Question

2) Consider two large, metal plates with equal and opposite charges as shown below: OV 300 V 3.0 mm 100 V 200 v a) Which plate, left or right, is the positively charged plate? How do you know? (1 pt) b) What is the magnitude and direction of the electric field at point A located halfway between the plates: y=0 m, x= 1.5mm? (2 pts) c) What is the magnitude and direction of the electric force felt by an electron placed at point B: y=0m, x= 0.75m? (2 pts) d) is the electric potential energy of an electron placed at point B greater than, less than, or equal to the electric potential energy of an electron placed at point A? Give a brief reason. (1 pts)

Answer 1

a) right plate is positively charged because right plate has higher potential.

(b) Electric field between the heater is given by ME E-OV 300 ixrosulm व 2003 Direction is from Right to left. Magnitude of force on electron is E = 20 = (1.681019) Lexros, FEZ 11641614 Direction to the right towards positive blate. Potential energy of electron at WA - QUA = (-1.6*10 19) (150) = UA= -2.4x10mg (d) A, at B, UB= q VB = (-1.6+60-ly (30000.25) UB= -1.28101? J ces Energy at B is higher. By energy conservation gūs ļ mave? U prave Tue 10.3410 m/s 2 et me 2 (1.6*10') (300) I al X (0-31 Ruske "y (300)