Two 2.5-cm-diameter electrodes with a 0.10-mm-thick sheet of Teflon between them

Question

Two 2.5-cm-diameter electrodes with a 0.10-mm-thick sheet of Teflon between them are attached to a 9.0 V battery. Without disconnecting the battery, the Teflon is removed.

Part A

What is the charge before the Teflon is removed?

Part B

What is the potential difference before the Teflon is removed?

Express your answer using two significant figures.

Part C

What is the electric field before the Teflon is removed?

Express your answer using two significant figures.

Part D

What is the charge after the Teflon is removed?

Express your answer using two significant figures.

Part E

What is the potential difference after the Teflon is removed?

Express your answer using two significant figures.

Part F

What are the electric field after the Teflon is removed?

Express your answer using two significant figures.

Explanation / Answer

Part A)

Q = CV and C = keA/d

k for Teflon, in by book, is 2.1 (If your book uses a different value, these answers will be off)

Q = (2.1)(8.85 X 10-12)(pi)(.0125)2(9)/(1 X 10-4)

Q = 8.2 X 10-10 C

Part B)

V = 9.0 Volts

Part C)

V = Ed

9 = E(1 X 10-4)

E = 9.0 X 104 V/m

Part D)

The original Q will be divided by 2.1

Q = 3.9 X 10-10 C

Part E)

V remains 9.0 Volts

Part F)

E remains 9.0 X 104 V/m

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