- CAPACITORS AND DIELECTRICS.

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

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Equation (27.1) shows that the potential of the sphere is proportional to the charge Q on the conductor. This is true in general for any configuration of conductors. This relationship can be written as

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where C is called the ** capacitance** of the system of conductors.
The unit of capacitance is the

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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

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Using the definition of the capacitance (eq.(27.2)), the capacitance of this system can be calculated:

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Equation (27.2) shows that the charge on a capacitor is proportional to the capacitance C and to the potential V. To increase the amount of charge stored on a capacitor while keeping the potential (voltage) fixed, the capacitance of the capacitor will need to be increased. Since the capacitance of the parallel plate capacitor is proportional to the plate area A and inversely proportional to the distance d between the plates, this can be achieved by increasing the surface area A and/or decreasing the separation distance d. These large capacitors are usually made of two parallel sheets of aluminized foil, a few inches wide and several meters long. The sheets are placed very close together, but kept from touching by a thin sheet of plastic sandwiched between them. The entire sandwich is covered with another sheet of plastic and rolled up like a roll of toilet paper.

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 ?

The problem will be solved under the assumption that the electric field
generated is that of an infinitely long line of charge. A schematic side view
of the tube is shown in Figure 27.1. The radius of the wire is r_{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

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According to Gauss' law, the flux [Phi] is equal to the enclosed charge divided
by [epsilon]_{0}. Therefore

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The electric field E(r) can be obtained using eq.(27.7):

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The potential difference between the wire and the cylinder can be obtained by integrating the electric field E(r):

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Using eq.(27.2) the capacitance of the Geiger tube can be calculated:

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Substituting the values for r_{w}, r_{c}, and L into eq.(27.10)
we obtain

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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

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Using eq.(27.12) the total charge on both capacitors can be calculated

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Equation (27.13) shows that the total charge on the capacitor system shown in
Figure 27.3 is proportional to the potential difference across the system. The
two capacitors in Figure 27.3 can be treated as one capacitor with a
capacitance C where C is related to C_{1} and C_{2} in the
following manner

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Figure 27.4 shows two capacitors, with capacitance C_{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

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and the voltage difference across C_{2} is equal to

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Equation (27.17) again shows that the voltage across the two capacitors, connected in series, is proportional to the charge Q. The system acts like a single capacitor C whose capacitance can be obtained from the following formula

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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 ?

The multiple capacitor shown in Figure 27.5 is equivalent to three identical capacitors connected in parallel (see Figure 27.6). The capacitance of each of the three capacitors is equal and given by

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The total capacitance of the multi-plate capacitor can be calculated using eq.(27.14):

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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' ?

The initial charges on each of the three capacitors, q_{1},
q_{2}, and q_{3}, are equal to

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After the three capacitors are connected, the charge will redistribute itself.
The charges on the three capacitors after the system settles down are equal to
Q_{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}:

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The voltage between P and P' can be expressed in terms of C_{3} and
Q_{3}, or in terms of C_{1}, C_{2}, Q_{1}, and
Q_{2}:

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and

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Using eq.(27.22) the following expressions for Q_{1} and Q_{2}
can be obtained:

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Substituting eq.(27.25) and eq.(27.26) into eq.(27.24) we obtain

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Combining eq.(27.27) and eq.(27.23), Q_{3} can be expressed in terms of
known variables:

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Substituting the known values of the capacitance and initial charges we obtain

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The voltage across P and P' can be found by combining eq.(27.29) and eq.(27.23):

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The charges on capacitor 1 and capacitor 2 are equal to

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

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where [kappa] is called the dielectric constant. Since the final electric
field E can never exceed the free electric field E_{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):

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The capacitance C of a system with a dielectric is inversely proportional to
the potential difference between the plates, and is related to the capacitance
C_{free} of a capacitor with no dielectric in the following manner

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Since [kappa] is larger than 1, the capacitance of a capacitor can be significantly increased by filling the space between the capacitor plates with a dielectric with a large [kappa].

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

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Assuming a charge density on the surface of the dielectric equal to
[sigma]_{bound}, the field generated by these bound charges is equal
to

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The electric field between the plates is equal to E_{free}/[kappa] and
thus

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Substituting eq.(27.36) and eq.(27.37) into eq.(27.38) gives

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or

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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.

a) Suppose the electric field in the capacitor without the dielectric is equal
to E_{0}. The electric field in the dielectric, E_{d}, is
related to the free electric field via the dielectric constant [kappa]:

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The potential difference between the plates can be obtained by integrating the electric field between the plates:

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The electric field in the empty region is thus equal to

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b) The electric field in the dielectric can be found by combining eq.(27.41) and (27.43):

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c) The free charge density [sigma]_{free} is equal to

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The bound charge density is related to the free charge density via the following relation

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Combining eq.(27.45) and eq.(27.46) we obtain

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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:

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If a dielectric is inserted between the plates, the electric field between the
plates will change (even though the charge on the plates is kept constant).
Obviously, Gauss' law, as stated in eq.(27.48), does not hold in this case.
The electric field E between the capacitor plates is related to the
dielectric-free field E_{free}:

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where [kappa] is the dielectric constant of the material between the plates. Gauss' law can now be rewritten as

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Gauss' law in vacuum is a special case of eq.(27.50) with [kappa] = 1.

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)

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The electric field in this region is therefore given by

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Using the electric field from eq.(27.52) and eq.(27.53) we can determine the potential difference [Delta]V between the inner and outer sphere:

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The capacitance of the system can be obtained from eq.(27.54) using the definition of the capacitance in terms of the charge Q and the potential difference [Delta]V:

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

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where V_{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

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This formula is also correct for a capacitor with a dielectric; the properties of the dielectric enters into this formula via the capacitance C.

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

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The potential difference across each capacitor will remain equal to 240 V after the capacitors are connected in series. The total potential difference across the ten capacitors is thus equal to

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If the two end terminals of the capacitor network are connected, a charge of 1.2 mC will flow from the positive terminal to the negative terminal (see Figure 27.11).

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The energy stored in each capacitor, after being charged to 240 V, is equal to

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Clearly no energy is lost in the process of changing the capacitor configuration from parallel to serial.

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 ?

Suppose the voltage across capacitor C_{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

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The charge on capacitor 1 is thus determined by the potential difference [Delta]V

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The voltage V_{23} across the capacitor (C_{2} + C_{3})
is related to the charge Q_{1}

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The charge on capacitor C_{2} is equal to

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The charge on capacitor C_{3} is equal to

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The electric potential energy stored in each capacitor is equal to

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For the three capacitors in this problem the electric potential energy is equal to

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