Chapter 25 Capacitance
25-3 Calculating the Capacitance
The goal is to calculate the capacitance of a capacitor from its geometry.
The plan is B (1) Assume a charge q on the plates B (2) calculate the electric field B E between the plates in terms of q, using Gauss' law B (3) knowing E, calculate the potential difference V between the plates from Eq. 24-18 B (4) calculate C from Eq.25-l.
Calculating the Electric Field Use Gauss' law:
Equation 25-3 B reduces to
Calculating the Potential Difference
Chapter 25 Capacitance
Path follows B an electric field line B from – to + plate.
This path B the vectors E and ds' B have opposite directionsB E· ds = - E ds.
The right side of Eq. 25-5 B be positive.
V = Vf - Vi, B Eq. 25-5 as
A Parallel-Plate Capacitor
Chapter 25 Capacitance
Fig. 25-5 B the plates of B parallel-plate capacitor B so large and so close together B neglect the fringing of the electric field at the edges of the plates,
Gaussian surface B encloses just the charge q on the positive plate B as in Fig. 25-5.
From Eq. 25-4 Ð
Substitute q B from Eq. 25-7 and V from Eq. 25-8 into Eq. 25-1 B q = CV
The capacitance B depends only on geometrical factors-namely B A Û d.
Chapter 25 Capacitance A Cylindrical Capacitor
Chapter 25 Capacitance
Figure 25-6 shows B in cross section B a cylindrical capacitor of length L formed by two coaxial cylinders of radii a Û b.
L >> b B neglect the fringing of the electric field B occurs at the ends of the cylinders.
Each plate contains B a charge of magnitude q.
Gaussian surface B a cylinder of length L and radius r, B closed by end caps and placed as is shown in Fig. 25-6.
Chapter 25 Capacitance
Depends only B on geometrical factors B L Û b Û a.
A Spherical Capacitor
Figure 25-6 B serve as a central cross section of a
capacitor B consists of two concentric spherical shells,
Chapter 25 Capacitance
This formula B and the others B involve the constant εo B multiplied by a quantity B has the dimensions of a
length.
