Answer
(a) $(-3.3\times 10^4N/C)\hat y $
(b) $(9.81m/s^2)\hat y $
Work Step by Step
(a) We know that
$ E=\frac{mg}{q}(-\hat y)$
We plug in the known values to obtain:
$ E=\frac{0.1176Kgm/s^2}{3.6\times 10^{-6}C}(-\hat y)$
$ E=(-3.3\times 10^4N/C)\hat y $
(b) We know that
$ F_q-F=ma $
$\implies (2q)E-mg=ma $
$\implies 2q(\frac{mg}{q})-mg=ma $
$\implies mg=ma $
$\implies g=a $
Thus, the required acceleration is $\vec{a}=(9.81m/s^2)\hat y $
