Question

**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?

Answer #1

57.Capacitance is given by

C=e_{o}A/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=e_{o}A/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)CV^{2}=(1/2)*1.5*12^{2}=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*V^{2}

106=(1/2)*1.5*V^{2}

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

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