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? Q = 5.6×1010 C SubmitMy AnswersGive Up Answer Requested Part B What is the potential difference before the Teflon is removed? V = 9.0 V SubmitMy AnswersGive Up Correct Part C What is the electric field before the Teflon is removed? E = 4.5×104 V/m SubmitMy AnswersGive Up Answer Requested Part D What is the charge after the Teflon is removed? Q = 2.8×1010 C SubmitMy AnswersGive Up Answer Requested Part E What is the potential difference after the Teflon is removed? V = 9.0 V SubmitMy AnswersGive Up Correct Part F What are the electric field after the Teflon is removed? E = 4.5×104 V/m SubmitMy AnswersGive Up Answer Requested

Explanation / Answer

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? Q = 5.6×1010 C

Part B What is the potential difference before the Teflon is removed? V = 9.0 V

Part C What is the electric field before the Teflon is removed? E = 4.5×104 V/m

Part D What is the charge after the Teflon is removed? Q = 2.8×1010 C

here,

radius of electrodes, r = 3/2 = 1.5 cm = 0.015 m
distance between the electrodes, d = 0.20mm = 0.2*10^-3 m

Area of Electrode,A = pi*r^2 = pi * 0.015^2 = 0.00070685 m^2

Capacitance of Electodes, C = eo*A/d = (8.85*10^-12 * 0.00070685)/(0.2*10^-3)
C = 3.128*10^-11 F

Part A:
Charge on Electrode before teflon is removed, Q = K*C*V ( k is dielecteric ocnstant of teflon)
Q = 2.1 * 3.128*10^-11 * 9
Q = 5.912*10^-10 C

Part b:
The potential difference is the battery voltage, 9V

Part C:
Electric field is given as:, E = V/Q   
E = 9 / (5.912*10^-10)
E = 1.522*10^10 V/m or N/C

Part D:
Now as teflon is removed there will no dielectric present, so
Q = C*V
Q = 3.128*10^-11 * 9
Q = 2.815*10^-10 C

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