Question

(a) Determine the electric field strength between two parallel conducting plates to see if it will exceed the breakdown strength for air (3 ? 106 V/m). The plates are separated by2.98 mm and a potential difference of 5575 V is applied.

(b) How close together can the plates be with this applied voltage without exceeding the breakdown strength?

Answer 1

(a) 1.87×10⁶ V/m

(b) 1.12 mm closer

Step-by-step explanation:

(a)

Electric Field = Electric potential/distance.

E = V/d ................... Equation 1

Where E = Electric Field, V = Electric potential, d = distance.

Given: V = 5575 V, d = 2.98 mm = 0.00298 m.

Substitute into equation 1

E = 5575/0.00298

E = 1.87×10⁶ V/m

(b)

without exceeding the breakdown strength,

make d the subject of equation 1

d = V/E.............. Equation 2

Given: E = 3×10⁶ V/m, V = 5575 V

Substitute into equation 2

d = 5575/3000000

d = 1.86 mm.

the plate will be = 2.98-1.86 = 1.12 mm closer