Two 3.0-cm-diameter electrodes with a 0.20-mm-thick sheet of Teflon between them

Question

Two 3.0-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

area of electrode is,

A = pi*3^2 / 4 = 7.065*10^(-4) m^2

k for teflon = 2.1

capacitance of capacitor without dielecric is,

C0 = e0*A / d = 8.85*10^(-12)*7.065*10^(-4) / 0.2*10^(-3)

C0 = 31.26*10^(-13) F

capacitance of capacitor with dielecric is

C = 2.1*31.26*10^(-13) = 65.65*10^(-13) F

(a)

q = C*V

q = 65.65*10^(-13) * 9 = 59.08*10^(-12) C

(b)

potential difference before the Teflon is removed V = 9 V

(c)

E = V / k*d

E = 9 / 2.1*0.2*10^(-3) = 2.14*10^4 N/C

(d)

charge after the Teflon is removed is

q = C0*V = 31.26*10^(-13)*9

q = 2.81*10^(-11) C

(e)

potential difference remains same .

V = 9 V

(f)

E = V / d

E = 9 / 0.2*10^(-3) = 4.5*10^4 N/C

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