Question

Two 2.7-cm-diameter electrodes with a 0.20-mm-thick sheet of Teflon between them are attached to a 9.0 V battery. Without disconnecting the battery, the Teflon is removed.

What is the charge before the Teflon is removed?

What is the potential difference before the Teflon is removed?

What is the electric field before the Teflon is removed?

What is the charge after the Teflon is removed?

What is the potential difference after the Teflon is removed?

What are the electric field after the Teflon is removed?

Answer 1

given

radius of each plate, r = 2.7/2 = 1.35 cm = 0.0135 m
Area of each plate, A = pi*r^2

= pi*0.0135^2

= 5.72*10^-4 m

d = 0.2 mm = 0.2*10^-3 m

we know,
dielctric constant of teflon, k = 2.1

Capacitance with dielctric, C = k*A*epsilon/d

= 2.1*5.72*10^-4*8.854*10^-12/(0.2*10^-3)

= 5.32*10^-11 F

Capacitance without dielctric, Co = C/k

= 5.32*10^-11/2.1

= 2.53*10^-11 F

a) the charge before the Teflon is removed, Q = C*V

= 5.32*10^-11*9

= 4.8*10^-10 C

b) V = 9.0 V

c) E = V/d

= 9/(0.2*10^-3)

= 4.5*10^4 N/c

d) Qo = Q/k

= 4.8*10^-10/2.1

= 2.3*10^-10 C

e) V = 9 V

f) E = Eo

= 4.5*10^4 N/c