Question

g A proton moves at 3.60 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.40 103 N/C. Ignore any gravitational effects. (a) Find the time interval required for the proton to travel 4.00 cm horizontally. ns (b) Find its vertical displacement during the time interval in which it travels 4.00 cm horizontally. (Indicate direction with the sign of your answer.) mm (c) Find the horizontal and vertical components of its velocity after it has traveled 4.00 cm horizontally.

Answer 1

Explanation:

It is given that,

Speed of the proton, $v=3.6\times 10^5\ m/s$

Electric field, $E=9.4\times 10^3\ N/C$

(a) Distance covered, d = 4 cm = 0.04 m

Let t is the time interval required for the proton to travel 4.00 cm horizontally. It can be calculated as :

$t=\dfrac{d}{v}$

$t=\dfrac{0.04}{3.6\times 10^5}$

$t=1.11\times 10^{-7}\ s$

or

t = 111 ns

(b) Since, initial speed = 0 in vertical direction. So,

So, q E = ma

$a_y=\dfrac{E_yq}{m}$

Displacement is given by :

$y=ut+\dfrac{1}{2}a_yt^2$

$y=\dfrac{1}{2}\dfrac{E_yq}{m}t^2$

$y=\dfrac{1}{2}\times \dfrac{9.4\times 10^3\times 1.67\times 10^{-19}}{1.67\times 10^{-27}}(1.11\times 10^{-7})^2$

$y=0.00579\ m$

(c) For vertical component of velocity, use equation of kinematics as :

$v_y^2-u^2=2a_yd$ (d = 4 cm)

$v=\sqrt{2\dfrac{E_yq}{m}d_y}$

$v=\sqrt{2\times \dfrac{9.4\times 10^3\times 1.67\times 10^{-19}}{1.67\times 10^{-27}}\times 0.04}$

$v=2.74\times 10^5\ m/s$

For horizontal component of velocity,

$v_x=\dfrac{d}{t}$

$v_x=\dfrac{0.04}{1.11\times 10^{-7}}$

$v_x=3.6\times 10^5\ m/s$

Hence, this is the required solution.