A parallel plate capacitor is constructed of two conducting plates spaced 2mm ap
ID: 1281236 • Letter: A
Question
A parallel plate capacitor is constructed of two conducting plates spaced 2mm apart. It is charged to a potential difference of 500 Volts. A proton (mass 1.67*10^-27 kg) is shot through a small hole in the negative plate with a speed of 3.0 * 10^5 m/s.
a)What is the total energy of the proton when it enters the plates?
b)What will the potential energy of the proton be when it reaches the 200 Volt equipotential?
c)What will the kinetic energy of the proton be when it reaches the 200 volt equipotential? Use conservation of energy.
d)What is the x coordinate of the proton when it comes to rest before reversing direction?
A parallel plate capacitor is constructed of two conducting plates spaced 2mm apart. It is charged to a potential difference of 500 Volts. A proton (mass 1.67*10^-27 kg) is shot through a small hole in the negative plate with a speed of 3.0 * 10^5 m/s.Explanation / Answer
a) E = 1/2 mv^2 = 0.5*1.67E-27*2.0E5^2= 3.34E-17 J
b) 200 eV = 200*1.6E-19= 3.2E-17 J
c) E = KE + PE
KE = E - PE = 3.34E-17-3.2E-17 = 0.14E-17 J = 1.4E-18 J
d)
so KE = 0
V = 500*x/2.0E-3
3.34E-17 = 1.6E-19*500*x/2.0E-3
x= 8.35E-4 m