# 27.1. Introduction

**A capacitor is an arrangement of conductors that is used to store electric charge. A very simple capacitor is an isolated metallic sphere. The potential of a sphere with radius R and charge Q is equal to**

_{ }

_{ }

**of the system of conductors. The unit of capacitance is the**

__capacitance__**(F). The capacitance of the metallic sphere is equal to**

__farad___{ }

## 27.2. The parallel-plate capacitor

**Another example of a capacitor is a system consisting of two parallel metallic plates. In Chapter 26 it was shown that the potential difference between two plates of area A, separation distance d, and with charges +Q and -Q, is given by**

_{ }

_{ }

### Example: Problem 27.7

The tube of a Geiger counter consists of a thin straight wire surrounded by a coaxial conducting shell. The diameter of the wire is 0.0025 cm and that of the shell is 2.5 cm. The length of the tube is 10 cm. What is the capacitance of a Geiger-counter tube ?**Figure 27.1. Schematic of a Geiger counter.**

_{w}, the radius of the cylinder is r

_{c}, the length of the counter is L, and the charge on the wire is +Q. The electric field in the region between the wire and the cylinder can be calculated using Gauss' law. The electric field in this region will have a radial direction and its magnitude will depend only on the radial distance r. Consider the cylinder with length L and radius r shown in Figure 27.1. The electric flux [Phi] through the surface of this cylinder is equal to

_{ }

_{0}. Therefore

_{ }

_{ }

_{ }

_{ }

_{w}, r

_{c}, and L into eq.(27.10) we obtain

_{ }

## 27.3. Capacitors in Combination

**The symbol of a capacitor is shown in Figure 27.2. Capacitors can be connected together; they can be connected in series or in parallel. Figure 27.3 shows two capacitors, with capacitance C**

_{1}and C

_{2}, connected in parallel. The potential difference across both capacitors must be equal and therefore

_{ }

**Figure 27.2. Symbol of a Capacitor.**

**Figure 27.3. Two capacitors connected in parallel.**

_{ }

_{1}and C

_{2}in the following manner

_{ }

_{1}and C

_{2}, connected in series. Suppose the potential difference across C

_{1}is [Delta]V

_{1}and the potential difference across C

_{2}is [Delta]V

_{2}. A charge Q on the top plate will induce a charge -Q on the bottom plate of C

_{1}. Since electric charge is conserved, the charge on the top plate of C

_{2}must be equal to Q. Thus the charge on the bottom plate of C

_{2}is equal to -Q. The voltage difference across C

_{1}is given by

_{ }

_{2}is equal to

_{ }

**Figure 27.4. Two capacitors connected in series.**

_{ }

_{ }

### Example: Problem 27.10

A multi-plate capacitor, such as used in radios, consists of four parallel plates arranged one above the other as shown in Figure 27.5. The area of each plate is A, and the distance between adjacent plates is d. What is the capacitance of this arrangement ?**Figure 27.5. A Multi-plate Capacitor.**

_{ }

_{ }

**Figure 27.6. Schematic of Multi-plate Capacitor shown in Figure 27.5.**

## Example: Problem 27.13

Three capacitors, of capacitance C_{1}= 2.0 uF, C

_{2}= 5.0 uF, and C

_{3}= 7.0 uF, are initially charged to 36 V by connecting each, for a few instants, to a 36-V battery. The battery is then removed and the charged capacitors are connected in a closed series circuit, with the positive and negative terminals joined as shown in Figure 27.7. What will be the final charge on each capacitor ? What will be the voltage across the points PP' ?

**Figure 27.7. Problem 27.13.**

_{1}, q

_{2}, and q

_{3}, are equal to

_{ }

_{1}, Q

_{2}, and Q

_{3}. Since charge is a conserved quantity, there is a relation between q

_{1}, q

_{2}, and q

_{3}, and Q

_{1}, Q

_{2}, and Q

_{3}:

_{ }

_{3}and Q

_{3}, or in terms of C

_{1}, C

_{2}, Q

_{1}, and Q

_{2}:

_{ }

_{ }

_{1}and Q

_{2}can be obtained:

_{ }

_{ }

_{ }

_{3}can be expressed in terms of known variables:

_{ }

_{ }

_{ }

_{ }

_{ }

## 27.4. Dielectrics

**If the space between the plates of a capacitor is filled with an insulator, the capacitance of the capacitor will chance compared to the situation in which there is vacuum between the plates. The change in the capacitance is caused by a change in the electric field between the plates. The electric field between the capacitor plates will induce dipole moments in the material between the plates. These induced dipole moments will reduce the electric field in the region between the plates. A material in which the induced dipole moment is linearly proportional to the applied electric field is called a**

**. In this type of materials the total electric field between the capacitor plates E is related to the electric field E**

__linear dielectric___{free}that would exist if no dielectric was present:

_{ }

_{free}, the dielectric constant [kappa] must be larger than 1.

The potential difference across a capacitor is proportional to the electric field between the plates. Since the presence of a dielectric reduces the strength of the electric field, it will also reduce the potential difference between the capacitor plates (if the total charge on the plates is kept constant):

_{ }

_{free}of a capacitor with no dielectric in the following manner

_{ }

The electric field between the two capacitor plates is the vector sum of the fields generated by the charges on the capacitor and the field generated by the surface charges on the surface of the dielectric. The electric field generated by the charges on the capacitor plates (charge density of [sigma]

_{free}) is given by

_{ }

_{bound}, the field generated by these bound charges is equal to

_{ }

_{free}/[kappa] and thus

_{ }

_{ }

_{ }

### Example: Problem 27.19

A parallel plate capacitor of plate area A and separation distance d contains a slab of dielectric of thickness d/2 (see Figure 27.8) and dielectric constant [kappa]. The potential difference between the plates is [Delta]V.a) In terms of the given quantities, find the electric field in the empty region of space between the plates.

b) Find the electric field inside the dielectric.

c) Find the density of bound charges on the surface of the dielectric.

**Figure 27.8. Problem 27.19.**

_{0}. The electric field in the dielectric, E

_{d}, is related to the free electric field via the dielectric constant [kappa]:

_{ }

_{ }

_{ }

_{ }

_{free}is equal to

_{ }

_{ }

_{ }

## 27.5. Gauss Law in Dielectrics

**The electric field in an "empty" capacitor can be obtained using Gauss' law. Consider an ideal capacitor (with no fringing fields) and the integration volume shown in Figure 27.9. The area of each capacitor plate is A and the charges on the plates are +/-Q. The charge enclosed by the integration volume shown in Figure 27.9 is equal to +Q. Gauss' law states that the electric flux [Phi] through the surface of the integration volume is related to the enclosed charge:**

_{ }

_{free}:

_{ }

_{ }

**Figure 27.9. Ideal Capacitor.**

### Example: Problem 27.25

A metallic sphere of radius R is surrounded by a concentric dielectric shell of inner radius R, and outer radius 3R/2. This is surrounded by a concentric, thin, metallic shell of radius 2R (see Figure 27.10). The dielectric constant of the shell is [kappa]. What is the capacitance of this contraption ?Suppose the charge on the inner sphere is Q

_{free}. The electric field inside the dielectric can be determined by applying Gauss' law for a dielectric (eq.(27.50)) and using as the integration volume a sphere of radius r (where R < r < 3R/2)

_{ }

_{ }

**Figure 27.10. Problem 27.25.**

_{ }

_{ }

_{ }

## 27.6 Energy in Capacitors

**The electric potential energy of a capacitor containing no dielectric and with charge +/-Q on its plates is given by**

_{ }

_{1}and V

_{2}are the potentials of the two plates. The electric potential energy can also be expressed in terms of the capacitance C of the capacitor

_{ }

### Example: Problem 27.40

Ten identical 5 uF capacitors are connected in parallel to a 240-V battery. The charged capacitors are then disconnected from the battery and reconnected in series, the positive terminal of each capacitor being connected to the negative terminal of the next. What is the potential difference between the negative terminal of the first capacitor and the positive terminal of the last capacitor ? If these terminals are connected via an external circuit, how much charge will flow around this circuit as the series arrangement discharges ? How much energy is released in the discharge ? Compare this charge and this energy with the charge and energy stored in the original, parallel arrangement, and explain any discrepancies.The charge on each capacitor, after being connected to the 240-V battery, is equal to

_{ }

_{ }

**Figure 27.11. Problem 27.40.**

_{ }

_{ }

### Example: Problem 27.39

Three capacitors are connected as shown in Figure 27.12. Their capacitances are C_{1}= 2.0 uF, C

_{2}= 6.0 uF, and C

_{3}= 8.0 uF. If a voltage of 200 V is applied to the two free terminals, what will be the charge on each capacitor ? What will be the electric energy of each ?

**Figure 27.12. problem 27.39.**

_{1}is V

_{1}, and the voltage across capacitor (C

_{2}+ C

_{3}) is V

_{2}. If the charge on capacitor C

_{1}is equal to Q

_{1}, then the charge on the parallel capacitor is also equal to Q

_{1}. The potential difference across this system is equal to

_{ }

_{ }

_{23}across the capacitor (C

_{2}+ C

_{3}) is related to the charge Q

_{1}

_{ }

_{2}is equal to

_{ }

_{3}is equal to

_{ }

_{ }

_{ }

_{ }

_{ }

## 0 comments:

## Post a Comment