Transcript Electric Potential for Point Charges

Electric Potential for Point Charges
So far, we have been talking about the energy of charges in an
electric field.
But electric fields can be related back to point charges.
In this section we introduce the electric potential for point
charges.
In the last lecture we saw that E = -V / x. In reality, this is
a derivative, and because E is actually a vector, this is really a
3 dimensional vector derivative.
We can use calculus (see next page) to integrate to get the
expression for the potential for a point charge, which is
OSE:
VQ = kQ/r = Q/40r.
We have taken V=0 at r= from the point charge Q. Note the
1/r dependence (for F and E the dependence was 1/r2). V is a
scalar and F and E are vectors, so working with V should be
much easier.
As with all equations involving charges, the sign on Q is
important.
As you might expect, the collection due to a sum of point
charges is equal to the sum of potentials. I’ll make this an
OSE.
OSE:
Vnet =  Vi.
We can use calculus to derive the expression (previous page)
for the potential for a point charge…
dV(xyz)
Ex = dx
dV(r)
Er = dr
Er dr= -dV(r)
Er  dr= -dV(r)
dV(r)= -Er  dr
noooo….
correctly including vector nature

b
a
b
dV(r)= -  Er  dr
a
b
Vb  Va= -  Er  dr
a
Vb  Va = - 
b
a
kq
dr
2
r
kq
Vr  V = -  2 dr
 r
r
for a point charge
remember, the r in the integral is
a “dummy” variable
r
kq
Vr  0=
r 
kq
V r  
r
in this derivation, I switch from
using Q to q for our point
charge—no particular reason
For a continuous distribution of charges, replace the sum by an
integral.
k dq
V r   
r
Example: What minimum work is required by an external
force to bring a charge q=3.00 C from a great distance
away (take r = ) to a point 0.500 m from a charge Q = 20.0
C?
This problem can be solved without a diagram, although you
may make one if it helps you.
OSE:
Wif = q Vif
Wif = q (Vf - Vi)
Wif = q (kQ/rf - kQ/ri)
Wif = q (kQ/rf - kQ/ri)
Wif = kqQ (1/rf - 1/ri)
0, because ri = 
Wif = kqQ /rf
Wif = (9x109)(3x10-6)(20x10-6) / (0.5)
Wif = 1.08 J
Example: Calculate the electric potential at point A in the
figure below due to the two charges shown.
y
30 cm
A
=30º
Q2=+50C
Q1=-50C
52 cm
OSE:
Vnet =  Vi.
VA = V1 + V2.
x
VA = V1 + V2
OSE:
VQ = kQ/r = Q/40r
VA = kQ1/r1 + kQ2 /r2
y
r1 = 30 cm
A
=30º
Q2=+50C
Q1=-50C
52 cm
x
VA = kQ1/r1 + kQ2 /r2
All the numbers are in SI units, making the calculation easy.
VA = (9x109)[(-50x10-6/0.6) + (+50x10-6/0.3)]
VA = 7.5x105 V.
Which would you rather work: this example, or the example
from lecture 22, which calculated the electric field, used
vectors, and took 6 slides?
Conceptual example. All charges in the figure have the same
magnitude.
+
-
(i)
+
-
(ii)
+
+
(iii)
Which set has a positive potential energy?
Which set has the most negative potential energy?
Which set requires the most work to separate the charges to ?
Electric Dipoles
An electric dipole is two charges +Q and –Q separated by a
distance l.
The figure shows electric field lines and equipotential lines for
an electric dipole.
Electric dipoles appear frequently in physics, chemistry, and
biology.
Potential due to a dipole.
-Q
sorry, klunky figure, but
I don’t feel like re-doing it
r
l
r

+Q
r
VP =
1
kQ
k(-Q)
1 
Δr
+
= kQ  =
kQ

r
(r +Δr)
r
(r+Δr)
r r+Δr 


P
-Q
r
l
r

+Q
r
P
If P is far from the charges, so that r>>l, then r = l cos  and
r >> r.
Δr
VP = kQ
r  r+Δr 
klunky script lowercase
becomes
V=
kQ l cosθ
.
2
r
letter l, but I don’t feel
like fixing it right now
-Q
r
l
r

+Q
r
P
The product Ql is called the dipole moment of the dipole.
OSE
kp cosθ
Vdipole =
r2
for r>>l .
P
C+
O-2
Example: dipole moment of C==O at point P (see text for
numbers).
kp cosθ
V=

2
r

9  109

 cos180   0.089 V .
 9  10 
8  10 30
10
2
The potential at P is much greater if you remove one of the
charges (makes sense; charges almost “cancel”).
Storing Electrical Energy: Capacitance
A capacitor is basically two parallel
conducting plates with insulating
material in between. The capacitor
doesn’t have to look like metal plates.
When a capacitor is connected to an
external potential, charges flow onto
the plates and create a potential
difference between the plates.
Capacitor for use in
high-performance
audio systems.
Capacitor plates
build up charge.
-
+-
The battery in this circuit has
some voltage V. We haven’t
defined what that means yet.
If the external potential is
disconnected, charges remain on the
plates, so capacitors are good for
storing charge (and energy).
+ +-
Capacitors are also very good at releasing
their stored charge all at once. The capacitors
in your TV are so good at storing energy that
touching the two terminals at the same time
can be fatal, even though the TV may not
have been used for months.
High-voltage capacitors like these are supposed to have
“bleeder resistors” that drain the charge away after the
circuit is turned off. I wouldn’t bet my life on it.
Graphic from http://www.feebleminds-gifs.com/.
assortment of
capacitors
The charge acquired by each plate of a capacitor is Q=CV
where C is the capacitance of the capacitor.
OSE:
Q = CV.
The unit of C is the farad but most capacitors have values
of C ranging from picofarads to microfarads (pF to F).
micro 10-6, nano 10-9, pico 10-12
The capacitance of an object depends only on the
materials it is made of and its geometry.
For a parallel plate capacitor with plates of area A
separated by a distance d, the capacitance is C=0A/d.
The material in between
the plates in this case is
“air.”
C=0A/d.*
0 is the permittivity of
free space (and
approximately of air).
d
*Not an OSE yet—not completely general.
area = A
If an insulating sheet (“dielectric”) is
placed between the plates, the
capacitance increases by a factor K,
which depends on the material in the
sheet. K is called the dielectric
constant of the material.
Thus C = K0A / d is true in general
(OSE) because K is 1 for a vacuum,
and approximately 1 for air. You can
also define  = K 0 and write C = A /
d (we won’t).
OSE:
C = K0A / d
dielectric
Example
(a) Calculate the capacitance of a capacitor whose plates are
20 x 3 cm and are separated by a 1.0 mm air gap.
OSE:
C = K0A / d
C = 1(8.85x10-12)(0.2x0.03) / 0.001
C =53x10-12 F
C = 53 pF
d = 0.001
area = 0.2 x 0.03
If you keep everything in SI (mks) units, the result is automatically in SI units.
(b) What is the charge on each plate of the capacitor is
connected to a 12 volt* battery?
0V
OSE:
Q = CV
Q = (53x10-12)(12)
V= 12
Q = 6.4x10-10 C
+12 V
*Remember, it’s the potential difference that matters.
If you keep everything in SI (mks) units, the result is automatically in SI units.
(c) What is the electric field between the plates?
OSE: E =
ΔVif
d
, away from +
V
E=
d
12 V
E=
0.001 m
0V
V= 12
E
d = 0.001
E = 12000 V/m, away from +
+12 V
If you keep everything in SI (mks) units, the result is automatically in SI units.
Anybody confused by this symbol “V” I’ve been using?
Maybe you should be!
V is the symbol for electrical potential, also called potential.
The units of V are volts, abbreviated V.
V is also the voltage of a battery, or the voltage in an
electrical circuit.
Actually, the V of a battery is really the potential difference,
measured in volts, between the terminals of a battery.
Nowhere have I called V an energy. The symbol V is often
used for potential energy, but I will not do that in this course.
I count 4 different meanings for V. You have to be aware of the context!
Dielectrics
The dielectric is the thin insulating sheet
in between the plates of a capacitor.
dielectric
Any reasons to use a dielectric (other than to make your
life more complicated)?
Lets you apply higher voltages (so more charge).
Lets you place the plates closer together (make d
smaller).
Increases the value of C because K>1.
OSE:
Q = CV
OSE:
C = K0A / d
Visit howstuffworks to read about capacitors and learn their
advantages/disadvantages compared to batteries!
V=0
Example
A capacitor connected as shown
acquires a charge Q.
V
While the capacitor is still connected
to the battery, a dielectric material is
inserted.
V
Will Q increase, decrease, or stay the same?
Why?
Storage of Electric Energy
The electrical energy stored in a capacitor is
OSE:
Ucapacitor = QV/2 = CV2/2 = Q2/2C
It is no accident that we use the symbol U for the energy
stored. This is another kind of potential energy. Use it in
your energy conservation equations just like any other form
of energy!
The derivation follows, for those who love calculus.
dW = V dq
q
dW =
dq
C

W
0
dW =

Q
0
q
dq
C
1 Q
W =  q dq
C 0
2
1q
W=
C 2
Q
0
1 Q2

2 C
work to move charge dq through potential V
(from last lecture)
from Q=CV
work to put charge Q on capacitor
C is constant
the other forms follow from definitions
Example: A camera flash unit stores energy in a 150 F
capacitor at 200 V. How much electric energy can be stored?
Ucapacitor = CV2/2
Ucapacitor = (150x10-6)(200)2 / 2
Ucapacitor = 3.0 J
Big concepts from this chapter:
● We defined electric potential. This lets us calculate electric
potential energies. A new component to add to your alreadyexisting conservation of energy toolbox.
● Electric field and potential are related. A new component
to add to your already-existing electric field toolbox.
● Capacitance. Yet another conservation of energy variation.
● Electron volt, electric dipoles—important, but applications
of fundamental concepts.
A big idea (and OSE) from mechanics:
Ef – Ei = (Wother )
if
Official Starting Equations:
Va = (PE)a /q
Wif = q Vif
PEif= q Vif
E=
ΔVif
d
,
away from +
VQ = kQ/r = Q/40r.
Vnet =  Vi.
Vdipole =
kp cosθ
r2
for r>>l.
Q = CV
C = K0A / d
Ucapacitor = QV/2 = CV2/2 = Q2/2C