capacitance - smu physics...represented by a single capacitance. the rules, based on conservation of...

Physics 1308: General Physics II - Professor Jodi Cooley

Micheal Faraday 1868 - 1953

Capacitance

Welcome Back to Physics 1308

by Thomas Phillips oil on canvas


Announcements• Assignments for Thursday, September 27th:

- Reading: Chapter 26.1 - 26.5

- Watch Video: https://youtu.be/tVwyfhgDVgQ — Lecture 10 - Electric Current and Resistance

• Homework 6 Assigned - due before class on Tuesday, October 2nd.

• Midterm Exam 1 is graded and will be returned as soon as all make-up exams are taken.

• A new department policy instituted by the department chair forbids the distribution of student work in the physics office. As such, your work will now be available for pick up during office hours and for 15 minutes immediately after class from Jasmine in the main lob of FOSC.


Review Key Concepts

Capacitance:A capacitor consists of two isolated conductors (the plates) with charges +q and -q. Its capacitance C is defined from

voltage difference ΔV between the plates


Review Key Concepts A simple type of capacitor is the “parallel plate capacitor”, where the capacitance can be calculated exactly:

Space between plates can be:

• “empty” - filled only with vacuum, containing no matter at all, in which case κ = 1

• filled with a material like plastic or glass, in which case κ > 1


Key ConceptsCapacitors in Parallel:

q1 = C1V q2 = C2V q3 = C3V

The total charge in parallel is then

q = q1 + q2 + q3 = (C1 + C2 + C3)V

The equivalent capacitance is thenCeq =

q

V= C1 + C2 + C3


Key Concepts Capacitors in Series:

V1 =q

C1V2 =

q

C2V3 =

q

C3

The total potential due to the battery is then

V = V1 + V2 + V3 = q

✓1

C1+

1

C 2+

1

C 3

◆

The equivalent capacitance is thenC =

1

V=

1

1/C1 + 1/C2 + 1/C3


Key ConceptsElectric Potential Energy of a Charged Capacitor:

U =q2

2C=

1

2CV 2

is equal to the work required to charge the capacitor. This energy can be associated with the capacitor’s electric field E.

Every electric field, in a capacitor or from any other source, has an associated stored energy. In vacuum, the energy density u (potential energy per unit volume) in a field of magnitude E is

u =1

2✏0E

2


Question 1Consider two capacitors, each having plate separation d. In each case, a slab of thickness d/3 is inserted between the plates. In case (a), the slab is not connected to either plate. In case (b), it is connected to the upper plate. The capacitance is higher for

A) case (a) B) case (b) C) The two capacitances are equal

The system in case (a) is equivalent to two capacitor in series, each with plate separation d/3. The system in case (b) is equivalent to a single capacitor with plate separation d/3. Adding the capacitances in case (a) and using the fact capacitance varies inversely with the plate separation, we find the capacitance is larger in case (b).

C = ✏0A

d1

C12=

1

C1+

1

C2�! C12 =

C1C2C1 + C2


Instructor Problem: A Network of Capacitors

The capacitors shown above have capacitances given by C1, C2, C3. C1 = 12 µF and 2C2 = 2C3 = C1.

A) What is the equivalent capacitance of the network? B) What is the charge stored on C2?


Part A:

Need to combine capacitors into equivalent capacitances, until all 3 capacitors can be represented by a single capacitance. The rules, based on conservation of charge and conservation of energy, are:

1. Capacitors in series: They have the same magnitude charge stored on all their plates, but their individual electric potential di ︎fferences may vary

1

Ctotal=

1

C1+

1

C2+ ...

2. Capacitors in parallel: They have the same electric potential di ︎fference across their plates, but their individual charge magnitudes on each plate may vary

Ctotal = C1 + C2 + ...

Find: CtotalGiven: C1 = 12 µF

2C2 = 2C3 = C1


From the diagram, it is clear that C2 and C3 are in parallel.

C23 = C2 + C3

=1

2C1 +

1

2C1

C23 = C1 …(1)

Now we are left with two capacitors: C23 and C1. We can redraw our simplified diagram.

It becomes obvious that the capacitors are in series.

=C212C1

=1

2C1 �! C123 = 6 µF

1

C123=

1

C1+

1

C23

=C1C23

C1 + C23C123


Part B:

Find: Charge (Q) stored on capacitor 3Given: C1 = 12 µF

2C2 = 2C3 = C1

Work backward from the total capacitance and its associated charge and voltage, down into the network, deconstructing the network as we go. The capacitance, voltage and charge are related to each other through the following equation.

Q = CV

V123 = 12V

Start at the beginning. We know the voltage of the entire network.

V123 = 12V

The total charge on the network is then

Q123 = C123V123 = 6 µF · 12V = 72 µC


Now unpack C1 and C23.

They are in series. So, by charge conservation, they carry the same charge magnitude on their plates.

Q1 = Q23 = Q123 = 72 µC

Thus,Q1 = C1V1 �! V1 =

Q1C1

=72 µC

12 µF= 6 V

Q23 = C23V23 = 6 V

Now we need to unpack C23 into capacitors 2 and 3.They are in series. So, they must have identical potential differences. By conservation of energy

V23 = V2 = V3 = 6 V

Thus,Q3 = C3V3 = 6 µF · 6 V = 36 µC

By similar arguments, we would find the same charge on C2. Together C2 and C3 sum to 72 µF — which is what we expect from conservation of charge.

�! V23 =Q23C23


Student Problem: Moar Networking!

The capacitors shown above have capacitances given by C1 = 2C2 = 2C3 = 12 µF. What is the equivalent capacitance of the network? What is the voltage across C2? What is the energy stored on C3?


Notice that capacitors 2 and 3 are in series and those two are combined in parallel with capacitor 1.

Part A:

Start by simplifying the circuit by combining C2 and C3

C23 =C2C3

C2 + C3=

(1/4)C21C1

= (1/4)C1 = 3 µF

We can then combine this in parallel with C1 to get the total capacitance.

C123 = C1 + C23 = 12 µF + 3 µF

C123 = 15 µF


Part B:

To get the voltage across C2 we need to find the properties of the total equivalent capacitor and work backwards, unpacking the network.

The voltage across the whole network V123 = 12 V. Thus,

Q123 = C123V123 = 15 µF · 12 V = 180 µC

We unpack C123 into C1 and C23, producing two capacitors in parallel with each other. Thus,

V1 = V23 = V123 = 12 V

From this we findQ1 = C1V1 = 12 µF · 12 V = 144 µC

Q23 = C23V23= 3 µF · 12 V = 36 µC

The sum of these is 180 µC — consistent with conservation of charge. So, we haven’t made a mistake yet!


Finally, unpack C23 to the series arrangement of C2 and C3. These are capacitors in series, so they have the same charge stored on their plates.

Q23 = Q2 = Q3 = 36 µC

=36 µC

6 µF

=36 µC

6 µF= 6 V

Sum of voltages equals 12 V as expected from conservations of energy!

V2 =Q2C2

Thus, we have= 6 V

V3 =Q3C3

As a check…


Part C:

The energy stored on C3 is given by

U3 =1

2C3V

23

=1

2(6 µF )(6 V )2

U3 = 108 µJ


The End for Today!

capacitance - smu physics...represented by a single capacitance. the rules, based on conservation of...

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