Prof. Anchordoqui
Problems set # 5
Physics 169
March 17, 2015
1. (i) Three capacitors are connected to a 12.0 V battery as shown in Fig. 1 Their capacitances
are C1 = 3.00 µF, C2 = 4.00 µF, and C3 = 2µF. Find the equivalent capacitance of this set of
capacitors. (ii) Find the charge on and the potential difference across each.
Solution (i) Using the rules for combining capacitors in series and in parallel, the circuit is
reduced in steps as shown in Fig. 2. The equivalent capacitor is hown to be a 2.00 µF capacitor.
(ii) From Fig. 2 (right panel) it follows that Qac = Cac (∆V )ac = 2.00 µF 12.0V = 24.0 µC. From
Fig. 2 (middle panel) it follows that Qab = Qbc = Qac = 24 µC and so the charge on the 3.00µF
ab
capacitor is Q3 = 24.0 µC. Continuing to use Fig. 2 (middle panel) it follows that (∆V )ab = Q
Cab =
24.0 µC
6.00 µF
24.0 µC
bc
= 4.00 V, and (∆V )3 = (∆V )bc = Q
Cbc = 3.00 µF = 8.00 V. Finally, from Fig. 2 (left panel)
it follows that (∆V )4 = (∆V )2 = (∆V )ab = 4.00 V, Q4 = C4 (∆V )4 = 4.00 µF 4.00 V = 16.0 µC,
and Q2 = C2 (∆V )2 = 2.00 µF4.00 V = 8.00 µC.
2. A dielectric rectangular slab has length s, width w, thickness d, and dielectric constant κ.
The slab is inserted on the right hand side of a parallel-plate capacitor consisting of two conducting
plates of width w, length L, and thickness d. The left hand side of the capacitor of length L − s is
empty, see Fig. 3. The capacitor is charged up such that the left hand side has surface charge densities ±σL on the facing surfaces of the top and bottom plates respectively and the right hand side
has surface charge densities ±σR on the facing surfaces of the top and bottom plates respectively.
The total charge on the entire top and bottom plates is +Q and −Q respectively. The charging
battery is then removed from the circuit. Neglect all edge effects. (i) Find an expression for the
magnitude of the electric field EL on the left hand side in terms of σL , σR , κ, s, w, L, 0 , and d as
needed. (ii) Find an expression for the magnitude of the electric field ER on the right hand side in
terms of σL , σR , κ, s, w, L, 0 , and d as needed. (iii) Find an expression that relates the surface
charge densities σL and σR in terms of κ, s, w, L, 0 , and d as needed. (iv) What is the total
charge +Q on the entire top plate? Express your answer in terms of σL , σR , κ, s, w, L, 0 , and d
as needed. (v) What is the capacitance of this system? Express your answer in terms of κ, s, w,
L, 0 , and d as needed. (vi) Suppose the dielectric is removed. What is the change in the stored
potential energy of the capacitor? Express your answer in terms of Q, κ, s, w, L, 0 , and d as needed.
σR
Solution: (i) Using Gauss’ law EL = σ0L . (ii) Using Gauss law for dielectrics ER = κ
. (iii) The
0
σL d
σR d
potential difference on the left side is EL d = 0 . On the right hand side it is ER d = κ0 . Since
these must be equal we must have σR /κ = σL . (iv) Q = σL (L − s)w + σR sw. h(v) The potential
i
σL (L−s)w+σR sw
Q
σR
0 w
difference is EL d = σL0d , so the capacitance is C = |∆V
=
=
(L
−
s)
+
s
=
d
σL
|
σL d/0
0 w
[(L
−
s)
+
κs].
(vi)
Since
the
battery
has
been
removed,
the
charge
on
the
capacitor
does
not
d
Q2
1 Q2
1 Q2
1
− C1
2oC0 − 2 Cn = 2
C0 o
(L−s)+κs−L
(κ−1)s
Q2 d
L[(L−s)+κs] = 20 w L[(L−s)+κs] .
change when we do this, and the change in the energy stored is =
Q2
2
n
d
0 Lw
−
d
0 w[(L−s)+κs]
o
=
Q2 d
20 w
h
1
L
−
1
(L−s)+κs
i
=
Q2 d
20 w
n
=
3. (i) Consider a plane-parallel capacitor completely filled with a dielectric material of dielectric
constant κ. What is the capacitance of this system? (ii) A parallel-plate capacitor is constructed
by filling the space between two square plates with blocks of three dielectric materials, as in Fig. 4.
You may assume that l d. Find an expression for the capacitance of the device in terms of the
plate area A, d, κ1 , κ2 , and κ3 .
Solution: (i) The capacitance is C = κd0 A = κC0 . (ii) The capacitor can be regarded as be0 A/2
0 A/2
, and C3 = κ3 d/2
, with C2 and
ing consisted of three capacitors, C1 = κ1 0dA/2 , C2 = κ2 d/2
C3 connected in series, and the combination connected in parallel with C1 . Thus, the equivalent
capacitance is C = C1 +
1
C2
+
1
C3
−1
= C1 +
C2 C3
C2 +C3
=
κ1 0 A/2
d
+
0 A
d
κ2 κ3
κ2 +κ3
=
0 A
d
κ1
2
+
κ2 κ3
κ2 +κ3
.
4. A model of a red blood cell portrays the cell as a spherical capacitor – a positively charged
liquid sphere of surface area A, separated by a membrane of thickness t from the surrounding
negatively charged fluid. Tiny electrodes introduced into the interior of the cell show a potential
difference of 100 mV across the membrane. The membranes thickness is estimated to be 100 nm
and its dielectric constant to be 5.00. (i) If an average red blood cell has a mass of 1.00 × 10−12 kg,
estimate the volume of the cell and thus find its surface area. The density of blood is 1, 100 kg/m3.
(ii) Estimate the capacitance of the cell. (iii) Calculate the charge on the surface of the membrane.
How many electronic charges does this represent?
Solution (i) The volume is V =
inner radius of the capacitor is r =
2/3
m
ρ
1.00×10−12 kg
1,100 kg/m3
1/3
=
3V
4π
2/3
= 9.09 × 10−16 m3 . Since V = 34 πr3 , the
= 6.01 × 10−6 m and the surface area is A = 4πr2 =
3
= 4π 4π
9.09 × 10−16 m3
= 4.54 × 10−10 m2 . (ii) The outer radius of the capacitor
4π 3V
4π
is R = r + t = 6.11 × 10−6 m, where t = 100 nm is the thickness of the membrane. The capacitance
Rr
is C = 4πκ0 R−r
= 2.04×10−14 F. (iii) Q = C ∆V = 2.04×10−14 F 100×10−3 V = 2.04×10−15 C,
and the number of electronic charges is n =
Q
e
=
2.01×10−14 C
1.60×10−19 C
= 1.27 × 104 .
5. A capacitor consists of two concentric spherical shells. The outer radius of the inner shell
is a = 0.1 m and the inner radius of the outer shell is b = 0.2 m. (i) What is the capacitance
C of this capacitor? (ii) Suppose the maximum possible electric field at the outer surface of the
inner shell before the air starts to ionize is Emax (a) = 3.0 × 106 V · m−1 . What is the maximum
possible charge on the inner capacitor? (iii) What is the maximum amount of energy stored in this
capacitor? (iv) What is the potential difference between the shells when E(a) = 3.0 × 106 V · m−1 ?
Solution: (i) The shells have spherical symmetry so we need to use spherical Gaussian surfaces.
Space is divided into three regions (I) outside r ≥ b, (II) in between a < r < b and (III) inside
~ = E r̂). In regions I and III these
r ≤ a. In each region the electric field is purely radial (that is E
Gaussian surfaces contain a total charge of zero, so the electric fields in these regions must be zero
as well. In region II, we choose the Gaussian sphere of radius r shown in Fig. 5. The electric flux on
v
~ · dA
~ = EA = E · 4πr2 . The enclosed charge is Qenc = +Q, and the electric field
the surface is E
is everywhere perpendicular to the surface. Thus Gauss law becomes E · 4πr2 = Q0 ⇒ E = 4πQ0 r2 .
That is, the electric field is exactly the same as that for a point charge. Summarizing:
(
~ =
E
Q
r̂
4π0 r2
~0
for a < r < b
.
elsewhere
We know the positively charged inner sheet is at a higher
potential
so
we shall calculate ∆V =
Ra
Ra Q
Q
Q a
1
1
~
V (a) − V (b) = − b E · d~s = − b 4π0 r2 dr = 4π0 r = 4π0 a − b > 0, which is positive as
b
Q
|∆V |
Q
=
( a1 − 1b )
F. Note that the units of capacitance are 0
we expect. We can now calculate the capacitance using the definition C =
=
Q
4π0
4π0
0.1 m 0.2 m
−11
0 ab
= 4π
b−a = 8.99×109 Nm2 /C2 0.1 m = 2.2 × 10
( a1 − 1b )
times an area ab divided by a length b − a, exactly the same units as the formula for a parallel-plate
capacitor C = 0 A/d. Also note that if the radii b and a are very close together, the spherical
capacitor begins to look very much like two parallel plates separated by a distance d = b − a and
area A ≈ 4π
a+b
2
2
≈ 4π
a+a 2
2
= 4πa2 ≈ 4πab. So, in this limit, the spherical formula is the
same at the plate one C = limb→a
4π0 ab
b−a
Therefore the maximum charge is Qmax =
stored is Umax =
Q2max
2C
=
(3.3×10−6 C)2
2·2.2×10−11 F
=
0 4πa2
d
Q
0 A
d . (ii) The electric field is E(a) = 40 a2 .
6 V·m−1 (0.1 m)2
. (iii) The energy
4π0 Emax (a)a2 = 3.0×10
8.99×109 N·m2 /C2
2.5 × 10−1 J. (iv) We can find the potential difference
≈
=
two different ways. Using the definition of capacitance we have that
|DeltaV Q
=
C
=
4π0 E(a)a2 (b−a)
4π0 ab
=
6
−1 (0.1 m)2
= 3.0×10 V·m
= 1.5 × 105 V. We already calculated the potential difference in
0.2
m
Q
Q
Q
1
1
part (i): ∆V = 4π
E(a)a2 Substitute this into
a − b . Recall that E(a) = 4π0 a2 or 4π0 =
0
b−a
our expression for potential difference yielding ∆V = E(a)a2 a1 − 1b = E(a)a2 b−a
ab = E(a)a b in
E(a)a(b−a)
b
agreement with our result above.
6. A parallel plate capacitor has capacitance C. It is connected to a battery until is fully charged,
and then disconnected. The plates are then pulled apart an extra distance d, during which the
measured potential difference between them changed by a factor of 4. What is the volume of the
dielectric necessary to fill the region between the plates? (Make sure that you give your answer only
in terms of variables defined in the statement of this problem, fundamental constants and numbers).
Solution How in the world do we know the volume? We must be able to figure out the crosssectional area and the distance between the plates. The first relationship we have is from knowing
the capacitance: C = 0xA where x is the original distance between the plates. Make sure you dont
use the more typical variable d here because that is used for the distance the plates are pulled
apart. Next, the original voltage V0 = Ex, which increases by a factor of 4 when the plates are
moved apart by a distance d, that is, 4V0 = E(x + d). From these two equations we can solve
for x: 4V0 = 4Ex = E(x + d) ⇒ x = d/3. Now, we can use the capacitance to get the area,
and multiply that by the distance
the plates (now x + d) to get the volume, i.e., volume
between
xC
dC d
4d2 C
= A(x + d) = 0 (x + d) = 30 3 + d = 90 .
7. Consider two nested cylindrical conductors of height h and radii a and b respectively. A
charge +Q is evenly distributed on the outer surface of the pail (the inner cylinder), −Q on the
inner surface of the shield (the outer cylinder). See Fig. 6. You may ignore edge effects. (i) Calculate the electric field between the two cylinders (a < r < b). (ii) Calculate the potential difference
between the two cylinders. (iii) Calculate the capacitance of this system, C = Q/∆V (iv) Numerically evaluate the capacitance, given: h ' 15 cm, a ' 4.75 cm and b ' 7.25 cm. (v) Find
the electric field energy density at any point between the conducting cylinders. How much energy
resides in a cylindrical shell between the conductors of radius r (with a < r < b), height h, thickness
dr, and volume 2πrhdr? Integrate your expression to find the total energy stored in the capacitor
and compare your result with that obtained using UE = 21 C(∆V )2 .
v
~ ·dA
~=
E
Solution: (i) For this we use Gauss law, with a Gaussian cylinder of radius r, height l:
Qinside
Q
1 Q
2πrlE = 0 = 0 h l ⇒ E(r) = 2πr0 h with a < r < b. (ii) The potential difference between the
Q
2π0 h
ln
b
a
.
(iii) C =
10−14 cm−1 · F ln(7.25
Q
∆V
=
Q
ln( ab )
Q
2π0 h
=
2π0 h
.
ln( ab )
(iv) C
Ra
Q
0
b 2πr0 0 h dr
2π0 h
= ln(b/a)
=
outer shell and the inner cylinder is ∆V = V (a) − V (b) = −
= −
Q
2π0 h
a
ln r0 =
b
15 cm · 2π · 8.85 ×
1
cm/4.75 cm) ' 20 pF. (v) The total energy density stored in the capacitor
2
2
Q2 dr
Q
Q
1
.
Then
dU
=
udV
=
2πrhdr = 4π
. Integrating we
is u = 12 0 E 2 = 21 0 2πr
2 0 2π0 rh
0h
0h r
Rb
R b Q2 dr
2
Q
find that U = a dU = a 4π0 h r = 4π
ln(b/a). From part (iii) C = 2π0 h/ ln(b/a), therefore
0h
Rb
R b Q2 dr
2
Q
Q2
U = a dU = a 4π0 h r = 4π0 h ln(b/a) = 2C = 12 C(∆V )2 , which agrees with that obtained above.
8. A capacitor is made of three sets of parallel plates of area A, with the two outer plates on the
left and the right connected together by a conducting wire as shown in Fig. 7. The outer plates are
separated by a distance d. The distance from the middle plate to the left plate is z. The distance
from the inner plate to the right plate is d − z. You may assume all three plates are very thin
compared to the distances d and z . Neglect edge effects. (i) The positive terminal of a battery
is connected to the outer plates. The negative terminal is connected to the middle plate. The
potential difference between the outer plates and inner plate is ∆V = V (z = 0) − V (z). Find the
capacitance of this system. (ii) Find the total energy stored in this system.
Solution When the battery is connected positive charges QL and QR appear on the outer plates
(on the inner facing surfaces) and negative charges −QL and −QR respectively appear on each side
of the inner plates. The plates on the left in Fig. 8 act as a capacitor arranged as show in Fig. 9.
This is a parallel plate capacitor with capacitance CL = QL /∆V . The plates on the right also act
as a capacitor with capacitance CR = QR /∆V . All the plates have the same area A so neglecting
edge effects we can use Gauss law to show that the magnitude of the electric field on the left is
EL = QL /0 A for parallel plates on the left. Because the electric field is uniform the potential
difference is ∆V = EL z = Q0LAz . Therefore the positive charge on the left satisfies QL = ∆V z0 A . A
similar argument applied to the plates on the right show that ∆V = ER (d − z) = Q0RA (d − z). Hence
0 A
the positive charge on the right satisfies QR = ∆V
d−z . The total positive charge at the higher
+QR
potential is the sum QL + QR capacitance of the capacitor is C = QL∆V
. Wecan nowsubstitute
0 A
∆V 0 A
0 Ad
1
1
the equations above and solve for the capacitance C = ∆V
∆V z + ∆V (d−z) = 0 A z + d−z = z(d−z) .
Note: Because the outer plates are at the same potential you can think of these capacitors as
QL
QR
connected in parallel. Therefore the equivalent capacitance is C = CL + CR = ∆V
+ ∆V
agreeing
with our result above. (ii) The stored energy is just U = 12 C(∆V )2 =
0 Ad(∆V )2
2z(d−z) .
9. Two flat, square metal plates have sides of length L, and thickness s/2, are arranged parallel
to each other with a separation of s, where s L so you may ignore fringing fields. A charge
Q is moved from the upper plate to the lower plate. Now a force is applied to a third uncharged
conducting plate of the same thickness s/2 so that it lies between the other two plates to a depth
x, maintaining the same spacing s/4 between its surface and the surfaces of the other two. The
configuration is shown in Fig. 10. (i) What is the capacitance of this system? (ii) How much energy
is stored in the electric field? (iii) If the middle plate is released, it starts to move. Will it move to
the right or left? [Hint: If the middle plate moves to the left by a small positive amount ∆x, the
change in potential energy is approximately ∆U = (dU/dx)∆x. Will the stored potential energy
increase or decrease? (iv)) For a small displacement ∆x in the direction you determined in part
(iii), find the horizontal force exerted by the charge distribution on the outer plates acting on the
charges on the middle plate that cause it to move.
Solution: (i) Divide the plates into left and right sides. The area on the left is AL = L(L − x)
and the area on the right is AR = Lx. The charge densities on the two sides are shown in Fig.11.
The charge on the left isQL = σL L(L − x). The charge on the right is QR = σR (Lx). The electric field on the two sides is shown in Fig. 11. We can calculate the electric field on both sides
using Gauss law (neglecting edge effects). We find EL = σ/0 , ER = σ/0 . Because the electric
field is uniform on both sides (neglecting edge effects) ∆VL = EL s = σL s/0 . Note that on the
σR s
right side the electric field is zero in the middle conductor so ∆VR = σRs/4
+ σRs/4
20 . The
0
0
potential difference on the two sides are the same because the upper and lower plates are held
at the same potential difference ∆V ≡ ∆VL = ∆VR . Therefore we can solve for a relationship
between the charge densities on the two sides by setting the above equations equal to each other
σL s
σR s
σR
0 = 20 ; hence σL = 2 . The capacitance of the system is the total charge divided by the
+QR
potential difference C = QL∆V
. Using our results for the charges and the potential difference
σL L(L−x)+σR Lx
we have that C =
. We can now substitute the value of σL and find that the
σL s/0
capacitance is C =
[L(L−x)+2Lx]
σL(L−x)/2+σR Lx
= 0 L(L+x)
. (ii)
= )
s
s
σR s/(20 )
2
2
Q
Q s
= 2C
= 20 L(L+x)
. (iii) If the middle plate moves
The energy stored in the
capacitor is then U
a positive distance ∆x to
the left resulting in a larger value of x, then the stored potential energy changes by an amount
Q2 s
∆U = dU
dx ∆x = − 20 L(L+x)2 ∆x. This change is negative hence the middle plate will move to the
left decreasing the stored potential energy of the system. Because the change in potential energy
Q2 s
is negative, the work done by the electric forces is positive W = 20 L(L+x)
2 ∆x. (iv) For a small
displacement in the positive x-direction, the work done is equal to W = F ∆x. Therefore the horizontal force exerted by the charges on the outer plates acting on the charges on the middle plate
Q2 s
is given by F = 20 L(L+x)
2.
10. A flat conducting sheet A is suspended by an insulating thread between the surfaces formed
by the bent conducting sheet B as shown in Fig. 12. The sheets are oppositely charged, the difference in potential, in volts, is ∆V . This causes a force F , in addition to the weight of A, pulling A
downward. (i) What is the capacitance of this arrangement of conductors as a function of y, the
distance that plate A is inserted between the sides of plate B ? (ii) How much energy is needed to
increase the inserted distance by ∆y? (iii) Find an expression for the difference in potential ∆V
(d) What If? If C 3 is increased, what happens to the
charge stored by each of the capacitors?
2
C1
C2
C3
Figure P26.22
Figure 1: Problem 1.
in terms of F and relevant dimensions shown in the figure.
theby circuit
inlargeFigure
P26.23,
where
23. Consider
Solution: a) We begin
assuming the shown
plates are very
and use Gauss
law to calculate
v
q
~ · dA
~ =
E
theC
electric
field between
theC
plates
choice
of Gaussian
surface
isCapacitor
shown in
F, Our
"
F,
!
3.00
"
and
#V
!
20.0
V.
v
1 ! 6.00
2
q
σA
~ · dA
~ = EAcap and
~ = σ î, beFig. 12. Then
E
= . Thus, Gauss’ law implies that E
C 1theisplates.
firstNote
charged
by the
closing
S 1negative
. Switch
S 1 is
tween
that the potential
difference
betweenof
theswitch
positive and
plates is
R x=0
(−)]
V (+)
− V (−)opened,
= − x=s Ex dx
= σs
So thecharged
surface chargecapacitor
density is equal to
= [V (+)−V
. to
. the
s
then
and
is σconnected
The area between the plates is yb. Note that there is a charge on inner surface of the surrounding
the
capacitor
by sheets
the isclosing
S 2. the
Calculate
sheet
equaluncharged
to σyb, so the total
charge on the outer
Q = 2σyb. of
Therefore
capaci2σyb
Q
2 yb
tance is C(y) = V (+)−V (−) = σs = s . (ii) If the inserted distance is increased by ∆y then
the initial charge acquired by C 1 and the final charge on
∆C = 2 s∆yb . Because the charge on the plates is fixed, the energy stored in the capacitor is given
Q
each
capacitor.
. Hence, when the inserted distance is increased by ∆y, the energy stored between
by U
(y) = 2C(y)
enc
0
enc
0
0
0
0
0
0
0
0
2
dU
dC ∆C
Q2
− 2C
2 ∆C
(−)]2 20 ∆yb
− [V (+)−V
.
2
s
the plates decreases by ∆U =
=
=
(iii) This decrease in
energy is used to pull the hanging plate in between the two positive charged plates. The work done
in pulling the hanging plate a distance ∆y is given by ∆W = Fy ∆y. By conservation of energy
(−)]2 20 ∆yb
0 = ∆U + ∆W = − [V (+)−V
+ Fy ∆y. We can solve this equation for the y-component of
2
s
the force Fy =
[V (+)−V (−)]2 20 b
2
s
∆V
or else find V (+) − V (−) =
C1
S1
q
sFy
0 b .
C2
S2
Figure P26.23
24. According to its design specification, the timer circuit
2
2
group of capacitors in Figure P16.31. (b) Find the charge on and the
potential difference across each.
Solution:
Problem 3 Dielectric
Using the rules for combining capacitors in series and in parallel,
the circuit is reduced in steps as shown below. The equivalent capacitor is
s , width w , thickness d , and dielectric
rectangular
2 .00length
µ F capacitor.
shown toslab
be a has
A dielectric
constant
! . The slab is inserted4.00
on µF
the right hand side of a parallel-plate capacitor consisting of
w , µF
two conducting plates of width 3.00
length L , and thickness d . The left hand side of the
6.00 µF 3.00 µF
2.00 µF
capacitor of length L ! s is empty. The capacitor
is charged up such
that the left hand
a
side has surface charge
densitiesb ±! cL on the afacing
of athe top and
bottom plates
b surfaces
c
c
respectively and the right
hand side has surface charge densities ±! R on the facing
2.00 µF
surfaces of the top and bottom plates respectively. The total charge on the entire top and
12.0 V
12.0 V
12.0 V
bottom plates is +Q and !Q
respectively.
The
charging
battery
is
then removed from the
Figure 1
Figure 2
Figure 3
of problem 1.
circuit. Neglect all(b)edge
From effects.
Figure 3: Figure
Q ac =2:CSolution
ac ( !V ) ac = ( 2 .00 µ F )( 12.0 V ) = 24.0 µ C
From Figure 2: Q ab = Q bc = Q ac = 24.0 µC
Thus, the charge on the 3.00 µF capacitor is Q 3 = 24.0 µ C
Continuing to use Figure 2, ( !V ) ab =
Q ab 24.0 µC
=
= 4.00 V ,
C ab 6.00 µF
Q bc 24.0 µ C
=
= 8.00 V
C bc 3.00 µF
From Figure 1, ( !V )4 = ( !V ) 2 = ( !V ) ab = 4.00 V
and Q 4 = C4 ( !V ) 4 = ( 4.00 µF ) ( 4.00 V ) = 16.0 µC
Q 2 = C2 ( !V ) 2 = ( 2.00 µF )( 4.00 V ) = 8.00 µC
and ( !V )3 = ( !V ) bc =
828
C H A Chapt16,
P T E R 2 6 •Problem-44:
Capacitance
and Dielectrics
Two capacitors C = 25.0 µF and C
1
2 = 5.00 µF are connected in p
and charged with a 100-V Figure
power supply.
(a)
Calculate
the
total
energy
stored in the two capacitors.
3: Problem 2.
potential
difference
would
be
required
across
the
same
two
capacitors
connected
in series
in order t
expression
for the magnitude of the electric field EL on the left hand
side
combination store the same energy as in (a)?
a) Find an
in terms ofS!
o lLu ,t i !
o nR: , ! , s , w , L , ! 0 , and d as needed.
!
(a) When connected in parallel, the energy stored is
b)
1
1
1
2
2
W = for
C 1( !V
)2magnitude
+ C 2 ( !V ) = of( C
1+ C
2 ) ( !V )
Find an expression
the
the
electric
field ER
2
2
2
w ,) !10
L , "6!F0 ,(100
side in terms of ! L , ! R= ,1 !(25.0
, s+, 5.00
andV )d2 =as0.150
needed.
J
2
[
]
κ
(b) When connected in series, the equivalent capacitance
is
2
on the right hand
d/2
κ
d
c) Find an
charge densities ! L and ! R in terms of
2
1
= 2 C eq (d!Vas
) ,needed.
the potential difference required to store the same energy as in part (a) ab
! , s , w , From
L , !W
0 , and
κ3
2W
2 ( 0.150 J )
!V =
=
= 268 V
#6
C
4.17
"10
F
eq
d) What is the total charge +Q on the entire top plate? Express your answer in terms
of ! L , ! R , ! , s , w , L , ! 0 , and d as needed.
'1
1 $
! 1
C eq = #
+
& µF = 4.17 µF
" 25.0relates
expression that
5.00 % 1the surface
!/2
Figure
P26.61
Figure 4: Problem
3.
4
e) What is the capacitance of this system? Express your answer in terms of ! , s ,
w , L , ! 0 , and d as needed.
Figure 5: Problem 5.
nclosed charge is Qenc = +Q , and the electric field is everywhere perpendicu
rface. Thus
Gauss’s Law becomes
Problem 10: Cylindrical Capacitor
Consider two nested cylindrical conductors of height h and radii a & b respectively. A
charge +Q is evenly distributed on the2 outerQsurface of the pailQ
(the inner cylinder), -Q on
E !(the
4"outer
r =cylinder).
$ You
E =may ignore
the inner surface of the shield
2edge effects.
#0
4"# 0 r
s, the electric field is exactly the same as that for a point charge.
marizing:
# Q
! %
r̂ for a < r < b
2
4!"
r 6: Problem
E=
$ between
a)
Calculate the electric
field
two cylinders
(a < r < b).
0 the
Figure
7.
%0
elsewhere
b)
Calculate the potential difference
between the
two cylinders:
&
c)
Calculate the capacitance of this system, C = Q/DV
d)
cm.
Numerically evaluate the capacitance, given: h ≅ 15 cm, a ≅ 4.75 cm and b ≅ 7.25
now the positively charged inner sheet is at a higher potential so we shall calcu
a
et creates a
magnitude
rying oppoall distance
ric field in
E " Q /A !.
o potential.
ial Q d/A!.
of plates is
the wire. Ignore the other conductors of the transmission
line and assume the electric field is everywhere purely
radial.
2d
d
57. Two large parallel metal plates are oriented horizontally
and separated by a distance 3d. A grounded conducting
wire joins them, and initially each plate carries no charge.
Now a third identical plate carrying charge Q is inserted
between the two plates, parallel to them and located a distance d from the upper plate, as in Figure P26.57.
(a) What induced charge appears on each of the two original plates? (b) What potential difference appears between
the middle plate and each of the other plates? Each plate
has area A.
Figure P26.57
58. A 2.00-nF parallel-plate capacitor is charged to an initial
potential difference &Vi " 100 V and then isolated. The
dielectric material between the plates is mica, with a
dielectric constant of 5.00. (a) How much work is required
to withdraw the mica sheet? (b) What is the potential
difference of the capacitor after the mica is withdrawn?
A parallel-plate capacitor is constructed using a
59.
dielectric material whose dielectric constant is 3.00
and whose dielectric strength is 2.00 ' 108 V/m. The
desired capacitance is 0.250 %F, and the capacitor must
withstand a maximum potential difference of 4 000 V. Find
the minimum area of the capacitor plates.
b) Find the total energy stored in this system.
150
125
100
75
50
P26.54, find
m, (b) the
ge on each
e group.
ach of area
stance d "
ected to the
sitive termice of 12.0 V.
what is the
Solution:
A capacitor is made of three sets of parallel plates of area A , with the two outer plates on
the left and the right connected together by a conducting wire.
outer plates
are positive charges Q and Q appear on the outer plates (on
When The
the battery
is connected
L
R
separated by a distance d . The distance from the middle plate to the left plate is z . The
the
inner
facing
surfaces)
and
negative
charges
and
respectively appear on
!Q
!Q
distance from the inner plate to the right plate is d ! z . You may assume all three plates
L
R
side of the inner plate.
are very thin compared to the distances d and z . Neglect edgeeach
effects.
5
Figure 7: Problem 8.
Problem 11 Capacitance of Multiple Plates
z
d
z
a) The positive terminal of a battery is connected to theThe
outer
plates.
Theleft
negative
plates
on the
in the above figure act as a capacitor arranged as show in the figure
terminal is connected to the middle plate. TheFigure
potential
difference
between
the
8:
Solution
of
problem
8.
below.
outer plates and inner plate is !V = V (z = 0) " V (z) . Find the capacitance of this
system.
The plates on the left in the above figure act as a capacitor arranged as show in the figure
below.
Problem
12: Capacitance, Stored Energy, and Electrostatic Work plates on right
plates
on left
This is
Two flat, square metal plates have sides of length L , and thickness s 2 , are
arranged
parallel
to each
with
athe
separation
of sof
,C
where
so you
may ignore
parallel plate
capacitor
with
. The
plates
= sQ<<L 8.L/ !V
Figure
9: other
More
oncapacitance
solution
problem
L
fringing fields. A charge Q is moved from the upper plate
to the lower plate. Now a
on the right
also act as aforce
capacitor
capacitance
. All
the splates
the same area
C R =plate
QRof/the
!V
is applied with
to a third
uncharged conducting
same
thickness
it
2 so thathave
lies between the other two plates to a depth x , maintaining the same spacing s 4
A so neglecting
edge effects we can use Gauss’s Law to show that the magnitude of the
between its surface and the surfaces of the other two.
electric field on the left is E L = QL / ! 0 A for parallel plates on the left. Because the
electric field is uniform the potential difference is
!V = E L z = QL z / " 0 A .
Therefore the positive
charge
on ofthe
satisfies
a) What is the
capacitance
thisleft
system?
Figure 10: Problem 9.
b) How much energy is stored in the electric field?
c) If the middle plate is released, it starts to move. Will it move to the right or left?
Hint: If the middle plate moves to the left by a small positive amount !x , the
change in potential energy is approximately !U ! (dU / dx)!x . Will the stored
The electric
field on the two sides is shown in the figure below.
potential energy increase
or decrease?
d) For a small displacement !x in the direction you determined in part c), find the
horizontal force exerted by the charge distribution on the outer plates acting on
the charges on the middle plate that cause it to move.
Solution:
a) What is the capacitance of this system?
We can calculate the electric field on both sides using Gauss’s Law (neglecting edge
The electric field on the two sidesDivide
is shown
the figure
below.
the in
plates
intoFigure
left
and11:
right sides. Theofarea
on the left
problem
9. is AL = L( L ! x) and the
effects.Solution
We find
area on the right is AR = Lx . The charge densities on the two sides are shown in the
= !charge
/ " 0 , onEthe
=!
/ "is
figure below. The charge on the left is QL = ! L L( L " x) . EThe
right
L
L
R
R
0
QR = ! LR (Lx) .
Because the electric field is uniform on both sides (neglecting edge effects)
!VL = E L s =
" Ls
(1)
nd Force in a Capacitor
t A is suspended by an insulating thread between the surfaces
ducting sheet B as shown in the figure on the left. The sheets are
difference in potential, in volts, is !V . This causes a force F , in
f A , pulling A downward.
Then
!
acitance of this arrangement of conductors as aE! function
!
d
a
= EAcapof y , the
"
""
te A is inserted
and between the sides of plate B ?
gy is
Figure 12: Problem 10.
1
1
q
=
" Acap .
enc !y ?
needed to increase the inserted distance
! 0 by
!0
Thus Gauss’s Law implies that
on for the difference in potential !V in terms of F and relevant
! 1
wn in the figure.
E = " î , between the plates .