Transcript 17-7 Capacitance - mrhsluniewskiscience

Chapter 17
Electric Potential
Objectives: The students will be
able to:
• Given the dimensions, distance between the plates, and
the dielectric constant of the material between the plates,
determine the magnitude of the capacitance of a parallel
plate capacitor.
• Given the capacitance, the dielectric constant, and either
the potential difference or the charge stored on the
plates of a parallel plate capacitor, determine the energy
and the energy density stored in the capacitor.
17.7 Capacitance
A capacitor consists of two conductors
that are close but not touching. A
capacitor has the ability to store electric
charge.
Capacitor
• Named for the capacity to store electric
charge and energy.
• A capacitor is two conducting plates
separated by a finite distance:
17.7 Capacitance
Parallel-plate capacitor connected to battery. (b)
is a circuit diagram.
17.7 Capacitance
When a capacitor is connected to a battery, the
charge on its plates is proportional to the
voltage:
(17-7)
The quantity C is called the capacitance.
Unit of capacitance: the farad (F)
1 F = 1 C/V
Figure 17-15
Key of a computer keyboard
Sample Problem: A 0.75 F capacitor is charged to a
voltage of 16 volts. What is the magnitude of the
charge on each plate of the capacitor?
Sample Problem: A 0.75 mF capacitor is charged to a
voltage of 16 volts. What is the magnitude of the
charge on each plate of the capacitor?
V = 16 V, C = 0.75 mF = 0.75 x 10-6 F
Q =?
C = Q/V or Q = CV
Q = (0.75 x 10-6)(16)
Q = 1.2 x 10-5 C
Q = CV
(17 - 7)
17.7 Capacitance
The capacitance does not depend on the
voltage; it is a function of the geometry and
materials of the capacitor.
For a parallel-plate capacitor:
(17-8)
We see that C depends only on geometric factors, A and d ,
and not on Q or V . We derive this useful relation in the
optional subsection at the end of this Section. The constant εo is
the permittivity of free space , which, as we saw in Chapter 16,
has the value 8.85 x 10–12 C2Nm2.
A simple type of capacitor is the parallel-plate
capacitor. It consists of two plates of area A
separated by a distance d.
By calculating the electric
field created by the
charges ±Q, we find
that the capacitance of a
parallel-plate capacitor
is:
The general properties of a
parallel-plate capacitor – that
the capacitance increases as
the plates become larger and
decreases as the separation
increases – are common to all
capacitors.
Capacitor Geometry
The capacitance of a
capacitor depends on
HOW you make it.
1
C  A C
d
A  area of plate
d  distance beteween plates
A
C
d
 o  constant of proportion ality
 o  vacuum permittivi ty constant
 o  8.85 x10 12
C2
Nm 2
o A
C
d
• Sample Problem: What is the
AREA of a 1 F capacitor that has a
plate separation of 1 mm?
C = 1 F, d = 1 mm = 0.001 m,
o = 8.85 x 10-12 C2/(Nm2)
Is this a practical capacitor to build?
NO! – How can you build this then?
A
C  o
D
A
1  8.85 x10
0.001
A  1.13 x 108 m2
The answer lies in
REDUCING the AREA. But
you must have a
CAPACITANCE of 1 F. How
can you keep the
capacitance at 1 F and
reduce the Area at the
same time?
Sides  10629 m
Add a DIELECTRIC!!!
12
Dielectric
A dielectric material
(dielectric for short)
is an electrical
insulator that can be
polarized by an
applied electric field.
Example 17-8 Capacitor calculations.
(a)Calculate the capacitance of a parallel-plate capacitor whose
plates are 20 cm x 3.0 cm and are separated by a 1.0-mm air gap.
(b) What is the charge on each plate if a 12-V battery is
connected across the two plates?
(c) What is the electric field between the plates?
(d) Estimate the area of the plates needed to achieve a
capacitance of 1 F, assuming the air gap d is 100
times smaller, or 10 microns (1 micron = 1 μm = 10-6 m).
Example 17-8 Capacitor calculations.
(a)Calculate the capacitance of a parallel-plate capacitor whose
plates are 20 cm x 3.0 cm and are separated by a 1.0-mm air gap.
(b) What is the charge on each plate if a 12-V battery is
connected across the two plates?
(c) What is the electric field between the plates?
(d) Estimate the area of the plates needed to achieve a
capacitance of 1 F, assuming the air gap d is 100
times smaller, or 10 microns (1 micron = 1 μm = 10-6 m).
Example 17-8 Capacitor calculations.
(a)Calculate the capacitance of a parallel-plate capacitor whose
plates are 20 cm x 3.0 cm and are separated by a 1.0-mm air gap.
Example 17-8 Capacitor calculations.
(b) What is the charge on each plate if a 12-V battery is
connected across the two plates?
Example 17-8 Capacitor calculations.
(c) What is the electric field between the plates?
Example 17-8 Capacitor calculations.
(d) Estimate the area of the plates needed to achieve a
capacitance of 1 F, assuming the air gap d is 100
times smaller, or 10 microns (1 micron = 1 μm = 10-6 m).
Elaboration
• Capacitors - pHET
Homework
• Problems in chapter 7
• 31, 34, 37, 38, 40
Closure
• When a battery is connected to a capacitor,
why do the plates acquire charges of the same
magnitude?
Closure
• When a battery is connected to a capacitor,
why do the plates acquire charges of the same
magnitude?