In the figure a proton is fired with a speed of 200000m/s from the midpoint of t

Question

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?? Please include all the equations you used and explain

Explanation / Answer

Suppose x is the distance where the proton turns around, the potential at this location is V = 500*x/0.015

So from conservation of energy

1/2 mv^2 = q V

0.5*1.67E-27*200000^2 = 1.6E-19*500*x/0.015

x=0.00626 m=6.26 mm

The potential at this location is V = 500*0.00626/0.015=209 V

The speed of the proton remains unaltered and it has same speed it initially had so 200000 m/s

1)

so escape speed means v = 0 at r = infinity

Ei = Efinal

1/2 mv^2 + sum of kqQ/r = 0

0.5*1.67E-27*v^2 + 2*9.0E9*-2.0E-9*1.6E-19/5.0E-3 = 0

v=1.17E6 m/s

Suppose x is the distance where the electron turns around, the potential at this location is V = 500*x/0.015

So from conservation of energy

1/2 mv^2 = q V

0.5*9.11E-31*200000^2 = 1.6E-19*500*x/0.015

x= 0.00000342 m

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