Part A Two 5 0mm × 5 0 mm electrodes are held 0.10 mm apart and are attached to

Question

Part A Two 5 0mm × 5 0 mm electrodes are held 0.10 mm apart and are attached to 8.5 V battery Without disconnecting the battery, a 0.10-mm-thick sheet of Mylar is inserted between the electrodes. What is the capacitor's potential difference before the Mylar is inserted? Express your answer with the appropriate units. | %-1 Value Units Submit My Answers Give Un Part B What is the capacitor's electric field before the Mylar is inserted? Express your answer with the appropriate units. E-Value Units Submit My Answers Give Up Part C What is the capacitor's charge before the Mylar is inserted? Express your answer with the appropriate units. Q-Value Units Submit My Answers Give Un Part D What is the capacitor's potential difference after the Mylar is inserted? Express your answer with the appropriate units.

Explanation / Answer

part A:

potential difference =potential of the battery=8.5 volts

part B:

electric field=potential difference/distance

=8.5 volts/0.1 mm

=85000 V/m

part C:

initial capacitance=epsilon*area/distance

=8.85*10^(-12)*0.005*0.005/(0.1*0.001)

=2.2125*10^(-12) F

then initial charge before Mylar is inserted

=capacitance*potential

=2.2125*10^(-12)*8.5

=1.8806*10^(-11) C

part D:
as the battery is still connected,

potential difference is still 8.5 volts

part E:

electric field=potential difference/distance

=8.5 volts/0.1 mm

=85000 V/m

part F:

dielectric constant of Mylar is 3.1

new capacitance after mylar is inserted =dielectric constant*capacitance before Mylar is inserted

=3.1*2.2125*10^(-12)

=6.85875*10^(-12) F

then new charge=capacitance*voltage

=5.83*10^(-11) C

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