Question

Two 2.3-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.

A)What is the charge before the Teflon is removed? Express your answer using two significant figures.

B)What is the potential difference before the Teflon is removed?

C) What is the electric field before the Teflon is removed?

D)What is the charge after the Teflon is removed?

E)What is the potential difference after the Teflon is removed?

F)What are the electric field after the Teflon is removed?

Answer 1

d = 2.3-cm= 2.3 x 10^-2

Before we begin solving the parts, let's calculate the area of elect5roides= r^2= (d/2)^2 = 3.14 ( 2.3 x 10^-2/ 2)^2 = Area = 4.15 x 10^-4 m^2 , d =thickness= 0.20-mm = 0.20 x 10^-3

K = dielectric constant for Teflon =2

C = K εA/d =2 ( 8.854 x 10^-12 ) ( 4.15 x 10^-4 ) / ( 0.20 x 10^-3 )= 36.7 x10^-12 F

a) Q = CV = 36.7 x10^-12 x 9 = 330.3 x 10^-12 C

b) V = 9V

c)E = v/d= 9/  0.20 x 10^-3 = 45 x 10^3 V/m

d) C ( after teflon is removed) = εA/d = 18.35 x 10^-12 F

Q = CV = 18.35 x 10^-12 F x 9 =165. 15 C

e_)V = 9v

f) E =  45 x 10^3 V/m