Question

# 16.3 Capacitance 57.     MC A capacitor is first connected to a 6.0-V battery and then disconnected and...

16.3 Capacitance

57.     MC A capacitor is first connected to a 6.0-V battery and then disconnected and connected to a 12.0-V battery. How does its capacitance change: (a) It increases, (b) it decreases, or (c) it stays the same?

58.     MC A capacitor is first connected to a 6.0-V battery and then disconnected and connected to a 12.0-V battery. How does the charge on one of its plates change: (a) It increases, (b) it decreases, or (c) it stays the same?

59.     MC A capacitor is first connected to a 6.0-V battery and then disconnected and connected to a 12.0-V battery. By how much does the electric field strength between its plates change: (a) two times, (b) four times, or (c) it stays the same.

60.     MC A capacitor has the distance between its plates cut in half. By what factor does its capacitance change: (a) It is cut in half, (b) it is reduced to one fourth its original value, (c) it is doubled, or (d) it is quadrupled?

61.     MC A capacitor has the area of its plates reduced. How would you adjust the distance between those plates to keep the capacitance constant: (a) increase it, (b) decrease it, or (c) changing the distance cannot ever make up for the plate area change?

65.     l How much charge flows through a 12-V battery when a capacitor is connected across its terminals?

66.      l A parallel-plate capacitor has a plate area of and a plate separation of 2.0 mm. What is its capacitance?

67.      l What plate separation is required for a parallel-plate capacitor to have a capacitance of if the plate area is

69.     ll A 12-V battery is connected to a parallel-plate capacitor with a plate area of and a plate separation of 5.0 mm. (a) What is the charge on the capacitor? (b) How much energy is stored in the capacitor?

72.     lll A 1.50-F capacitor is connected to a 12.0-V battery for a long time, and then is disconnected. The capacitor briefly runs a 1.00-W toy motor for 2.00 s. After this time, (a) by how much has the energy stored in the capacitor decreased? (b) What is the voltage across the plates? (c) How much charge is stored on the capacitor? (d) How much longer could the capacitor run the motor, assuming the motor ran at full power until the end?

57.Capacitance is given by

C=eoA/d

so it doesn't depend on Voltage ,so

It stays same

58.

Increases

Charge

Q=CV

So ,if Voltage is increased it Increases.

59.

Two times

Electric field

E=V/d

Electric field is directly proportional to Potential difference ,so increasing Potential difference 2 times,will result in 2 times electric field

60.

It is doubled

Capacitance

C=eoA/d

Capacitance is inversly proportional to distance ,so reducing the distance by 1/2 times will result 2 time increase in capacitance

61.

Decrease it

65 ,66,67,69 data was incomplete.

72.

Energy stored in Capacitor

E=(1/2)CV2=(1/2)*1.5*122=108 J

Energy Used by capacitor

E=P*t=1*2=2 J

So energy remaining in the capacitor

E=108-2=106 J

b)

E=(1/2)*C*V2

106=(1/2)*1.5*V2

V=11.888 Volts

c)

Charge on capacitor is

Q=CV=11.888*1.5

Q=17.832 C

d)

time =108*1

t=108 s

#### Earn Coins

Coins can be redeemed for fabulous gifts.