Two 2.7-cm-diameter electrodes with a 0.20-mm-thick sheet of Teflon between them

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.

Part A

What is the charge before the Teflon is removed?

Part B

What is the potential difference before the Teflon is removed?

Part C

What is the electric field before the Teflon is removed?

Part D

What is the charge after the Teflon is removed?

Part E

What is the potential difference after the Teflon is removed?

Part F

What are the electric field after the Teflon is removed?

Explanation / Answer

a) C = 0A/d = 8.85*10^-12*3.14*0.0135^2/2*10^-4 = 25.32 pF

q = cv = 227.9 *10^-12 C

b) v = 9 V

c) E = V/d =  45*10^3 V/m

d) c' = kc = 2*25.32*10^-12 = 50.64 pF

q' = 2* 227.9 *10^-12 C = 455.8 *10^-12 C

e) v = 9V

f) E' = E/k = 22.5*10^3 V/m

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