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?
b) What is the potential difference before the Teflon is removed?
c) What is the electric field before the Teflon is removed?
Note: The dielectric constant of Teflon is 2.0.
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?
2 Answers
-
a)
C = εA/d
Calculate the area in meters:
A = π*r^2 = π*(.023/2)^2 = 4.15*10^-6 m^2
Calculate the plate separation distance in meters:
d = .20/1000 = 200*10^-6 m
Put it all together:
C = 2.0*8.854*10^-12*4.15*10^-6/200*10^-6 = 36.8*10^-12 = 36.8 pF
Q = C*V = 36.8*10^-12 * 9 = 331*10^-12 C
b)
The potential difference is the battery voltage, 9V
c)
The electric field is the voltage divided by the separation distance:
E = V/d = 9/200*10^-6 = 45*10^3 V/m
-
Teflon Is
Source(s): https://shrinkurl.im/baAzq