Question

a proton is traveling horizontally to the right at 4.50×10⁶ m/s. (a) find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.20 cm. (b) how much time does it take the proton to stop after entering the field? (c) what minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)?

Explanation:

Step1: Find the acceleration of the proton

Using the kinematic - equation $v^{2}=v_{0}^{2}+2ax$. Given $v = 0$, $v_{0}=4.50\times10^{6}\ m/s$, and $x = 3.20\times10^{- 2}\ m$.
$0=(4.50\times10^{6})^{2}+2a(3.20\times10^{-2})$
$a=\frac{- (4.50\times10^{6})^{2}}{2\times3.20\times10^{-2}}=\frac{-20.25\times10^{12}}{6.4\times10^{-2}}=- 3.164\times10^{14}\ m/s^{2}$

Step2: Calculate the electric - field magnitude for the proton

The force on a charged particle in an electric field is $F = qE$, and from Newton's second law $F = ma$. For a proton, $q = 1.6\times10^{-19}\ C$ and $m = 1.67\times10^{-27}\ kg$.
$E=\frac{ma}{q}=\frac{1.67\times10^{-27}\times3.164\times10^{14}}{1.6\times10^{-19}}=\frac{5.284\times10^{-13}}{1.6\times10^{-19}} = 3.30\times10^{6}\ N/C$
The direction of the electric field is opposite to the direction of the proton's motion (to the left) since the proton is positively charged and we want to decelerate it.

Step3: Find the time for the proton to stop

Using the kinematic - equation $v = v_{0}+at$. Since $v = 0$, $v_{0}=4.50\times10^{6}\ m/s$ and $a=-3.164\times10^{14}\ m/s^{2}$
$t=\frac{v - v_{0}}{a}=\frac{0 - 4.50\times10^{6}}{-3.164\times10^{14}}=1.42\times10^{-8}\ s$

Step4: Calculate the electric - field magnitude for the electron

For an electron, $q=-1.6\times10^{-19}\ C$ and $m = 9.11\times10^{-31}\ kg$. Using $a=-3.164\times10^{14}\ m/s^{2}$ (same acceleration as in part (a) for the same stopping - distance condition) and $F = qE=ma$
$E=\frac{ma}{q}=\frac{9.11\times10^{-31}\times3.164\times10^{14}}{1.6\times10^{-19}}=\frac{28.82\times10^{-17}}{1.6\times10^{-19}}=1.80\times10^{3}\ N/C$
The direction of the electric field is in the same direction as the electron's motion (to the right) since the electron is negatively charged and we want to decelerate it.

Answer:

(a) Magnitude: $3.30\times10^{6}\ N/C$, Direction: to the left
(b) $1.42\times10^{-8}\ s$
(c) Magnitude: $1.80\times10^{3}\ N/C$, Direction: to the right