Two oppositely charged parallel plates are 20.0 cm apart and have an electric field strength of 82000N/C. If an electron is placed halfway between the plates, what is the speed that it reaches the positive plate?
1 month ago
Answered By Giampiero ("GP") I
Use conservation of energy to answer this question, together with the value of the voltage half way between the plates, which is the half the field times the plate separation. So V = 1/2 Ed = 1/2 82000 N/C x 0.200 m = 8200 V. Multiply this value by the electron's charge and you get its potential energy, or E = Vq = 8200 V x 1.60 x 10-19 C = 1.312 x 10-15 Joules. Now use conservation of energy: this potential energy will all be converted to kinetic energy once the electron reaches the positive plate. We get E = 1/2 m v2. Isolating the velocity yields v = 5.37 x 107 m/s.
1 month ago
Answered By Giampiero ("GP") I
Use conservation of energy to answer this question, together with the value of the voltage half way between the plates, which is the half the field times the plate separation. So V = 1/2 Ed = 1/2 82000 N/C x 0.200 m = 8200 V. Multiply this value by the electron's charge and you get its potential energy, or E = Vq = 8200 V x 1.60 x 10-19 C = 1.312 x 10-15 Joules. Now use conservation of energy: this potential energy will all be converted to kinetic energy once the electron reaches the positive plate. We get E = 1/2 m v2. Isolating the velocity yields v = 5.37 x 107 m/s.