Answer
(a) $F = (6.4\times 10^{-17}~\hat{i}+~1.6\times 10^{-17}~\hat{j})~N$
(b) $F = (-6.4\times 10^{-17}~\hat{i}-~1.6\times 10^{-17}~\hat{j})~N$
(c) $a = 4.0\times 10^{10}~m/s^2$
(d) $a = 7.3\times 10^{13}~m/s^2$
Work Step by Step
(a) We can find the force on the proton.
$F = E~q$
$F = [(400~\hat{i}+~100~\hat{j})~N/C]~(1.6\times 10^{-19}~C)$
$F = (6.4\times 10^{-17}~\hat{i}+~1.6\times 10^{-17}~\hat{j})~N$
(b) We can find the force on the electron.
$F = E~q$
$F = [(400~\hat{i}+~100~\hat{j})~N/C]~(-1.6\times 10^{-19}~C)$
$F = (-6.4\times 10^{-17}~\hat{i}-~1.6\times 10^{-17}~\hat{j})~N$
(c) We can find the magnitude of the force.
$F = \sqrt{F_x^2+F_y^2}$
$F = \sqrt{(6.4\times 10^{-17}~N)^2+(1.6\times 10^{-17}~N)^2}$
$F = 6.6\times 10^{-17}~N$
We can find the magnitude of the proton's acceleration.
$a = \frac{F}{m}$
$a = \frac{6.6\times 10^{-17}~N}{1.67\times 10^{-27}~kg}$
$a = 4.0\times 10^{10}~m/s^2$
(d) We can find the magnitude of the force.
$F = \sqrt{F_x^2+F_y^2}$
$F = \sqrt{(-6.4\times 10^{-17}~N)^2+(-1.6\times 10^{-17}~N)^2}$
$F = 6.6\times 10^{-17}~N$
We can find the magnitude of the electron's acceleration.
$a = \frac{F}{m}$
$a = \frac{6.6\times 10^{-17}~N}{9.1\times 10^{-31}~kg}$
$a = 7.3\times 10^{13}~m/s^2$
