capacitance - smu physics · 2018-09-18 · physics 1308: general physics ii - professor jodi...

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 Tuesday, September 25th:

- Reading: Chapter 25.3 - 25.5

- Watch Video: https://youtu.be/Am4tuzpRjLw — Lecture 9 - Using Capacitors

• Homework 5 Assigned - due before class on Tuesday, September 25th.

• Midterm Exam 1 will be in class on Thursday, September 20th.


Review Question 1Two electrons are separated by a distance R. If the distance between the charges is increased to 2R, what happens to the total electric potential energy of the system?

A) The total electric potential energy of the system would increase to four times its initial value.

B) The total electric potential energy of the system would increase to two times its initial value.

C) The total electric potential energy of the system would remain the same. D) The total electric potential energy of the system would decrease to one half

its initial value. E) The total electric potential energy of the system would decrease to one

fourth its initial value.


Key Concepts Review

• Electric charges exert forces on one another via the electric field, which is always present whether or not a charge is there to by acted upon. The electric field is a conservative force field.

• The electric field comes along with an associated energy per unit charge, its “electric potential.” By understanding how charges move in an electric potential, we can understand how energy is stored in the field.


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


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


Capacitor Demonstrations


Question 1In calculating both the electric field and the capacitance of two closely spaced conducting plates, it is frequently assumed that the area of the plates is somewhat larger than the distance between the plates. Why is this assumption made?

A)The capacitance is too small to calculate if the plates are too far apart.

B) The electric field near the edges of the plates is not uniform.

C) The charge would otherwise be too small to generate a significant electric field.

D)Coulomb’s law would not otherwise apply. E) Gauss’ law would not otherwise apply.


Instructor Problem: Defibrillator RevisitedA portable defibrillator has to store a large charge. The charge can be released across the heart to restart it.

The device has to store 145 µC in its capacitor, when exposed to a potential difference of ΔV = 2.3 kV.

a)What is the capacitance required to achieve this?

b)If we assume that the capacitor is a parallel-plate capacitor and has plates that are 50.0 cm by 10.0 cm in area, what separation is needed if the gap is empty space? Is this feasible? Discuss.


a) First, take account of what is known and what is asked for in the problem.

Given:�V = 2.3 kV = 2.3⇥ 103 V

Q = 145 µC = 145⇥ 10�6 C

= 1.45⇥ 10�4 C

Find:

“capacitance”

What Capacitance is needed?

We know that for every capacitor, the capacitor equation relates Q and ΔV.

This equation gives us the units of capacitance.

=C

V

capacitance

q = C�V

C =q

�V“Coulombs per Volt” is a “Farad” named after Micheal Faraday (F).

= F


Apply to our case:

C =q

�V

C = 6.3⇥ 10�8 F = 63 nF

b) Now assume this is a parallel-plate capacitor filled with empty space. Now, our capacitance equation becomes

C = ✏0A

d

permittivity of free spacecapacitance per meter of the vacuum✏0 = 8.85⇥ 10�12 C2/N ·m2

Aside:N ·m = J = C · 1

(J/C)· 1

m=

C

V· 1

m=

F

mJ

C= V,

C2

N ·m2= C · 1

(N ·m/C)· 1

m

=1.43⇥ 10�4 C

2.3⇥ 103 V


C = ✏0A

dTake account of our knowns

In part a) we found the capacitance — so we have that too.

Given:l = 50.0 cm = 0.500 mw = 10.0 cm = 0.100 m

Find:separation gap = d

Start with out equation for a parallel plate capacitor and solve.

C = ✏0A

d

d =✏0A

C=

✏0lw

C

=(8.85⇥ 10�12 C2/N ·m2)(0.500 m)(0.100 m)

63⇥ 10�9 F

d = 7.0⇥ 10�6 m = 7.0 um


Does 7 µm seem practical? Discuss.

What is the width of a human hair?

Ans: typically 100 µm

What is the diameter of a human cell?

Ans: typically 10 µm

Is it possible to engineer a capacitor to this tolerance?

Ans: probably not

How do we improve out ability to build a capacitor like this?

increase A, increase ε0 (charge properties of space between plates) put a material in the gap (a “dielectric”), a nonconductive material with dipole properties


The dipole weakens the E inside the capacitor and allows more Q to be placed on the capacitor for the same external ΔV.

Q = CV = ✏0A

dV

Keep V, A constant —> increase ε0 and increase d to hold Q constant.

✏0 �! ✏ = ✏0

e.g. put some ceramic inside the capacitor with κ = 3000 - 6000,

If κ = 4500, then

d = ✏0A

C= 0.03 m = 3 cm

Thats better! We can tune κ to d.


Cell MembraneLipid bi-layer is CLEARLY not a vacuum (empty space).

Rather than ϵ0, one has to employ a modified ϵ that takes into account the presence of matter:

ϵ = κϵ0

where κ is the dielectric constant. For a cell wall, this is about 3.0.


Student Problem: Cell Membrane ThicknessThe membrane thickness of the biological cell was first measured using its capacitance! An experiment determines that the electric potential difference inside and outside the cell is 70 mV. It also finds that to maintain this potential, the cell has to move 0.17 pC of Sodium from inside to outside.

a) What is the capacitance of the cell membrane? Cells can be seen with microscopes and have a radius of about 5.0µm.

b) What is the thickness of the cell membrane? HINT: treat the cell as a parallel-plate capacitor whose area is that of a sphere.


First, take account of what is known and what is asked for in the problem.

Given:

�V = 70 mV = 7.0⇥ 10�2 V

q = 0.17 pC = 0.17⇥ 10�12 C

a) Find the capacitance of the cell membrane.

C =q

�V =1.7⇥ 10�13 C

7.0⇥ 10�2 V

C = 2.4 pF


b) If the cells can be treated as spheres, each has a radius of rc = 5.0 µm.

Treat the cell membrane as a parallel-plate capacitor of area A. Then, we have

A = 4⇡r2

We also know that we have to employ our knowledge of dielectrics in this case.

✏ = ✏0

Putting it all together:

d = ✏0A

C= ✏0

4⇡r2

C= (8.85⇥ 10�12 C2/N ·m2)(3.0)

4⇡(5.0⇥ 10�6 m)2

(2.4⇥ 10�12 F )

d = 3.4 nm


Hugo FrickeDuring his productive career in research, combining physics and biology, led to a much deeper understanding of many s u b j e c t s , i n c l u d i n g l a y i n g t h e groundwork for understanding human tolerance to extreme radiation doses (critical for developing radiation therapies), the first indirect measurement of the thickness of the animal cell membrane (by studying electrical properties of aqueous systems), and the effect of chemical agents like anesthetics and stimulants on the human brain.

Hugo Fricke 1892-1972


References• Fricke, Hugo (1925). "The electrical capacity of suspensions with special reference to blood".

Journal of General Physiology 9 (2): 137–52.

• Ohki S. Dielectric constant and refractive index of lipid bilayers. J Theor Biol. 1968 Apr;19(1):97–115.

• Sjöstrand FS, Andersson-Cedergren E, Dewey MM (April 1958). "The ultrastructure of the intercalated discs of frog, mouse and guinea pig cardiac muscle". J. Ultrastruct. Res. 1 (3): 271–87.

• Robertson JD (1960). "The molecular structure and contact relationships of cell membranes". Prog. Biophys. Mol. Biol. 10: 343–418.

• Robertson JD (1959). "The ultrastructure of cell membranes and their derivatives". Biochem. Soc. Symp. 16: 3–43.


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capacitance - smu physics · 2018-09-18 · physics 1308: general physics ii - professor jodi...

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