The plates of a parallel plate capacitor are seperated by 5mm. The electric field between them is 4.2 x 10^5 V/m. A proton isreleased with an initial velocity of zero at the positive plate.Assume there is a vacuum between the plates. A) What is the kinetic energy, in J, when it reaches thenegative plate? B) Express the kinetic energy of the proton in eV The plates of a parallel plate capacitor are seperated by 5mm. The electric field between them is 4.2 x 10^5 V/m. A proton isreleased with an initial velocity of zero at the positive plate.Assume there is a vacuum between the plates. A) What is the kinetic energy, in J, when it reaches thenegative plate? B) Express the kinetic energy of the proton in eV
Explanation / Answer
d= 5 mm E= 4.2 x 10^5 V/m Initial Kinetic Energy = 0 charge of a proton = 1.6 E -19 a) E= change in V (potential difference)/ d E x d = Potentialdifference Potential difference =2100 KE= q x V = 2100 x (1.6 e-19) KE = 3.36 e -16 b) 1J = 1.6 e -19 eV (3.36 e -16)/(1.6 e -19) =2100 eV