Work Step by Step
We can use Equation (22-26) to find the magnitude of the electric field:
$E = \frac{\sigma}{2\epsilon_0}(1 - \frac{z}{\sqrt{z^2+R^2}})$
$E = \frac{\sigma}{2\epsilon_0}(1 - \frac{R/100}{\sqrt{(R/100)^2+R^2}})$
$E = \frac{\sigma}{2\epsilon_0}(1 - 0.01)$
$E = \frac{4.00\times 10^{-6}~C/m^2}{(2)(8.854\times 10^{-12}~C^2/N~m^2)}\cdot (0.99)$
$E = 2.236\times 10^5~N/C$
We can find the magnitude of the force on the electron:
$F = \vert q \vert E$
$F = (1.6\times 10^{-19}~C)(2.236\times 10^5~N/C)$
$F = 3.58\times 10^{-14}~N$
We can find the magnitude of the acceleration:
$F = ma$
$a = \frac{F}{m}$
$a = \frac{3.58\times 10^{-14}~N}{9.109\times 10^{-31}~kg}$
$a = 3.93\times 10^{16}~m/s^2$
