Summary
Electric potential: PE per unit charge, V=PE/q
 Potential difference: work done to move the charge
from one point to another,
Vab=Vb-Va=DPE/q=Wab/q or Wab=qVab
 For uniform electric field: Vab=-Ed
 For electric field due to a point charge:
V=kQ/r, (V=0 at r=∞)

Electric Potential due to a Point Charge
V=kQ/r (V=0 at r=∞)
V
V
r
+
r
r
r
Three point charges –Q, –Q, and +3Q are arranged along a line
as shown in the sketch.
What is the electric potential at the point P?
(a) +kQ/R
(c) –1.6kQ/R
(b) –2kQ/R
(d) +1.6kQ/R X
(e) +4.4kQ/R
The sketch shows cross sections of equipotential surfaces between two charged conductors
shown in solid black. Points on the equipotential surfaces near the conductors are labeled A,
B, C, ..., H.
1. What is the magnitude of the potential difference between points A and H?
(a) 100 V
(b) 200 V
(c) 400 V
X(d) 600 V
(e) 700 V
2. What is the direction of the electric field at point E?
(a) toward B
X(b) toward G
(c) toward H
(d) toward C
(e) toward F
3. How much work is required to move a +6.0 µC point charge from B to F to D to A?
(a) -1.2x10–3 JX
(b) +1.2x10–3 J (c) +3.6x10–3 J (d) –3.6x10–3 J (e) zero joules
The sketch below shows cross section of equipotential surfaces between two charged
conductors that are shown below in solid black. Various points on the equipotential
surfaces near the conductors are labeled A,B,C…,I.
At which of the labeled points will the electric field have the greatest magnitude?
(a)
G
(c)
A
(e)
D
X(b)
I
(d)
H
At which of the labeled points will an electron have the greatest potential energy?
(a)
A
(c)
G
(e)
I
(b)
D
H
X(d)
What is the potential difference between points B and E?
(a)
10 V
(c)
40 V
(e)
60 V
(b)
30 V
50 V
X(d)
Capacitance
Capacitor:
A Device to
Store Energy
in the Form of
an Electric
Field
Capacitance
A capacitor is a device used in a variety of
electric circuits
 The capacitance, C, of a capacitor is
defined as the ratio of the magnitude of
the charge on either conductor (plate) to
the magnitude of the potential difference
between the conductors (plates)

Capacitance, cont


Q
C
DV
Units: Farad (F)
1
F=1C/V
 A Farad is very large

Often will see µF or pF
Parallel-Plate Capacitor
The capacitance of a device depends on
the geometric arrangement of the
conductors
 For a parallel-plate capacitor whose plates
are separated by air:

A
C  o
d
Q
+
+
+
+
+
+
+
+
d
-Q
+ -
Capacitance C=Q/V
Unit: farad, F. 1 F=1 coulomb/volt
Parallel-plate Capacitor
Q=AE/4πk=εoAE, where A is area of the plate, and εo is the
permittivity of free space, 8.85x10-12 C2/N-m2
Or E=s (charge density)/εo
V=Ed
C=Q/V= εoAE/Ed=εoA/d
Properties of Capacitance C:
Constant for a given capacitor, independent of
Q or V
C depends on the geometry of the capacitor and the
material between the plates.
Question: The energy content of a
charged capacitor resides in its
(a) plates
(b) potential difference
(c) charge
(d) electric field
Answer: d
Question: The plates of a parallel-plate
capacitor of capacitance C are brought
together to one-third their original
separation. The capacitance is now
(a) C/9
(b) C/3
(c) 3C
(d) 9C
Answer: c
QUICK QUIZ 16.5
You charge a parallel-plate capacitor, remove it from the
battery, and prevent the wires connected to the plates from
touching each other. When you pull the plates farther
apart, do the following quantities increase, decrease, or
stay the same? (a) C; (b) Q; (c) E between the plates; (d)
DV.
QUICK QUIZ 16.5 ANSWER
(a) C decreases
(b) Q stays the same
(c) E stays the same
(d) DV increases
Dielectric
Insulating materials
 Capacitance increased by a factor K

Dielectric Constant, K
A measure of how effective it is in reducing an electric
field across the plates
Q
-Q
Vo
Co=Q/Vo
Q
-Q
+
+
+
V=Vo/K
C=Q/V=KQ/Vo=KCo
QUICK QUIZ 16.6
A fully charged parallel-plate capacitor remains
connected to a battery while you slide a dielectric
between the plates. Do the following quantities
increase, decrease, or stay the same? (a) C; (b) Q; (c)
E between the plates; (d) DV.
QUICK QUIZ 16.6 ANSWER
(a) C increases
(b) Q increases
(c) E stays the same
(d) DV remains the same
Electric Field Energy
Work must be done to separate positive and
negative charges against the Coulomb forces
attracting them together. This work is stored as
potential energy or electric field energy in a
capacitor.
e
-
+
Vinitial
The average potential
difference during the charge
transfer is:
<V>=(Vfinal+Vinitial)/2
V
=(Vf+0)/2=Vf/2
+
-Q +
+
+
e
V
-Q
-
The total charge transferred
is Q
Vfinal
The work done is
W=U=Q<V>=(1/2) QV
=(1/2) CV2=(1/2) Q2/C
(C=Q/V)
Energy Stored in a Capacitor
Energy stored = ½ Q ΔV
 From the definition of capacitance, this
can be rewritten in different forms

2
1
1
Q
Energy  QDV  CDV 2 
2
2
2C
A Summary of the various formulas for charge Q,
potential difference V, capacitance and energy W
Known Quantities
Unknown
Quantity
Q=
V=
C, V C, Q Q, V
(2WC)1/2
CV
(2W/C)1/2
Q/C
C=
W=
W, C
Q/V
CV2/2 Q2/2
C
QV/2
W, V W, Q
2W/V
2W/
Q
2W/V2 Q2/2W