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(A) The proton is moving in a uniform electric field directed downwards. The force on the proton is $F = qE$ downwards. Since the gravitational force

Vedclass · Accepted Answer

$1.35 	imes 10^{-2} \, s$

Vedclass · Answer

(A) The proton is moving in a uniform electric field directed downwards. The force on the proton is $F = qE$ downwards. Since the gravitational force $mg$ is negligible compared to the electric force $qE$,the effective acceleration is $a = \frac{qE}{m}$ downwards.The initial velocity components are $u_x = u \cos 	heta$ and $u_y = u \sin 	heta$.The motion is similar to projectile motion where the acceleration due to gravity $g$ is replaced by $a = \frac{qE}{m}$.The time of flight $T$ is given by:$T = \frac{2 u_y}{a} = \frac{2 u \sin 	heta}{\left(\frac{q E}{m}ight)}$Given:$u = 150 \, m/s$,$	heta = 60^\circ$,$E = 2 	imes 10^{-4} \, N/C$,$q = 1.6 	imes 10^{-19} \, C$,$m = 1.67 	imes 10^{-27} \, kg$.$T = \frac{2 	imes 150 	imes \sin(60^\circ) 	imes 1.67 	imes 10^{-27}}{1.6 	imes 10^{-19} 	imes 2 	imes 10^{-4}}$$T = \frac{300 	imes 0.866 	imes 1.67 	imes 10^{-27}}{3.2 	imes 10^{-23}}$$T = \frac{433.938 	imes 10^{-27}}{3.2 	imes 10^{-23}} \approx 135.6 	imes 10^{-4} \, s = 1.356 	imes 10^{-2} \, s$.Comparing with the options,the closest value is $1.35 	imes 10^{-2} \, s$.

$A$ proton is fired at an initial velocity of $150 \, m/s$ at an angle of $60^\circ$ above the horizontal into a uniform electric field of $2 \times 10^{-4} \, N/C$ between two charged parallel plates as shown in the figure. The total time the particle is in motion is:

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