Answer
$1.33\times10^{14}\;m/s^2$
Work Step by Step
Assume that this is the only force acting on the electron. Use equation 16–3 to calculate the magnitude of the force, and apply Newton’s second law.
$$\vec{E}=\frac{\vec{F}}{q}$$
$$\vec{F}=q\vec{E}=m\vec{a}$$
$$a=\frac{|q|}{m}E=\frac{(1.602\times10^{-19}C)}{9.11\times10^{-31}kg} (756\;N/C)$$
$$= 1.33\times10^{14}\;m/s^2$$
The electron is negatively charged, so the force acting on it is opposite to the field, and so is the acceleration.
