1) A parallel-plate capacitor is charged by a 8.00Vbattery, then the battery is

Question

1) A parallel-plate capacitor is charged by a 8.00Vbattery, then the battery is removed.

What is the potential difference between the plates after the battery is disconnected?

What is the potential difference between the plates after a sheet of Teflon is inserted between them?

2)A capacitor charging circuit consists of a battery, an uncharged 20 ?F capacitor, and a 5.0k? resistor. At t = 0 s the switch is closed; 0.15 s later, the current is 0.46mA.

What is the battery's emf?

3)The capacitor in (Figure 1) is initially uncharged and the switch, in position c, is not connected to either side of the circuit. The switch is now flipped to position a for 15ms ,then to position b for 15ms , and then brought back to position c. Suppose that E = 9V . What is the final potential difference across the capacitor?

1) A parallel-plate capacitor is charged by a 8.00Vbattery, then the battery is removed. What is the potential difference between the plates after the battery is disconnected? What is the potential difference between the plates after a sheet of Teflon is inserted between them? 2)A capacitor charging circuit consists of a battery, an uncharged 20 ?F capacitor, and a 5.0k? resistor. At t = 0 s the switch is closed; 0.15 s later, the current is 0.46mA. What is the battery's emf? 3)The capacitor in (Figure 1) is initially uncharged and the switch, in position c, is not connected to either side of the circuit. The switch is now flipped to position a for 15ms ,then to position b for 15ms , and then brought back to position c. Suppose that E = 9V . What is the final potential difference across the capacitor?

Explanation / Answer

(A)

Parallel plate cpacitor is conneted with 8 V battery . The capcitor will get full charged and gain voltage equal to 8V.

so when the battery is disconnected the potential difference across the parallel plate capacitor is 8V. (ANS)

We have relation Q=CV

where Q is charge stored at capacitor , C is capacitance of capacitor and V is voltage gained .

so

V=Q/C

also when Teflon is inserted between the parallel plate capacitor the voltage decreases , given by

V=Q/(KC)

where K is permitivity . Value of K for Teflon is =2.1 F/m

V=8/2.1

V=3.81 V (ANS)

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