Our study of electricity has been kept fairly abstract and idealized up to now. This chapter deals with a simple, but highly practical and pervasive electronic device, the capacitor.

A capacitor is made by placing two conducting plates close,but not touching each other. (Two sheets of foil separated by a sheet of waxed paper works nicely.) If we then apply a potential by connecting a battery across the two plates, a charge will be removed from one plate and placed on the other.

uncharged	charging	charged

It can be shown experimentally that the greater the potential, the greater the charge developed. If we plot potential versus charge, we will get a direct relationship.

We may represent this mathematically as: q = CV where C is a constant called the capacitance of the device.

Thus C = q/V has units of coul/volt, or farad (after Michael Faraday).

A farad is, in practice, an extremely large capacitance, so µF (microfarad or 10^-6 farad) or even µµF (micro-microfarad, or picofarad, or 10^-12 farad) are used in practice.

1. What is the capacitance of a capacitor which stores 4 * 10^-3 coulombs when 500 V is applied to it?

C = q/V C = (4 * 10-3 coul)/(500 volt) C = 8 * 10-6 coul/volt C = 8 µF

2. What charge is deposited on a 40 µF capacitor when 15 V is applied to it?

q = CV q = (40 * 10-6)(15) q = 6000 * 10-6 C q = 6 * 10-4 coul

3. How much potential is required to place a 7 * 10^-3 C charge on a 3000 µF capacitor?

V = q/C V = (7 * 10-3)/(3000 * 10-6) V = 2.33 V

4. 500 V will deposit 9 * 10^-7 C when it is applied across a capacitor. What potential will be required to deposit 12 * 10^-8 C on the same capacitor?

C = q/V C = (9 * 10-7)/(500) C = 1.80 * 10-9 farad V' = q'/C V' = (12 * 10-8)/(1.8 * 10-9) V' = 66.7 V note: You may prefer to note that q is reduced to 1.2/9 of its previous value, so V' = V(q'/q) = (500)(1.2/9) = 66.7 V

5. Two capacitor plates are separated by 3 * 10^-3 m. With a potential of 4,000 V applied to the plates, what electric field will be generated in the space between?

V = W/q V = (F * d)/q V = (F/q) * d V = E * d E = V/d E = (4000)/(3 * 10-3) E = 1.33 * 106 V/m note: (V/m) = (nt/C) work it out!

6. Two plates contain a field of 33,000 nt/C and are separated by 7 * 10^-5 m. What is the potential between the plates?

V = E * d V = (3.3 * 104 nt/C)(7 * 105) V = 2.33 V

7. Two plates of a capacitor are charged to a potential V and then disconnected from the battery. The distance between the plates is then doubled.
a) What effect does this have on the charge?
b) What effect does this have on the E-field between the plates?
c) What effect does this have on the potential between the plates?
d) If the old capacitance was C, what is the new capacitance?

a) None, because each plate is insulated.

b) None, because the field is still confined to the space between the plates.

c) Potential is doubled: V' = E(2d) V' = 2(Ed) V' = 2V

d) C = q/V C' = q/V' C' = q/2V C' = (1/2)C

8. If the area of the capacitor plates were doubled in a particular situation, what effect would this have on the capacitance? (Think in terms of the ability to hold a charge with a given potential.)

By doubling the plate area, twice as much charge may be stored at the same potential. IMAGE IMAGE C = q/V C' = q'/V C' = 2q/V C' = 2C note: capacitance is doubled

A small digression is in order. Everyone is familiar with wires we use them around the house and we see them in electronic equipment. One of the major purposes of a wire is to transfer an electric potential from one point to another. The ideal wire (commonly used by physicists) allows the free flow of electrons along its length. Since this movement occurs without effort, the potential (work/charge) must be uniform over the entire length.

Now back to the main issue. There are two ways to connect capacitors together. One is to connect them in series:

The other way is to connect them in parallel:

Here are some things to know about these circuits.

1) In a series connection the charge on C₁ will equal the charge on C₂. It HAS to there is nowhere for the charge forced off C₁ to go, but onto C₂. (No charge-eating monsters have been reported in the literature.)
2) In a parallel circuit the potential across C₁ will equal the potential across C₁. It HAS to - wires connect the leads, and wires have equal potential along their entire length.

9. The charge on the 5 µF capacitor is 3 * 10^-5 coulombs. What is the charge on the 200 µF capacitor?

By rule (1): q = 3 * 10-5 coul

10. What is the potential across the 200 µF capacitor in problem 9?

q = CV V = q/C V = (3 * 10-5)/(200 * 10-6) V = .150 V

11. What is the potential across the 8 µF? What is the charge on the 22 µF?

a) V = 12 V note: see rule 2

b) q = CV q = (22 * 10-6)(12) q = 2.64 * 10-4 coul

12. If the potential on the 4 µF is 10 V, what is the potential on the 8 µF capacitor?

q4 = CV q4 = (4 * 10-5)(10) q4 = 4 * 10-5 C q8 = q4 q8 = 4 * 10-5 V = q/C V = (4 * 10-5)/(8 * 10-6) V = 5.0 V

13. For the circuit shown, write
a) the equivalent charges
b) the equivalent potentials.

a) q2 = q3 q1 = q1 + q4 q1 = q3 + q4 q1 = q5

b) V2 + V3 = V4 V = V1 + V4 + V5 V = V1 + V2 + V3 + V5

14. For the circuit shown, write
a) the equivalent charges.
b) the equivalent potentials.

a) q1 = q2 q1 q2 + q3 + q4 = q5 + q6 = q7

b) V1 + V2 = V3 = V4 V5 = V6 V = V7 + V5 V6 + V1 + V2 V3 V4

15. For the circuit shown, write
a) the equivalent charges.
b) the equivalent potentials.

a) q7 = q8 q2 = q3 = q6 + q7 q8 q1 = q2 q3 + q4 q5

b) V6 = V7 + V8 V4 + V5 = V2 + V3 + V6 V7 +V8 V = V1 + V4 + V5