Φ21 Fall 2006
HW7 Solutions
1 Potential Dierence and Potential Near a Charged Sheet
Let
A = (x1 , y1 )
and
B = (x2 , y2 )
be two points near and on the same
side of a charged sheet with surface charge density
+σ
(Figure 1). The
E = 2²σ0
everywhere, and the eld points away from the sheet, as shown in the
electric eld
~
E
due to such a charged sheet has magnitude
diagram.
Part A. What is the potential dierence
VAB = VA − VB
between points
A and B? Express your answer in terms of some or all of
and
y2 .
E , x1 , y1 , x2 ,
R
~ · d~s. (Note
VAB = VA − VB = − E
(x2 , y2 ) to (x1 , y1 ) because VA − VB = Vfinal −
Solution: The potential change is
that the integral is from
Vinitial
makes the path in that direction.) Since any path is allowed, take
the integral to be a path straight across to
(x1 , y2 ) then down to the nal
Figure 1: A surface charge
its
~
E
σ
and
eld.
point (Not the blue curve). The rst part of the integral is zero because
~ ⊥ d~s.
E
The second part of the integral simplies because
dot product becomes a simple product) and because
the second segment,
ds = dy .
Z
VAB = −
~
E
~ k d~s
E
(the
is uniform (it comes out of the integral). Also, along
Z
~ · d~s = −E
E
dy = −E∆y = −E (y1 − y2 )
Recall that the potential dierence, the quantity asked for in Part A, is a well-dened quantity for any
situation. The potential, however, is only dened once you pick a point as the zero-potential point. Dierent
choices simply change the potential by an additive constant, so the potential dierence will stay the same,
regardless of what point you designate as having zero potential.
Part B. If the potential at
some positive distance
Solution: Substitute
y1
y = ±∞
is taken to be zero, what is the value of the potential at a point
y2 = ∞
to obtain
Solution: Substitute
y1
at
VA = −E (y1 − ∞) = ∞.
Part C. Now take the potential to be zero at
some positive distance
VA
from the surface of the sheet?
y=0
instead of innity. What is the value of
VA
at point A
from the sheet?
y2 = 0
to obtain
VA = −Ey1 .
2 Introduction to Capacitance
Learning Goal: To understand the meaning of capacitance and ways of calculating capacitance
When a positive charge
q
is placed on a conductor that is insulated from ground, an electric eld emanates
from the conductor to ground, and the conductor will have a nonzero potential
V
relative to ground. If more
charge is placed on the conductor, this voltage will increase proportionately. The ratio of charge to voltage
is called the capacitance
C
of this conductor:
C = q/V .
Capacitance is one of the central concepts in electrostatics, and specially constructed devices called capacitors
are essential elements of electronic circuits. In a capacitor, a second conducting surface is placed near the
rst (they are often called electrodes), and the relevant voltage is the voltage between these two electrodes.
This tutorial is designed to help you understand capacitance by assisting you in calculating the capacitance
of a parallel-plate capacitor, which consists of two plates each of area
A separated by a small distance d with
air or vacuum in between. In guring out the capacitance of this conguration of conductors, it is important
to keep in mind that the voltage dierence is the line integral of the electric eld between the plates.
1
Part A. What property of objects is best measured by their capacitance?
◦
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•
◦
ability to conduct electric current
ability to distort an external electrostatic eld
ability to store charge
ability to store electrostatic energy
Capacitance is a measure of the ability of a system of two conductors to store electric charge and energy.
Q
Capacitance is dened as C =
V . This ratio remains constant as long as the system retains its geometry
and the amount of dielectric does not change. Capacitors are special devices designed to combine a large
capacitance with a small size. However, any pair of conductors separated by a dielectric (or vacuum) has
some capacitance. Even an isolated electrode has a small capacitance. That is, if a charge
V with respect to ground will change, and the ratio VQ is its capacitance C .
Q
is placed on it,
its potential
Part B. Assume that charge
−q
is placed on the top plate, and
+q
is
placed on the bottom plate. What is the magnitude of the electric eld
E between the plates?
Solution: Use Gauss's Law and the fact that
E = 0 outside a capacitor.
a enclosing only
Draw a small Gaussian can with cross-sectional area
the positive plate. Remember that the surface charge on the capacitor is
σ = q/A where A is the area of the capacitor. The ux integral simplies
~ is zero or is parallel to the Gaussian surface everywhere except
E
~ is perpendicular
for a surface area a inside the capacitor. In this region, E
because
to the surface, pointing out, and is uniform.
portion of the surface charge
The enclosed charge is a
σ.
Figure
I
~ · dA
~ =
E
Ea =
E
=
2:
A
parallel-plate
Qenc
²0
σa
²0
σ
q
=
²0
A²0
simple
capacitor.
air-lled
The
blue Gaussian surface has crosssectional area
a.
Part C. What is the voltage V between the plates of the capacitor?
Solution: The voltage is the electric potential dierence, calculated from a known
~
E
by doing a line integral.
The integral is between the plates along any path, so we choose a straight line along an electric eld line to
make it easy.
~ k d~s
E
so the dot product turns into a simple product, and
Z
the integral.
∆V = −
is uniform so it comes out of
Z
~ · d~s = −
E
The minus sign means the initial point (the
Part D. Now nd the capacitance
E
C
in the introduction and constants like
+
E ds = −E∆s = −Ed = −
plate) is at higher potential than the nal point.
of the parallel-plate capacitor. Express
²0 .
Solution: The capacitance is simply
C=
qd
A²0
q
q
²0 A
=
=
V
qd/A²0
d
2
C
in terms of quantities given
Part E. Consider an air-lled charged capacitor. How can its capacitance be increased? (Hint: What does
capacitance depend on? Capacitance depends on the inherent properties of the system of conductors, such
as its geometry and the presence of dielectric, not on the charge placed on the conductors.)
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•
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Increase the charge on the capacitor.
Decrease the charge on the capacitor.
Increase the spacing between the plates of the capacitor.
Decrease the spacing between the plates of the capacitor.
Increase the length of the wires leading to the capacitor plates.
Part F. Consider a charged parallel-plate capacitor.
Which combination of changes would quadruple its
capacitance?
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Double the charge and double the plate area.
Double the charge and double the plate separation.
Halve the charge and double the plate separation.
Halve the charge and double the plate area.
Halve the plate separation; double the plate area.
Double the plate separation; halve the plate area.
3 Problem 30.64
What are the charge on and the potential dierence across each capacitor
in Figure 3?
Solution: There are two ways to two this problem. (1) Systematically
reduce the capacitors to series and parallel equivalents then start solving,
or (2) Use Kircho 's Laws. This solution uses Kircho 's Laws. Kircho 's
Current Law (Junction Law) states that the current going into a junction
equals the current leaving. Since the charge on the capacitors all
1 comes
from the current of the circuit (by integrating), this law applies to the
capacitor charges as well.
Figure 3: A circuit with capacitors.
C1 , C2 ,
X
=
Iout
Kircho 's Current Law at the point where
X
Iin
Q1 + Q2
C1 V1 + C2 V2
and
C3
join:
= Q3
= C3 V3
Kircho 's loop law clockwise around the loop with the battery,
C3 :
X
∆V
= 0
9 V − V1 − V 3
= 0
C1 ,
and
loop
Kircho 's loop law clockwise around the loop with
C1
and
C2
is simply the rule that Voltages are equal
across parallel components.
V1 = V2
We now have 3 equations for 3 unknowns.
V2
and
V3
are both solved for in terms of
the rst equation.
C1 V1 + C2 V1
V1
1 We're
= C3 (9 − V1 )
9C3
=
= 1V
C1 + C2 + C3
saying that the capacitors are uncharged when they are inserted into the circuit.
3
V1 ,
so substitute into
The rest of the unknowns now fall into place.
V2 = V1 = 1
Q1 = C1 V1 = 4 µC
V3 = 9 − V1 = 8 V
Q2 = C2 V2 = 12 µC
Q3 = C3 V3 = 16 µC
4 Problem 30.23
(Note, the numbers in this problem have been randomized.)
µF capacitor to 47.0 µC?
¢
¡
¢
V = Q/C = 47 × 10−6 C / 2.30 × 10−6 F = 20.4 V
What is the emf of a battery that will charge a 2.30
Solution:
Q = CV
¡
5 Problem 30.45
(Note: The numbers in this problem are randomized.)
The electric potential in a region of space is
What is the strength of the electric eld at
eld at
V = (210x2 − 200y 2 ) V, where x and y
(3.00 m, 3.00 m)? Part B. What is the
are in meters. Part A.
direction of the electric
(3.00 m, 3.00 m)?
Solution: The electric eld is the gradient of the electric potential. In two dimensions:
= −∇V
µ
¶
dV
dV
= −
î +
ĵ
dx
dy
h
i
~ 3) = − 420xî − 400y ĵ
E(3,
x=3,y=3
³
´
=
−1260î + 1200ĵ N/C
~
E
The magnitude is
¯ ¯ q
√
¯~¯
E = ¯E
¯ = Ex2 + Ey2 = 1260 + 12002 = 1740
V/m.
The direction relative to the x-axis is found from the dot product.
E cos θ
θ
~ · î = Ex
= E
= cos−1 (Ex /E)
= cos−1 (1260/1740) = 0.761 rad = 43.6◦
4