Understanding Capacitors and Dielectrics: Theory and Examples

1. TOPIC 5Capacitors and Dielectrics

2. Capacitors • Capacitors are a means of storing electric charge (and electric energy) • It takes energy to bring charge together • A capacitor allows more charge to be stored for a given energy • It does this by reducing the potential at which the charge is stored • It can do this by bringing an opposite charge into close proximity, to reduce the overall repulsion

3. Capacitance Capacitance (C) is charge per unit potential difference C = Q/V Unit is Farad (F): 1 F = 1 Coulomb/Volt Typical capacitances measured in F (10–6 F) or pF (10–12 F)

4. Parallel Plate Capacitor Two plates, area A, separation d, carrying charge Q. Gauss’s Law (using dotted Gaussian surface shown) E A = Q/0  E = Q/0A

5. Example 1 – Parallel Plate Capacitor A parallel plate capacitor has plates with dimensions 3 cm by 4 cm, separated by 2 mm. The plates are connected across a 60 V battery. Find: (a) the capacitance; (b) the magnitude of charge on each plate; (c) the energy stored in the capacitor – see later!

6. Example 2 – Cylindrical Capacitor What is the capacitance of a long cylindrical (coaxial) cable of inner radius a, outer radius b and length L as shown?

7. Example 3 – Spherical Capacitor What is the capacitance of two concentric spherical conducting shells of inner radius a and outer radius b?

8. Capacitors in Parallel Capacitors connected as shown, with terminals connected together, are said to be in parallel. They behave as a single capacitor with effective capacitance C. Total charge Q = Q1 + Q2 = C1V + C2V Therefore C = Q/V = C1 + C2

9. Capacitors in Series Capacitors connected together as shown, sharing one common terminal, are said to be in series. They behave as a single capacitor with effective capacitance C. The external charge stored is Q. The voltages across the capacitors Vi= Q/Ci must add up to V. Therefore V = Q/C1 + Q/C2 = Q/C

10. Example 4 –Capacitor Network If each of the individual capacitors in the network below has a capacitance C, what is the overall effective capacitance?

11. Energy stored in a Capacitor Adding an increment of charge dq to a capacitor requires work dW = V dq = q/C dq This is obviously the increase in (potential) energy stored of the capacitor U The total energy required to charge a capacitor from zero charge to Q is therefore Since Q = CV, we can express this in other ways:

12. Example 1 – Parallel Plate Capacitor A parallel plate capacitor has plates with dimensions 3 cm by 4 cm, separated by 2 mm. The plates are connected across a 60 V battery. Find: (a) the capacitance; (b) the magnitude of charge on each plate; (c) the energy stored in the capacitor – see later! Previously: C = 5.3 pF Q = 3.210–10 C

13. Energy stored in a Capacitor (2) The energy stored in the capacitor can also be considered as the energy stored in its electric field. We have For the parallel plate capacitor we also have V = Ed and So But A d is the volume where the electric field exists, so the energy density is This is a general result for the energy density in a field.

14. Dielectrics A conductor contains free charges that can move through the material. A dielectric contains bound charges, which cannot move freely but will displace through small distances when affected by an electric field. This leaves excess bound charges on the surface of the material. This reduces the electric field within the bulk of the material.

15. Dielectrics and Capacitors The factor by which the electric field is reduced is known as the dielectric constantk (or r). If the gap between the plates of a capacitor is filled with dielectric material, the voltage between the plates for a given charge will also be reduced by the factor k. Since C = Q / V, this means that C is increased by k. For the parallel plate capacitor, we therefore have

16. Example 5 Demonstrate that the energy stored in a spherical capacitor is consistent with an energy density stored in the field of