Answer
$\sigma = 2.9\times 10^{-6}~C/m^2$
Work Step by Step
We can use the graph to find the rate of deceleration:
$a = \frac{\Delta v}{\Delta t}$
$a = \frac{-2.0\times 10^5~m/s}{7.0\times 10^{-12}~s}$
$a = -2.857\times 10^{16}~m/s^2$
We can find the sheet's surface charge density $\sigma$:
$F = ma$
$E~q = ma$
$\frac{\sigma q}{2\epsilon_0} = ma$
$\sigma = \frac{2\epsilon_0~ma}{q}$
$\sigma = \frac{(2)(8.854\times 10^{-12}~F/m)~(9.109\times 10^{-31}~kg)(-2.857\times 10^{16}~m/s^2)}{-1.6\times 10^{-19}~C}$
$\sigma = 2.9\times 10^{-6}~C/m^2$
