Need help due tomorrow... In the figure a proton is fired with a speed of 200000

Question

Need help due tomorrow...

In the figure a proton is fired with a speed of 200000m/s from the midpoint of the capacitor toward the positive plate as shown. The plates are 1.5cm apart. the voltage difference is 500 volts...

1) Find the location where the proton stops and turns around (use x=0 as the location of negative plate).

2)what is the potential at that location

3)what is the protons speed as it reacher 100v?

4) If an electron is fired from the positive plate with the same velocity(200000m/s) how far from the 0V plate will it stop??

Explanation / Answer

initial speed of proton, u = 2*10^5 m/s

V = 500 volts

E = v/d

= 500/0.015

= 33333 N/C

a = -q*E/m

= 1.6*10^-19*33333/(1.67*10^-27)

= -3.19*10^12 m/s^2

1) Apply, v^2 - u^2 = 2*a*s

s = (v^2-u^2)/(2*a)

= (0^2 - (2*10^5)^2)/(2*(-3.19*10^12))

= 0.627 cm

so location, x = 0.75 + 0.627

= 1.377 cm <<<<<<<<<---------------Answer

2) potential at x = 1.377 is = 500*1.377/1.5

= 459 volts <<<<<<<<<---------------Answer

3) Workdone = q*V

0.5*m*(v^2-u^2) = 1.6*10^-19*(250-100)

v = sqrt((2*10^5)^2 + 2*1.6*10^-19*150/(1.67*10^-27) )

= 2.62*10^5 m/s <<<<<<<<<---------------Answer

4) a = -q*E/m

= -1.6*10^-19*33333/(9.1*10^-31)

= 5.86*10^15 m/s^2

Apply, v^2 - u^2 = 2*a*s

s = -u^2/(2*a)

= -(2*10^5)^2/(2*5.86*10^15)

= 3.41*10^-6 m <<<<<<<<<---------------Answer

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