Answer
$F = (6.4\times 10^{-19}~\hat{i})~N+(8.0\times 10^{-19}~\hat{k})~N$
Work Step by Step
We can find the force on the proton due to the magnetic field:
$F_B = qv\times B$
$F_B = (1.6\times 10^{-19}~C)(2000\hat{j}~m/s)\times (-2.50\hat{i}~mT)$
$F_B = (8.0\times 10^{-19}~\hat{k})~N$
We can find the force on the proton due to the electric field:
$F_E = q~E$
$F_E = (1.6\times 10^{-19}~C)(4.00\hat{i}~V/m)$
$F_E = (6.4\times 10^{-19}~\hat{i})~N$
We can find the net force acting on the proton:
$F = F_E+F_B$
$F = (6.4\times 10^{-19}~\hat{i})~N+(8.0\times 10^{-19}~\hat{k})~N$
