Two 2.5-cm-diameter electrodes with a 0.20-mm-thick sheet of Teflon between them

Question

Two 2.5-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?
Express your answer using two significant figures.
Q = ______C

Part B
What is the potential difference before the Teflon is removed?
Express your answer using two significant figures.
V = _______V

Part C
What is the electric field before the Teflon is removed?
Express your answer using two significant figures.
E = ________V/m

Part D
What is the charge after the Teflon is removed?
Express your answer using two significant figures.
Q = _______C

Part E
What is the potential difference after the Teflon is removed?
Express your answer using two significant figures.
V = ________V

Part F
What are the electric field after the Teflon is removed?
Express your answer using two significant figures.

Explanation / Answer

given diameter of electrodes d = 2.5cm = 0.025m    radius of electrodes r = d/2 = 0.025m/2 = 1.25* 10^-2m thick ness of teflon   t = 0.20* 10^-3m    voltage across the two plates v0 = 9.0v dielectric constant of teflon k = 2.1 A)   charge before teflon is removed        Q =k 0Av0/t           = (2.1)8.85 * 10^-12 C^2/N.m^2 ( 3.14) ( 1.25 * 10^-2m)^2 * 9.0v/ 0.20 * 10^-3m           = 410.3 * 10^-12C B) potential difference before teflon is removed is same because battery is not disconted         V =             = 9.0v C) intial electric field           V = E0t/k          E0 = Vk/t             = (2.1)9.0v/0.20 * 10^-3m = 94500V/m    electric field before teflon is removed E = E0/k = 94500V/m/2.1 = 45000V/m D) charge after teflon is removed Q= Cv0    = 0Av0/t    =8.85 * 10^-12 C^2/N.m^2 ( 3.14) ( 1.25 * 10^-2m)^2 * 9.0v/ 0.20 * 10^-3m    = 195.3 * 10^-12C E)potential difference afterteflon is removed is same because battery is not disconted F) electric field E = V/t                              = 9.0v/0.20 * 10^-3m                              =45000V/m    = 0Av0/t    =8.85 * 10^-12 C^2/N.m^2 ( 3.14) ( 1.25 * 10^-2m)^2 * 9.0v/ 0.20 * 10^-3m    = 195.3 * 10^-12C E)potential difference afterteflon is removed is same because battery is not disconted F) electric field E = V/t                              = 9.0v/0.20 * 10^-3m                              =45000V/m                   

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