Capacitors! Physics 102: Lecture 4, Slide 1 SPH4UW

Capacitors!

Physics 102: Lecture 4, Slide 1

SPH4UW

Capacitance: The ability to store separated charge C=Q/V

A B

E=E0

d

V = VA – VB = +E0 d

+

+

+

+

+

-

-

-

-

-

A B

d

+

+

+

+

+

-

-

-

-

-

+

+

+

+

+

-

-

-

-

-

E=2E0

V = VA – VB =+2E0d

Potential difference is proportional to charge: Double Q Double V

Q = CV

Charge Q on plates Charge 2Q on plates

Capacitor

Any pair conductors separated by a small distance. (e.g. two metal plates)

Capacitor stores separated charge Positive Q on one conductor, negative Q on

otherNet charge is zero

E

d

+

+

+

+

+

-

-

-

-

-

Q=CV

U =(½) Q V • Stores Energy

Capacitance Practice

How much charge is on a 0.9 F capacitor which has a potential difference of 200 Volts?

Q = CV

How much energy is stored in this capacitor?

U = ½ Q V

= (0.9)(200) = 180 Coulombs

= ½ (180) (200) = 18,000 Joules!

Capacitance of Parallel Plate Capacitor

V = Ed AND E = Q/(e0A) (Between two large plates)

So: V =

Recall: CQ/V So: (For parallel plate capacitor)

A

d

A

E+ -

V

e0=1/(4pk)=8.85x10-12 C2/Nm2

C = e0A/d

Qd/(e0A)

Parallel Plate CapacitorCalculate the capacitance of a parallel plate capacitor made from two large square metal sheets 1.3 m on a side, separated by 0.1 m.

A

d

A

101.5 10 F

12(8.85 10 )(1.3)(1.3)

(0.1)C

C = e0A/d

DielectricPlacing a dielectric

between the plates increases the capacitance.

Capacitance with dielectric

Dielectric constant (k > 1)

Capacitance without dielectric

C = k C0 The dielectric allows the potential difference between the plates to obtain a higher value and thus store more energy, therefore the capacitance is greater

A parallel plate capacitor is given a charge q. The plates are then pulled a small distance further apart. What happens to the charge q on each plate of the capacitor?

Understanding: Parallel Plates

1) Increases 2) Constant 3) Decreases

Remember charge is real/physical. There is no place for the charges to go.

+q-q+

+

+

+

-

-

-

-

d

pullpull

A parallel plate capacitor is given a charge q. The plates are then pulled a small distance further apart. Which of the following apply to the situation after the plates have been moved?

1)The capacitance increases True False

2)The electric field increases True False

3)The voltage between the plates increases True False

4)The energy stored in the capacitor increases True False

Understanding

C = e0A/d C decreases!

E= Q/(e0A) Constant

V= Ed

U = ½QV

+q-q+

+

+

+

-

-

-

-

d

pullpull

Two identical parallel plate capacitors are shown in end-view in A) of the figure. Each has a capacitance of C.

A) B)

If the two are joined as in (B) of the figure, forming a single capacitor, what is the final capacitance?

1) 2C

2) C

3) C/2

Understanding

0ACd

Voltage in Circuits

Elements are connected by wires. Any connected region of wire has the same

potential.

Vwire 1= 0 V Vwire 2= 5 V Vwire 3= 12 V Vwire 4= 15 V

VC1= 5 V VC3= 3 VVC2= 7 VC1 C2 C3

• The potential difference across an element is the element’s “voltage.”

Capacitors in ParallelBoth ends connected together by wire

C1 C2 Ceq

• Share Charge: Qeq = Q1 + Q2

• Total Cap: Ceq = (Q1 + Q2)/V = C1 + C2

= Veq • Same voltage: V1 = V2

15 V

10 V

15 V

10 V

15 V

10 V

Parallel PracticeA 4 m F capacitor and 6 m F capacitor are connected in parallel and charged to 5 volts. Calculate Ceq, and the charge on each capacitor.


C4 C6 Ceq

5 V

0 V

5 V

0 V

5 V

0 V

Ceq = C4+C6

Q4 = C4 V4

Q6 = C6 V6

Qeq = Ceq Veq

= 4 mF+6 mF = 10 mF

= (4 mF)(5 V) = 20 mC

= (6 mF)(5 V) = 30 mC

= (10 mF)(5 V) = 50 mC = Q4+ Q6

V = 5 V

Capacitors in SeriesConnected end-to-end with NO other exitsSame Charge: Q1 = Q2 = Qeq


Ceq

C1

C2

+

-

+Q

-Q

+Q

-Q

+Q

-Q

1 2

1 1 1

eqC C C

• Share Voltage: V1+V2=Veq

+ ++ ++

++ +++

+

+

-- - - -

- - - -

Series PracticeA 4 mF capacitor and 6 mF capacitor are connected in series and charged to 5 volts. Calculate Ceq, and the charge on the 4 mF

capacitor.


Ceq

C4

C6

+

-

+

-

+

-

+Q

-Q

+Q

-Q

+Q

-Q

Q = CV

11 1

2.44 6

FF F

mm m

4 6 eqQ Q Q eqC V

1

eq4 6

1 1C

C C

2.4 5 12F V Cm m

5 V

0 V

5 V

0 V


Comparison:Series vs. Parallel

Series• Can follow a wire from one

element to the other with no branches in between.

Parallel• Can find a loop of wire

containing both elements but no others (may have branches).

C1

C2

C2C1

Comparison:Capacitors vs. Resistors

Capacitors store energy as separated charge: U=1/2QV Capacitance: ability to store separated charge:

C = ke0A/d Voltage determines charge: V=Q/C

Resistors dissipate energy as power: P=VI Resistance: how difficult it is for charges to get through:

R = r L /A Voltage determines current: V=IR

Don’t mix capacitor and resistor equations!

Electromotive Force

BatteryMaintains potential difference V Not constant powerNot constant currentDoes NOT produce or supply charges, just

“pushes” them.

+

-

Understanding A circuit consists of three initially uncharged capacitors C1, C2, and C3, which are then connected to a battery of emf E. The capacitors obtain charges q1, q2,q3, and have voltages across their plates V1, V2, and V3. Which if any of the following are true?

C2

C3C1E +

-

-q2

+q1

-q1

+q3 -q3

+q2

V1

V2

V3

1) q1 = q2

2) q2 = q3

3) V2 = V3

4) E = V1

5) V1 < V2

6) Ceq > C1

C2

C3C1E +

-

-q2

+q1

-q1

+q3 -q3

+q2

V1

V2

V3

1) q1 = q2 Not necessarily C1 and C2 are NOT in series.

2) q2 = q3 Yes! C2 and C3 are in series.


C2

C3C1E +

-

-q2

+q1

-q1

+q3 -q3

+q2

V1

V2

V3

3) V2 = V3 Not necessarily, only if C2 = C3

4) E = V1

10V

0V

7V??

Yes! Both ends are connected by wires


C2

C3C1E +

-

-q2

+q1

-q1

+q3 -q3

+q2

V1

V2

V3

5) V1 < V2 Nope, V1 > V2. (E.g. V1 = 10-0, V2 =10-7

6) Ceq > C1

10V

0V

7V??

Yes! C1 is in parallel with C23 (Ceq = C1 + C23)


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Capacitors! Physics 102: Lecture 4, Slide 1 SPH4UW