Answer
(a) $(5.0\times 10^6\frac{N}{C})\hat{x}$
(b) $(-0.2T)\hat{z}$
Work Step by Step
(a) We know that
$\vec{E}=\frac{F_{E}}{e}$
We plug in the known values to obtain:
$\vec{E}=\frac{8.0\times 10^{-13}}{1.6\times 10^{-19}}$
$\vec{E}=(5.0\times 10^6\frac{N}{C})\hat{x}$
(b) As we know that
$B=\frac{F_E-F_{net}}{qv}$
We plug in the known values to obtain:
$B=\frac{8.0\times 10^{-13}-7.5\times 10^{-13}}{1.6\times 10^{-19}(1.5\times 10^6)}=(-0.2T)\hat{z}$
