Question

1. My Notes -11 points OSColPhys2016 19.5.WA.040. The electric field in the region between the plates of a parallel plate capacitor has a magnitude of 7.7 x 105 V/m. If the plate separation is 0.51 mm, determine the potential difference between the plates. CV Additional Materials Reading 2. -/2 points OSColPhys2016 19.5.WA.043. My Notes (a) Determine the electric field strength between two parallel conducting plates to see if it will exceed the breakdown strength for air (3 x 106 V/m). The plates are separated by 3.18 mm and a potential difference of 5350 V is applied. OV/m (b) How close together can the plates be with this applied voltage without exceeding the breakdown strength? mm

Answer 1

o E = 7.7 XLOS Vilm , d = 0·51X163 mm Potential difference between the plates is given by - DV = Exd = 7.7x105x0.51x103 = 3.921 Ov=392.7 volt Kd 12@ AV=5350V d = 3.18X103m I Electric field between the plates given by - = OV = 5350 es → E = 1.68X106 Vilm | so we can see that this value is smaller than breakdown strength of air. 2(6 Let separation between the blates is "d" for which electric field between them is just equal to breakdown strength of In this situation, E=3x106m AV = 5350 V = AV d= AV = 5350 E 3X106 = 1783.33x10 m d=178x103m. d=178mm