17. (a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air (3.0*106 V/m) if the plates are separated by 2.00 mm and 5.0*103 V a potential difference of is applied? (b) How close together can the plates be with this applied voltage?

Distance between plates d = 2 x 10⁻³m

Potential diff applied = 5 x 10³ V

Electric field = Potential diff applied / d

= 5 x 10³ / 2 x 10⁻³

= 2.5 x 10⁶ V/m

This is less than breakdown strength for air 3.0*10⁶ V/m

b) Let the plates be at a separation of d. so

5 x 10³ / d = 3.0*10⁶ (break down voltage)

d = 5 x 10³ / 3.0*10⁶

= 1.67 x 10⁻³ m

= 1.67 mm.