Transcript Lecture 8 - UCF Physics

Capacitance
Chapter 26
(Continued)
Capacitors
Q
+
-Q
V
-
If the plates are held at an electric
potential difference V then get
charges Q and -Q on the plates with
Q=CV
V should be really be written ∆V, but we often don’t bother.
The battery’s ability to push charge is called its “electromotive
force” or emf. A 6V battery has an emf of 6V.
We often refer to electric potential, potential difference, and emf
simply and sloppily as “voltage,” because all have units of volts.
Dielectrics in Capacitors
• Suppose we fill the space between the plates of a capacitor
with an insulating material (a “dielectric”):
+Q
-+
-+ -+
-+
-+
-+ -+
-+
-+
-+ -+
-+
-+
-+ -+
-+
-Q
• The material will be “polarized” - charge -e electrons are
pulled to the left, stretched away from the positively charged
atomic cores.
• Consequently the E field within the capacitor will be reduced.
Dielectrics in Capacitors
+Q
-+
-+ -+
-+
-+
-+ -+
-+
-+
-+ -+
-+
-+
-+ -+
-+
-Q
• At a point in the interior some electrons shift left. But other
electrons shift over to replace them (on average). So there is
no net charge density in the bulk.
• However, some electrons are pushed towards the left plate,
and there is a deficit of electrons at the right plate.
• Hence there is some induced charge -Q’ at the very left side of
the insulator, and an induced charge +Q’ at the very right side.
Dielectrics in Capacitors
+Q
+
+
+
+
+
+
+
+
+
-
-Q’
+ + +Q’ -+ + -
-Q
It turns out that Q’ is proportional to Q. It is convenient to
write the constant of proportionality this way:
 1
Q' 
Q
  
 is called the dielectric constant. Its value depends on the
dielectric (whether it is plastic, glass, etc.) For a vacuum =1;
for air  is close to 1. Except for a vacuum  >1.
Dielectrics in Capacitors
+Q
Gaussian Surface
(a 3D box)
+
+
+
+
+
+
+
+
+
-
-Q’
+ + +Q’ -+ + -
-Q
Suppose the field inside is E0 without the dielectric. Let us
calculate the new field E when the dielectric is present.
Use Gauss’s Law in both cases:
E0 = Q / (0 A) and E = (Q – Q’) / (0 A)
E0 / E = Q / (Q – Q’)  E0 / E =  or E = E0/
Thus the field is reduced by factor 
Dielectrics in Capacitors
• We see that a dielectric reduces the electric field by a
factor (E=E0/)
• Hence for a given charge Q on the metal plate the
potential difference V = Ed is also reduced by  (V=V0/).
• Thus C= Q/V is increased by  [C=C0 where C0=Q/V0 is
the capacitance without the dielectric].
C 
oA
d
parallel plate capacitor
with dielectric.
• Adding a dielectric increases the capacitance.
Dielectrics & Gauss’s Law
With a dielectric present, Gauss’s Law can be
rewritten from
+q
to
0  E  dA  q  q'
-q’
0   E  dA  q
Instead of having to think about the confusing
induced charge q’, we can simply use the free

charge q. But E is replaced by E.
Parallel and Series
Parallel
Series
Capacitors in Circuits
+Q
-Q
(Symbol for
a capacitor)
C
V
A piece of metal in equilibrium has a constant value of potential.
Thus, the potential of a plate and attached wire is the same.
The potential difference between the ends of the wires is V,
the same as the potential difference between the plates.
Capacitors in Parallel
• Suppose there is a potential
difference V between a and b.
• Then q1=C1V & q2=C2V
• We want to replace C1 and C2 with an
equivalent capacitance for which q=CV,
where the charge on C is q = q1 + q2
a
q1
q2
C1
b
C2
V
a
C-q
• From the above, q=q1 + q2=(C1 + C2)V. That is, q=CV with
C = C1 + C2
• This is the equation for capacitors in parallel.
• Increasing the number of capacitors increases the capacitance.
b
Capacitors in Series
a
C1
C2
-q +q -q +q
V1
C
b
a -q
V2
+q
b
V
• The capacitors have the same q but different potential differences.
Here the total potential difference between a and b is V = V1 + V2
• Again we want to find q=CV with C an equivalent capacitance.
• Use V1 = q/C1 and V2 = q/C2
• Then V = V1 + V2 = q /C1 +q /C2 = [(1/C1) + (1/C2)]q
• or, V = q/C with
1 / C = 1 / C1
+ 1 / C2
• This is the equation for capacitors in series.
• Increasing the number of capacitors decreases the capacitance.
Equivalent Capacitors
Parallel
C1
C2
C = C1 + C2
Series
For each there is an
equivalent single capacitor
C for which q=CV. Here q
is the total charge to flow
down the main wire, and
V the potential difference
across the unit.
C1
C2
1 / C = 1 / C1
+ 1 / C2