1. Capacitance & Dielectrics
Chapter 24 opener. Capacitors come in a wide range of sizes and shapes, only a few of which are shown here. A capacitor is basically two conductors that do not touch, and which therefore can store charge of opposite sign on its two conductors. Capacitors are used in a wide variety of circuits, as we shall see in this and later Chapters.
Various
Capacitors

2. Capacitors - Definition
Capacitor  Any configuration of 2
conductors that are close but not touching.
A Capacitor has the ability to store electric
charge.
A “Parallel Plate”
Capacitor
Figure Capacitors: diagrams of (a) parallel plate, (b) cylindrical (rolled up parallel plate).

3. Makeup of a Capacitor A capacitor always consists of 2 conductors.
These are called plates.
When the capacitor is charged, the plates carry charges of equal magnitude & opposite sign.
A potential difference
exists between the plates
due to the charge.

4. Parallel Plate Capacitor
Each plate is connected to a
battery terminal.
The battery is a source of potential difference.
If the capacitor is initially
uncharged, the battery establishes
an electric field in the wires.
This field applies a force on
electrons in the wire just
outside of the plates.
The force causes the electrons
to move onto the negative plate.

5. Parallel Plate Capacitor, Continued
This process continues until
equilibrium is achieved.
The plate, the wire & the terminal
are then all at the same potential.
At this point, there is no field in the
wire & the movement of the
electrons stops.
The plate is now negatively charged.
A similar process occurs at the other plate,
electrons moving away from the plate &
leaving it positively charged. In its final
configuration, the potential difference
across the capacitor plates is the same as
between the battery terminals.

6. (a) Parallel-Plate Capacitor connected to a battery.
(b) A Capacitor in a circuit diagram.
Parallel Plate
Capacitor
Capacitor
in a circuit
Figure (a) Parallel-plate capacitor connected to a battery. (b) Same circuit shown using symbols.

7. Experiment Shows That 1 F = 1 C/V Capacitance C:
When a capacitor is connected to a
battery, the charge Q on its plates is
proportional to the battery voltage V, with
the proportionality constant equal to the
Capacitance C:
This is The Definition of Capacitance.
The SI unit of capacitance is the Farad (F)
1 F = 1 C/V

8. Definition of Capacitance
As we just said, the capacitance, C, of a capacitor is
the ratio of the magnitude of the charge on one plate to
the potential difference between the plates. As we also
said, the SI capacitance unit is
The Farad (F)
The farad is a large unit, typically you will see
microfarads (μF) & picofarads (pF).
Capacitance is:
Always a positive quantity.
Constant for a given capacitor.
A measure of the capacitor’s ability to store charge
The amount of charge the capacitor can store per unit
of potential difference.

9. Calculating Capacitance: General Procedure or “Recipe”
For a given geometry
1. Let the plates (A & B) have charges Q & -Q.
2. Calculate the electric field E between the two
plates using techniques of previous chapters
3. Calculate the potential difference V between
the plates using (parallel plate capacitor).
4. Use C = Q/V to calculate the capacitance.
Figure Parallel-plate capacitor, each of whose plates has area A. Fringing of the field is ignored.

10. Example: Parallel Plate Capacitor
Earlier Result: The magnitude of the electric
field E between 2 closely space charged
plates with charge Q & area A is
For plates of area A, the surface charge density
is: σ = (Q/A). So, E between the plates is:
E = Q/(ε0A). The relation between V & E:
Combining these gives the
potential difference
VB - VA = (Qd)/(ε0A).
Figure Parallel-plate capacitor, each of whose plates has area A. Fringing of the field is ignored.
So, the capacitance of a parallel plate capacitor is:

11. Example: Parallel Plate Capacitor
The charge density on the plates σ = (Q/A). A is
the area of each plate. Q is the charge on each
plate, equal with opposite signs.
The electric field is uniform between the
plates & zero elsewhere.
So, the capacitance is proportional to the area of
its plates & inversely proportional to the distance
between the plates.

12. Example: Calculate
(a) The capacitance C of a parallel-plate capacitor whose plates are 20 cm × 3.0 cm & are separated by a 1.0-mm air gap.
Solution: a. C = 53 pF.
b. Q = CV = 6.4 x C.
c. E = V/d = 1.2 x 104 V/m.
d. A = Cd/ε0 = 108 m2.

13. C = 53 pF Example: Calculate Answer
(a) The capacitance C of a parallel-plate capacitor whose plates are 20 cm × 3.0 cm & are separated by a 1.0-mm air gap.
Answer
Solution: a. C = 53 pF.
b. Q = CV = 6.4 x C.
c. E = V/d = 1.2 x 104 V/m.
d. A = Cd/ε0 = 108 m2.
C = 53 pF

14. Example: Calculate (b) The charge Q on each plate if a 12-V
battery is connected across the 2 plates.
Solution: a. C = 53 pF.
b. Q = CV = 6.4 x C.
c. E = V/d = 1.2 x 104 V/m.
d. A = Cd/ε0 = 108 m2.

15. Answer Q = CV = 6.4  10-10 C Example: Calculate
(b) The charge Q on each plate if a 12-V
battery is connected across the 2 plates.
Answer
Q = CV = 6.4  C
Solution: a. C = 53 pF.
b. Q = CV = 6.4 x C.
c. E = V/d = 1.2 x 104 V/m.
d. A = Cd/ε0 = 108 m2.

16. Example: Calculate (c) The electric field E between the plates.
Solution: a. C = 53 pF.
b. Q = CV = 6.4 x C.
c. E = V/d = 1.2 x 104 V/m.
d. A = Cd/ε0 = 108 m2.

17. Answer E = V/d = 1.2  104 V/m Example: Calculate
(c) The electric field E between the plates.
Answer
E = V/d = 1.2  104 V/m
Solution: a. C = 53 pF.
b. Q = CV = 6.4 x C.
c. E = V/d = 1.2 x 104 V/m.
d. A = Cd/ε0 = 108 m2.

18. Example: Calculate
(d) An estimate of the area A of the plates needed to achieve a capacitance of
C = 1 F, given the same air gap d.
Solution: a. C = 53 pF.
b. Q = CV = 6.4 x C.
c. E = V/d = 1.2 x 104 V/m.
d. A = Cd/ε0 = 108 m2.

19. A = (Cd/ε0) = 108 m2 !!! = 100 km2  36 square miles!!
Example: Calculate
(d) An estimate of the area A of the plates needed to achieve a capacitance of
C = 1 F, given the same air gap d.
Answer
A = (Cd/ε0) = 108 m2 !!!
= 100 km2
 36 square miles!!
Solution: a. C = 53 pF.
b. Q = CV = 6.4 x C.
c. E = V/d = 1.2 x 104 V/m.
d. A = Cd/ε0 = 108 m2.

20. Capacitors can now be made with capacitances of
C = 1 F or more, but these are NOT parallel-plate
capacitors. They are usually made from activated
carbon, which acts a capacitor on a small scale. The
capacitance of 0.1 g of activated carbon is about 1 F.
Some computer keyboards use capacitors; depressing the
key changes the capacitance, which is detected in a circuit
Figure Key on a computer keyboard. Pressing the key reduces the capacitor spacing thus increasing the capacitance which can be detected electronically.

21. Example: Cylindrical Capacitor
See figure. A Cylindrical Capacitor
consists of a cylinder (or wire) of radius
Rb surrounded by a coaxial cylindrical
shell of inner radius Ra.
Both cylinders have length ℓ, which is
assumed to be much greater than the
separation of the cylinders, so “end
effects” can be neglected.
The capacitor is charged (by connecting it to a
battery) so that one cylinder has charge +Q
(say, the inner one) the other one charge -Q.
Formula for the capacitance C (using calculus) is:
Figure (a) Cylindrical capacitor consists of two coaxial cylindrical conductors. (b) The electric field lines are shown in cross-sectional view.
Solution: We need to find the potential difference between the cylinders; we can do this by integrating the field (which was calculated for a long wire already). The field is proportional to 1/R, so the potential is proportional to ln Ra/Rb. Then C = Q/V.

22. Capacitance of Cylindrical Capacitor Results
Notation Change from previous slide!
Rb  a, Ra  b Sorry!
The potential difference is
DV = -2keln (b/a)
 = (Q/ℓ)
So the capacitance is

23. Example: Spherical Capacitor
See Figure
A Spherical Capacitor consists of
2 concentric spherical, conducting
shells of radii ra & rb. The inner
shell carries a uniformly
distributed charge Q on its surface.
The outer shell carries an equal &
opposite charge -Q.
Formula for the capacitance C (using calculus) is:
Figure Cross section through the center of a spherical capacitor. The thin inner shell has radius rb and the thin outer shell has radius ra.
Solution: As in the previous example, we find the potential difference by integrating the field (that of a point charge) from rb to ra. Then C = Q/V.

24. Capacitance of a Spherical Capacitor Results
Notation Change from previous slide!
rb  a, ra  b Sorry!
The potential
difference is: l
So, the capacitance is:

25. Capacitance of 2 Long Parallel Wires
Example:
Capacitance of 2 Long Parallel Wires
Estimate the capacitance per unit length C/ℓ of 2 very long straight parallel wires, each of radius R, carrying uniform charges +Q & –Q, separated by a distance d which is large compared to R (d >> R). The line charge density is λ = Q/ℓ
Solution: The field between the wires is the superposition of the fields due to each wire. Integrating gives the potential; then C = Q/V. See text for details. Since the charge is a charge density (charge per unit length), the capacitance is also quoted as C/L – capacitance per unit length.

26. Capacitance of 2 Long Parallel Wires E fields from Gauss’s Law:
At x from wire a: Ea = [λ/(20x) Note again: λ = Q/ℓ
At (d – x) from wire b: Eb = [λ/(20{d – x})]
Solution: The field between the wires is the superposition of the fields due to each wire. Integrating gives the potential; then C = Q/V. See text for details. Since the charge is a charge density (charge per unit length), the capacitance is also quoted as C/L – capacitance per unit length.

27. Capacitance of 2 Long Parallel Wires E fields from Gauss’s Law:
At x from wire a: Ea = [λ/(20x) Note again: λ = Q/ℓ
At (d – x) from wire b: Eb = [λ/(20{d – x})]
Total: E = Ea + Eb
Solution: The field between the wires is the superposition of the fields due to each wire. Integrating gives the potential; then C = Q/V. See text for details. Since the charge is a charge density (charge per unit length), the capacitance is also quoted as C/L – capacitance per unit length.

28. Capacitance of 2 Long Parallel Wires E fields from Gauss’s Law:
At x from wire a: Ea = [λ/(20x) Note again: λ = Q/ℓ
At (d – x) from wire b: Eb = [λ/(20{d – x})]
Total: E = Ea + Eb
Potential Difference: V = Vb – Va = - (Eb - Ea)x.
Solution: The field between the wires is the superposition of the fields due to each wire. Integrating gives the potential; then C = Q/V. See text for details. Since the charge is a charge density (charge per unit length), the capacitance is also quoted as C/L – capacitance per unit length.

29. Capacitance of 2 Long Parallel Wires E fields from Gauss’s Law:
At x from wire a: Ea = [λ/(20x) Note again: λ = Q/ℓ
At (d – x) from wire b: Eb = [λ/(20{d – x})]
Total: E = Ea + Eb
Potential Difference: V = Vb – Va = - (Eb - Ea)x..
Limit: R  d – R. Simplifying with the assumption: d >> R gives:
V  [Q/(0ℓ )] ln(d/R). Capacitance: C = Q/V so
C  (0ℓ)/ln(d/R)
Solution: The field between the wires is the superposition of the fields due to each wire. Integrating gives the potential; then C = Q/V. See text for details. Since the charge is a charge density (charge per unit length), the capacitance is also quoted as C/L – capacitance per unit length.

30. Capacitance of 2 Long Parallel Wires E fields from Gauss’s Law:
At x from wire a: Ea = [λ/(20x). Note again: λ = Q/ℓ
At (d – x) from wire b: Eb = [λ/(20{d – x})]
Total: E = Ea + Eb
Potential Difference: V = Vb – Va = - (Eb - Ea)x.
Limit: R  d – R. Simplifying with the assumption: d >> R gives:
V  [Q/(0ℓ )] ln(d/R). Capacitance: C = Q/V so
C  (0ℓ)/ln(d/R)
Solution: The field between the wires is the superposition of the fields due to each wire. Integrating gives the potential; then C = Q/V. See text for details. Since the charge is a charge density (charge per unit length), the capacitance is also quoted as C/L – capacitance per unit length.