Two 2.1-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. Required:
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?
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?

a. Q = 1881.73 x C

b. As battery is not removed so, potential difference will remain same.

c. E = 21.42 x V/m

d. Q = 895.5 x C

e. Again the potential difference will not change it will remain same as 9 V

f. E = 45 x V/m

Explanation:

Solution:

Here, Teflon is used so, the dielectric constant of the Teflon K = 2.1

Diameter = 2.1 cm

Radius = 2.1/2 cm

Radius = 1.05 cm

Radius = 0.015 m

Now, we need to find the area of each plate:

A =

A = (3.14) ()

A = 0.000225

A = 2.25 x

We are given the thickness of the plate which equal to the distance between the two plates.

d = 0.20 mm = 0.2 x m

d = 0.2 x  m = distance between two plates.

Hence, the capacitance of the dielectric without the dielectric

C =

Putting up the values we get,

E = 8.85 x

C =    

C = 99.5

If dielectric is included then,

= K C

= (2.1) ( 99.5 x )

= 209.08 x F

As we know the voltage of the battery V = 9V So,

a) Charge before the Teflon is removed:

Q = CV

Q = V

Q = (209.08 x F) (9V)

Q = 1881.73 x C

b) Potential Difference before the Teflon is removed = ?

As battery is not removed so, potential difference will remain same.

c) Electric Field =?

As we know,

E = V/(K.d)

E = 9V/(2.1 x 0.2 x )

E = 21.42 x V/m

d) After the Teflon is removed

Q = CV

Q = (99.5 ) ( 9)

Q = 895.5 x C

e) Again the potential difference will not change it will remain same as 9 V

f) Electric Field = ?

E =  (Teflon is removed)

E = 9/0.2 x

E = 45 x V/m

did yall get answers for this yet