Answer
$a = (-2.1\times 10^{13}~m/s^2)~\hat{j}$
Work Step by Step
We can find the magnitude of the force on the electron:
$F = \vert q \vert E$
$F = (1.6\times 10^{-19}~C)(120~N/C)$
$F = 1.92\times 10^{-17}~N$
We can find the magnitude of deceleration:
$F = ma$
$a = \frac{F}{m}$
$a = \frac{1.92\times 10^{-17}~N}{9.109\times 10^{-31}~kg}$
$a = 2.1\times 10^{13}~m/s^2$
The direction of the electric field is in the +y direction and an electron's charge is negative. Therefore, the direction of the electron's acceleration is in the -y direction.
We can express the electron's acceleration in unit-vector notation:
$a = (-2.1\times 10^{13}~m/s^2)~\hat{j}$
