prof. anchordoqui march 17, 2015

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

capacitor is Q3 = 24.0 µC. Continuing to use Fig. 2 (middle panel) it follows that (∆V )ab = QabCab

=24.0 µC6.00 µF = 4.00 V, and (∆V )3 = (∆V )bc = Qbc

Cbc= 24.0 µC

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 den-

sities ±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, ε0, 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, ε0, and d as needed. (iii) Find an expression that relates the surface

charge densities σL and σR in terms of κ, s, w, L, ε0, 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, ε0, and d

as needed. (v) What is the capacitance of this system? Express your answer in terms of κ, s, w,

L, ε0, 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, ε0, and d as needed.

Solution: (i) Using Gauss’ law EL = σLε0

. (ii) Using Gauss law for dielectrics ER = σRκε0

. (iii) The

potential difference on the left side is ELd = σLdε0

. On the right hand side it is ERd = σRdκε0

. Since

these must be equal we must have σR/κ = σL. (iv) Q = σL(L − s)w + σRsw. (v) The potential

difference is ELd = σLdε0

, so the capacitance is C = Q|∆V | = σL(L−s)w+σRsw

σLd/ε0= ε0w

d

[(L− s) + σR

σLs]

=ε0wd [(L− s) + κs]. (vi) Since the battery has been removed, the charge on the capacitor does not

change when we do this, and the change in the energy stored is = 12Q2

C0− 1

2Q2

C = Q2

2

(1C0− 1

C

)=

Q2

2

{d

ε0Lw− d

ε0w[(L−s)+κs]

}= Q2d

2ε0w

[1L −

1(L−s)+κs

]= Q2d

2ε0w

{(L−s)+κs−LL[(L−s)+κs]

}= Q2d

2ε0w

{(κ−1)s

L[(L−s)+κs]

}.

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 = κε0Ad = κC0. (ii) The capacitor can be regarded as be-

ing consisted of three capacitors, C1 = κ1ε0A/2d , C2 = κ2ε0A/2

d/2 , and C3 = κ3ε0A/2d/2 , with C2 and

C3 connected in series, and the combination connected in parallel with C1. Thus, the equivalent

capacitance is C = C1 +(

1C2

+ 1C3

)−1= C1 + C2C3

C2+C3= κ1ε0A/2

d + ε0Ad

(κ2κ3κ2+κ3

)= ε0A

d

(κ12 + κ2κ3

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 = mρ = 1.00×10−12 kg

1,100 kg/m3 = 9.09 × 10−16 m3. Since V = 43πr

3, the

inner radius of the capacitor is r =(

3V4π

)1/3= 6.01 × 10−6 m and the surface area is A = 4πr2 =

4π(

3V4π

)2/3= 4π

(3

4π9.09× 10−16 m3)2/3

= 4.54× 10−10 m2. (ii) The outer radius of the capacitor

is R = r+ t = 6.11×10−6 m, where t = 100 nm is the thickness of the membrane. The capacitance

is C = 4πκε0RrR−r = 2.04×10−13 F. (iii) Q = C ∆V = 2.04×10−13 F 100×10−3 V = 2.01×10−14 C,

and the number of electronic charges is n = Qe = 2.01×10−14 C

1.60×10−19 C= 1.27× 105.

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

r ≤ a. In each region the electric field is purely radial (that is ~E = Er). In regions I and III these

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

the surface isv~E · d ~A = EA = E · 4πr2. The enclosed charge is Qenc = +Q, and the electric field

is everywhere perpendicular to the surface. Thus Gauss law becomes E · 4πr2 = Qε0⇒ E = Q

4πε0r2.

That is, the electric field is exactly the same as that for a point charge. Summarizing:

~E =

{Q

4πε0r2r for a < r < b

~0 elsewhere.

We know the positively charged inner sheet is at a higher potential so we shall calculate ∆V =

V (a) − V (b) = −∫ ab~E · d~s = −

∫ ab

Q4πε0r2

dr = Q4πε0r

∣∣∣ab

= Q4πε0

(1a −

1b

)> 0, which is positive as

we expect. We can now calculate the capacitance using the definition C = Q|∆V | = Q

Q4πε0

( 1a− 1b )

=

4πε0( 1a− 1b )

= 4πε0abb−a = 0.1 m 0.2 m

8.99×109 Nm2/C2 0.1 m= 2.2× 10−11 F. Note that the units of capacitance are ε0

times an area ab divided by a length b−a, exactly the same units as the formula for a parallel-plate

capacitor C = ε0A/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→a4πε0abb−a ≈ ε04πa2

d = ε0Ad . (ii) The electric field is E(a) = Q

4ε0a2.

Therefore the maximum charge is Qmax = 4πε0Emax(a)a2 = 3.0×106 V·m−1 (0.1 m)2

8.99×109 N·m2/C2 . (iii) The energy

stored is Umax = Q2max2C = (3.3×10−6 C)2

2·2.2×10−11 F= 2.5 × 10−1 J. (iv) We can find the potential difference

two different ways. Using the definition of capacitance we have that |∆V | = QC = 4πε0E(a)a2(b−a)

4πε0ab=

E(a)a(b−a)b = 3.0×106 V·m−1(0.1 m)2

0.2 m = 1.5 × 105 V. We already calculated the potential difference in

part (i): ∆V = Q4πε0

(1a −

1b

). Recall that E(a) = Q

4πε0a2or Q

4πε0= E(a)a2 Substitute this into

our expression for potential difference yielding ∆V = E(a)a2(

1a −

1b

)= E(a)a2 b−a

ab = E(a)a b−ab in

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 cross-

sectional area and the distance between the plates. The first relationship we have is from knowing

the capacitance: C = ε0Ax 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 between the plates (now x + d) to get the volume, i.e., volume

= A(x+ d) = xCε0

(x+ d) = dC3ε0

(d3 + d

)= 4d2C

9ε0.

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) Calcu-

late 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) Nu-

merically 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 = 12C(∆V )2.

Solution: (i) For this we use Gauss law, with a Gaussian cylinder of radius r, height l:v~E ·d ~A =

2πrlE = Qinsideε0

= 1ε0Qh l ⇒ E(r) = Q

2πrε0hwith a < r < b. (ii) The potential difference between the

outer shell and the inner cylinder is ∆V = V (a) − V (b) = −∫ ab

Q2πr′ε0h

dr′ = − Q2πε0h

ln r′∣∣∣ab

=

Q2πε0h

ln(ba

). (iii) C = Q

∆V = QQ

2πε0hln( ba)

= 2πε0hln( ba)

. (iv) C = 2πε0hln(b/a) = 15 cm · 2π · 8.85 ×

10−14 cm−1 · F 1ln(7.25 cm/4.75 cm) ' 20 pF. (v) The total energy density stored in the capacitor

is u = 12ε0E

2 = 12ε0

(Q

2πrε0h

)2. Then dU = udV = 1

2ε0(

Q2πε0rh

)22πrhdr = Q2

4πε0hdrr . Integrating we

find that U =∫ ba dU =

∫ ba

Q2

4πε0hdrr = Q2

4πε0hln(b/a). From part (iii) C = 2πε0h/ ln(b/a), therefore

U =∫ ba dU =

∫ ba

Q2

4πε0hdrr = Q2

4πε0hln(b/a) = Q2

2C = 12C(∆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/ε0A for parallel plates on the left. Because the electric field is uniform the potential

difference is ∆V = ELz = QLzε0A

. Therefore the positive charge on the left satisfies QL = ∆V ε0Az . A

similar argument applied to the plates on the right show that ∆V = ER(d−z) = QRε0A

(d−z). Hence

the positive charge on the right satisfies QR = ∆V ε0Ad−z . The total positive charge at the higher

potential is the sum QL +QR capacitance of the capacitor is C = QL+QR∆V . We can now substitute

the equations above and solve for the capacitance C = ∆V ε0A∆V z + ∆V ε0A

∆V (d−z) = ε0A(

1z + 1

d−z

)= ε0Ad

z(d−z) .

Note: Because the outer plates are at the same potential you can think of these capacitors as

connected in parallel. Therefore the equivalent capacitance is C = CL + CR = QL∆V + QR

∆V agreeing

with our result above. (ii) The stored energy is just U = 12C(∆V )2 = ε0Ad(∆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 = σLL(L − x). The charge on the right is QR = σR(Lx). The elec-

tric 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/ε0, ER = σ/ε0. Because the electric

field is uniform on both sides (neglecting edge effects) ∆VL = ELs = σLs/ε0. Note that on the

right side the electric field is zero in the middle conductor so ∆VR = σRs/4ε0

+ σRs/4ε0

σRs2ε0

. The

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 otherσLsε0

= σRs2ε0

; hence σL = σR2 . The capacitance of the system is the total charge divided by the

potential difference C = QL+QR∆V . Using our results for the charges and the potential difference

we have that C = σLL(L−x)+σRLxσLs/ε0

. We can now substitute the value of σL and find that the

capacitance is C = σL(L−x)/2+σRLxσRs/(2ε0) =

ε)[L(L−x)+2Lx]

s = ε0L(L+x)s . (ii) The energy stored in the

capacitor is then U = Q2

2C = Q2s2ε0L(L+x) . (iii) If the middle plate moves a positive distance ∆x to

the left resulting in a larger value of x, then the stored potential energy changes by an amount

∆U = dUdx∆x = − Q2s

2ε0L(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

is negative, the work done by the electric forces is positive W = Q2s2ε0L(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 hor-

izontal force exerted by the charges on the outer plates acting on the charges on the middle plate

is given by F = Q2s2ε0L(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 differ-

ence 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

824 C H A P T E R 2 6 • Capacitance and Dielectrics

22. Three capacitors are connected to a battery as shown inFigure P26.22. Their capacitances are C 1 ! 3C, C 2 ! C,and C 3 ! 5C. (a) What is the equivalent capacitance ofthis set of capacitors? (b) State the ranking of the capaci-tors according to the charge they store, from largest tosmallest. (c) Rank the capacitors according to the poten-tial differences across them, from largest to smallest.(d) What If? If C 3 is increased, what happens to thecharge stored by each of the capacitors?

line with capacitance 29.8 "F between A and B. What addi-tional capacitor should be installed in series or in parallelin that circuit, to meet the specification?

A group of identical capacitors is connected first in seriesand then in parallel. The combined capacitance in parallelis 100 times larger than for the series connection. Howmany capacitors are in the group?

26. Consider three capacitors C1, C 2, C 3, and a battery. If C1 isconnected to the battery, the charge on C1 is 30.8 "C. NowC1 is disconnected, discharged, and connected in serieswith C 2. When the series combination of C 2 and C1 is con-nected across the battery, the charge on C 1 is 23.1 "C. Thecircuit is disconnected and the capacitors discharged.Capacitor C 3, capacitor C1, and the battery are connectedin series, resulting in a charge on C1 of 25.2 "C. If, afterbeing disconnected and discharged, C 1, C 2, and C3 areconnected in series with one another and with the battery,what is the charge on C 1?

27. Find the equivalent capacitance between points a and b forthe group of capacitors connected as shown in FigureP26.27. Take C1 ! 5.00 "F, C2 ! 10.0 "F, and C 3 ! 2.00 "F.

25.

6.00 µF

20.0 µF

3.00 µF15.0 µF

a b

µ µ

µ

µFigure P26.21

C2 C3

C1

Figure P26.22

C1 C2

S2S1

∆V

Figure P26.23

C2 C2

C1 C1

C2 C2

C3

b

a

Figure P26.27 Problems 27 and 28.

ba

6.0 µF

5.0 µF

7.0 µF

4.0 µFµ

µ

µ

µ

Figure P26.29

Consider the circuit shown in Figure P26.23, whereC 1 ! 6.00 "F, C 2 ! 3.00 "F, and #V ! 20.0 V. CapacitorC 1 is first charged by the closing of switch S1. Switch S1 isthen opened, and the charged capacitor is connected tothe uncharged capacitor by the closing of S2. Calculatethe initial charge acquired by C 1 and the final charge oneach capacitor.

23.

24. According to its design specification, the timer circuitdelaying the closing of an elevator door is to have a capaci-tance of 32.0 "F between two points A and B. (a) Whenone circuit is being constructed, the inexpensive butdurable capacitor installed between these two points isfound to have capacitance 34.8 "F. To meet the specifica-tion, one additional capacitor can be placed between thetwo points. Should it be in series or in parallel with the34.8-"F capacitor? What should be its capacitance?(b) What If? The next circuit comes down the assembly

28. For the network described in the previous problem, if thepotential difference between points a and b is 60.0 V, whatcharge is stored on C3?

29. Find the equivalent capacitance between points a and b inthe combination of capacitors shown in Figure P26.29.

30. Some physical systems possessing capacitance continuouslydistributed over space can be modeled as an infinite arrayof discrete circuit elements. Examples are a microwavewaveguide and the axon of a nerve cell. To practice analy-

Figure 1: Problem 1.

in terms of F and relevant dimensions shown in the figure.

Solution: a) We begin by assuming the plates are very large and use Gauss law to calculate

the electric field between the platesv~E · d ~A = qenc

ε0Our choice of Gaussian surface is shown in

Fig. 12. Thenv~E · d ~A = EAcap and qenc

ε0= σA

ε0. Thus, Gauss’ law implies that ~E = σ

ε0i, be-

tween the plates. Note that the potential difference between the positive and negative plates is

V (+) − V (−) = −∫ x=0x=s Exdx = σs

ε0. So the surface charge density is equal to σ = ε0[V (+)−V (−)]

s .

The area between the plates is yb. Note that there is a charge on inner surface of the surrounding

sheet equal to σyb, so the total charge on the outer sheets is Q = 2σyb. Therefore the capaci-

tance is C(y) = QV (+)−V (−) = 2σybε0

σs = 2ε0ybs . (ii) If the inserted distance is increased by ∆y then

∆C = 2ε0∆ybs . Because the charge on the plates is fixed, the energy stored in the capacitor is given

by U(y) = Q2

2C(y) . Hence, when the inserted distance is increased by ∆y, the energy stored between

the plates decreases by ∆U = dUdC∆C = − Q2

2C2 ∆C = − [V (+)−V (−)]2

22ε0∆yb

s . (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

0 = ∆U + ∆W = − [V (+)−V (−)]2

22ε0∆yb

s +Fy∆y. We can solve this equation for the y-component of

the force Fy = [V (+)−V (−)]2

22ε0bs or else find V (+)− V (−) =

√sFyε0b

.

Physics 112 Homework 2 (solutions) (2004 Fall)

4

Chapt16, Problem-31: (a) Find the equivalent capacitance of thegroup of capacitors in Figure P16.31. (b) Find the charge on and thepotential difference across each.

Solut ion:

Using the rules for combining capacitors in series and in parallel,

the circuit is reduced in steps as shown below. The equivalent capacitor is

shown to be a 2 .00 µF capacitor.

a b c

Figure 2

12.0 V

3.00 µF6.00 µF

a c

Figure 3

12.0 V

2.00 µF

a b c

4.00 µF

2.00 µF

3.00 µF

Figure 1

12.0 V

(b) From Figure 3: Qac = Cac !V( )ac = 2 .00 µF( ) 12.0 V( ) = 24.0 µC

From Figure 2: Qab =Qbc = Qac = 24.0 µC

Thus, the charge on the 3.00 µF capacitor is

Q3 = 24.0 µC

Continuing to use Figure 2, !V( )ab =

Qab

Cab

=24.0 µC

6.00 µF= 4.00 V ,

and !V( )

3= !V( )bc =

Qbc

Cbc

=24.0 µC

3.00 µF=

8.00 V

From Figure 1, !V( )

4= !V( )

2= !V( )

ab=

4.00 V

and Q4 = C4 !V( )

4= 4.00 µF( ) 4.00 V( ) =

16.0 µC

Q2 = C2 !V( )

2= 2.00 µF( ) 4.00 V( )=

8.00 µC

Chapt16, Problem-44: Two capacitors C1 = 25.0 µF and C2 = 5.00 µF are connected in paralleland charged with a 100-V power supply. (a) Calculate the total energy stored in the two capacitors. (b) Whatpotential difference would be required across the same two capacitors connected in series in order that thecombination store the same energy as in (a)?

Solut ion:

(a) When connected in parallel, the energy stored is

W =1

2C1 !V( )2

+1

2C2 !V( )2

=1

2C1+C2( ) !V( )2

=1

225.0+ 5.00( )!10

"6 F[ ] 100 V( )2

= 0.150 J

(b) When connected in series, the equivalent capacitance is

Ceq =

1

25.0+

1

5.00

! " #

$ % &

'1

µF = 4.17 µF

From W = 1

2Ceq !V( )2

, the potential difference required to store the same energy as in part (a) above is

!V =2W

Ceq

=2 0.150 J( )

4.17 "10#6

F=

268 V

Figure 2: Solution of problem 1.

Problem 3 Dielectric A dielectric rectangular slab has length s , width w , thickness d , and dielectric constant ! . 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. The capacitor is charged up such that the left hand side has surface charge densities ±! L on the facing surfaces of the top and bottom plates respectively and the right hand side has surface charge densities ±! R 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.

a) Find an expression for the magnitude of the electric field EL on the left hand side in terms of ! L , ! R , ! , s , w , L , !0 , and d as needed.

b) Find an expression for the magnitude of the electric field ER on the right hand

side in terms of ! L , ! R , ! , s , w , L , !0 , and d as needed.

c) Find an expression that relates the surface charge densities ! L and ! R in terms of ! , s , w , L , !0 , and d as needed.

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.

e) What is the capacitance of this system? Express your answer in terms of ! , s , w , L , !0 , and d as needed.

f) Suppose the dielectric is removed. What is the change in the stored potential energy of the capacitor? Express your answer in terms of Q , ! , s , w , L , !0 , and d as needed.


828 C H A P T E R 2 6 • Capacitance and Dielectrics

62. A 10.0-!F capacitor is charged to 15.0 V. It is nextconnected in series with an uncharged 5.00-!F capacitor.The series combination is finally connected across a 50.0-Vbattery, as diagrammed in Figure P26.62. Find the newpotential differences across the 5-!F and 10-!F capacitors.

sent the potential difference. (c) Find the direction andmagnitude of the force exerted on the dielectric, assuminga constant potential difference "V. Ignore friction. (d) Obtain a numerical value for the force assuming that! # 5.00 cm, "V # 2 000 V, d # 2.00 mm, and the dielec-tric is glass ($ # 4.50). (Suggestion: The system can be con-sidered as two capacitors connected in parallel.)

65. A capacitor is constructed from two square plates of sides !and separation d, as suggested in Figure P26.64. You mayassume that d is much less than !. The plates carry charges% Q 0 and & Q 0. A block of metal has a width !, a length !,and a thickness slightly less than d. It is inserted a distancex into the capacitor. The charges on the plates are notdisturbed as the block slides in. In a static situation, ametal prevents an electric field from penetrating inside it.The metal can be thought of as a perfect dielectric, with$ : '. (a) Calculate the stored energy as a function of x.(b) Find the direction and magnitude of the force thatacts on the metallic block. (c) The area of the advancingfront face of the block is essentially equal to !d. Consider-ing the force on the block as acting on this face, find thestress (force per area) on it. (d) For comparison, expressthe energy density in the electric field between the capaci-tor plates in terms of Q 0 , !, d , and (0.

66. When considering the energy supply for an automobile,the energy per unit mass of the energy source is an impor-tant parameter. Using the following data, compare theenergy per unit mass ( J/kg) for gasoline, lead–acid batter-ies, and capacitors. (The ampere A will be introduced inthe next chapter as the SI unit of electric current.1 A # 1 C/s.)Gasoline: 126 000 Btu/gal; density # 670 kg/m3.Lead–acid battery: 12.0 V; 100 A ) h; mass # 16.0 kg.Capacitor: potential difference at full charge # 12.0 V;capacitance # 0.100 F; mass # 0.100 kg.

An isolated capacitor of unknown capacitance has beencharged to a potential difference of 100 V. When thecharged capacitor is then connected in parallel to anuncharged 10.0-!F capacitor, the potential differenceacross the combination is 30.0 V. Calculate the unknowncapacitance.

68. To repair a power supply for a stereo amplifier, an elec-tronics technician needs a 100-!F capacitor capable ofwithstanding a potential difference of 90 V between theplates. The only available supply is a box of five 100-!Fcapacitors, each having a maximum voltage capability of50 V. Can the technician substitute a combination of thesecapacitors that has the proper electrical characteristics? Ifso, what will be the maximum voltage across any of thecapacitors used? (Suggestion: The technician may not haveto use all the capacitors in the box.)

A parallel-plate capacitor of plate separation d is chargedto a potential difference "V0. A dielectric slab of thicknessd and dielectric constant $ is introduced between theplates while the battery remains connected to the plates.(a) Show that the ratio of energy stored after the dielectricis introduced to the energy stored in the empty capacitoris U/U0 # $. Give a physical explanation for this increasein stored energy. (b) What happens to the charge on thecapacitor? (Note that this situation is not the same as in

69.

67.

d/2

!/2

d

!

1κ2κ

3κ

Figure P26.61

5.00 Fµ

50.0 V

∆Vi = 15.0 V

–+10.0 Fµ

Figure P26.62

xd

!

κ

Figure P26.64 Problems 64 and 65.

63. (a) Two spheres have radii a and b and their centers are adistance d apart. Show that the capacitance of this system is

provided that d is large compared with a and b. (Suggestion:Because the spheres are far apart, assume that the poten-tial of each equals the sum of the potentials due to eachsphere, and when calculating those potentials assume thatV # keQ /r applies.) (b) Show that as d approaches infinitythe above result reduces to that of two spherical capacitorsin series.

64. A capacitor is constructed from two square plates of sides !and separation d. A material of dielectric constant $is inserted a distance x into the capacitor, as shown inFigure P26.64. Assume that d is much smaller than x.(a) Find the equivalent capacitance of the device. (b) Cal-culate the energy stored in the capacitor, letting "V repre-

C #4*(0

1a

%1b

&2d


The enclosed charge is Qenc = +Q , and the electric field is everywhere perpendicular to the surface. Thus Gauss’s Law becomes

E !4"r 2 = Q#0

$ E = Q4"#0r

2

That is, the electric field is exactly the same as that for a point charge. Summarizing:

!E =

Q4!"0r

2 r for a < r < b

0 elsewhere

#

$%

&%

We know the positively charged inner sheet is at a higher potential so we shall calculate

!V =V (a) "V (b) = "

!E # d!s

b

a

$ = " Q4%&0r

2 drb

a

$ = Q4%&0r b

a

= Q4%&0

1a" 1

b'()

*+,> 0

which is positive as we expect. We can now calculate the capacitance using the definition

C = Q!V

= QQ

4"#0

1a$ 1

b%&'

()*

=4"#0

1a$ 1

b%&'

()*

=4"#0abb $ a

C =

4!"0abb # a

= (0.1 m)(0.2 m)(9 $109 N %m2 %C2 )(0.1 m)

= 2.2 $10#11 F .

Note that the units of capacitance are !0 times an area ab divided by a length b ! a , exactly the same units as the formula for a parallel-plate capacitor C = !0A / d . Also note


Problem 10: Cylindrical Capacitor Consider two nested cylindrical conductors of height h and radii a & 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). You may ignore edge effects.

a) Calculate the electric field between the two cylinders (a < r < b). b) Calculate the potential difference between the two cylinders: c) Calculate the capacitance of this system, C = Q/DV d) Numerically evaluate the capacitance, given: h ≅ 15 cm, a ≅ 4.75 cm and b ≅ 7.25 cm. e) 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 2!rh dr ? Integrate your expression to find the total energy stored in the capacitor and compare your result with that obtained using U E = (1 / 2)C(!V )2 .


0 25 50 75 100 125 150

0

25

50

75

100

125

150

Problems

827

53.T

he general form of G

auss’s law describes how

a chargecreates an electric field in a m

aterial, as well as in vacuum

.It is

where !

"#

!0is the perm

ittivity of the material. (a) A

sheet with charge Q

uniformly distributed over its area A

issurrounded by a dielectric. Show

that the sheet creates auniform

electric field at nearby points, with m

agnitudeE

"Q

/2A

!.(b) Two large sheets of area A

, carrying oppo-site

charges of equal magnitude Q

, are a small distance

dapart. Show

that they create uniform electric field in

thespace

between

them,

with

magnitude

E"

Q/A

!.(c)

Assum

e that the negative plate is at zero potential.Show

that

the positive

plate is

at potential

Qd/A

!.(d)

Show that the capacitance of the pair of plates is

A!/d

"#A

!0 /d.

Additio

nal Pro

blem

s

54.For the system

of capacitors shown in Figure P26.54, find

(a) the

equivalent capacitance

of the

system,

(b) the

potential across each capacitor, (c) the charge on eachcapacitor, and (d) the total energy stored by the group.

! E

$dA

"q!

capacitance of the three-plate system P

1 P2 P

3 ? (b) What is

the charge on P2 ? (c) If P

4is now

connected to the positiveterm

inal of the battery, what is the capacitance of the four-

plate system P

1 P2 P

3 P4 ? (d) W

hat is the charge on P4 ?

56.O

ne conductor of an overhead electric transmission line is

a long aluminum

wire 2.40

cm in radius. Suppose that at a

particular mom

ent it carries charge per length 1.40%

C/m

and is at potential 345kV. Find the potential 12.0

m below

the wire. Ignore the other conductors of the transm

issionline and assum

e the electric field is everywhere purely

radial.

57.Tw

o 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.N

ow a third identical plate carrying charge Q

is insertedbetw

een the two plates, parallel to them

and located a dis-tance

dfrom

the

upper plate,

as in

Figure P26.57.

(a) What induced charge appears on each of the tw

o origi-nal plates? (b) W

hat potential difference appears between

the middle plate and each of the other plates? E

ach platehas area A

.

4.00 µF2.00 µF

6.00 µF3.00 µF90.0 V

µµ

µµ

Figure P

26.54

12.0 V

P2

P3

P4

P1

dd

d

Figure P

26.55

2d d

Figure P

26.57

55.Four parallel m

etal plates P1 , P

2 , P3 , and P

4 , each of area7.50

cm2,

are separated

successively by

a distance

d"

1.19m

m, as show

n in Figure P26.55. P1

is connected to thenegative term

inal of a battery, and P2

to the positive termi-

nal. The battery m

aintains a potential difference of 12.0V.

(a)If P

3is connected to the negative term

inal, what is the

58.A

2.00-nF parallel-plate capacitor is charged to an initialpotential difference &

Vi "

100V

and then isolated. The

dielectric m

aterial betw

een the

plates is

mica,

with

adielectric constant of 5.00. (a) H

ow m

uch work is required

to withdraw

the mica sheet? (b) W

hat is the potentialdifference of the capacitor after the m

ica is withdraw

n?

A

parallel-plate capacitor

is constructed

using a

dielectric m

aterial w

hose dielectric

constant is

3.00and

whose

dielectric strength

is 2.00

'10

8V

/m.

The

desired capacitance is 0.250%

F, and the capacitor must

withstand a m

aximum

potential difference of 4000

V. Findthe m

inimum

area of the capacitor plates.

60.A

10.0-%F capacitor has plates w

ith vacuum betw

een them.

Each plate carries a charge of m

agnitude 1000

%C

. Aparticle w

ith charge (3.00

%C

and mass 2.00

'10

(16

kgis fired from

the positive plate toward the negative plate

with an initial speed of 2.00

'10

6m

/s. Does it reach the

negative plate? If so, find its impact speed. If not, w

hatfraction of the w

ay across the capacitor does it travel?

61.A

parallel-plate

capacitor is

constructed by

fillingthe

space between tw

o square plates with blocks of three

dielectric materials, as in Figure P26.61. You m

ay assume

that !)

)d. (a) Find an expression for the capacitance of

the device in terms of the plate area A

and d, #1 , #

2 , and#

3 . (b)

Calculate

the capacitance

using the

valuesA

"1.00

cm2,

d"

2.00m

m,

#1

"4.90,

#2

"5.60,

and#

3"

2.10.

59.

zd � z

Figure 7: Problem 8.Problem 11 Capacitance of Multiple Plates 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. 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.

a) 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.

b) 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 plate.

The plates on the left in the above figure act as a capacitor arranged as show in the figure below.

plates on left plates on right

This is 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’s 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 = ELz = QLz / "0 A . Therefore the positive charge on the left satisfies


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 plate.

The plates on the left in the above figure act as a capacitor arranged as show in the figure below.

plates on left plates on right

This is 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’s 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 = ELz = QLz / "0 A . Therefore the positive charge on the left satisfies

Figure 9: More on the solution of problem 8.

Problem 12: Capacitance, Stored Energy, and Electrostatic Work 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.

a) What is the capacitance of this system? 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 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?

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 the figure below. The charge on the left is QL = ! L L(L " x) . The charge on the right is

QR = ! LR (Lx) .


The electric field on the two sides is shown in the figure below.

We can calculate the electric field on both sides using Gauss’s Law (neglecting edge effects. We find

EL = ! L / "0 , ER = ! R / "0 Because the electric field is uniform on both sides (neglecting edge effects)

!VL = ELs =

" Ls#0

(1)

Note that on the right side the electric field is zero in the middle conductor so

!VR =

" R (s / 4)#0

+" R (s / 4)

#0

=" Rs2#0

. (2)

The 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 Eq. (1) equal to Eq. (2)

The electric field on the two sides is shown in the figure below.

We can calculate the electric field on both sides using Gauss’s Law (neglecting edge effects. We find

EL = ! L / "0 , ER = ! R / "0 Because the electric field is uniform on both sides (neglecting edge effects)

!VL = ELs =

" Ls#0

(1)

Note that on the right side the electric field is zero in the middle conductor so

!VR =

" R (s / 4)#0

+" R (s / 4)

#0

=" Rs2#0

. (2)

The 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 Eq. (1) equal to Eq. (2)


Problem 13 Energy and Force in a Capacitor A flat conducting sheet A is suspended by an insulating thread between the surfaces formed by the bent conducting sheet B as shown in the figure on the left. 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.

a) 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 ? b) How much energy is needed to increase the inserted distance by !y ?

c) Find an expression for the difference in potential !V in terms of F and relevant

dimensions shown in the figure. Solution: a) We begin by assuming the plates are very large and use Gauss’s Law to calculate the electric field between the plates

!E ! d !a""" = 1

#0

qenc .

Our choice of Gaussian surface is shown in the figure below.

Then

!E ! d !a""" = EAcap

and

1!0

qenc =1!0

" Acap .

Thus Gauss’s Law implies that

!E = 1

!0

" i , between the plates .

Note that the potential difference between the positive and negative plates is

V (+) !V (!) = ! Ex dx

x= s

x=0

" =1#0

$s .

So the surface charge density is equal to

! =

"0 (V (+) #V (#))s

The area between the plates is yb . Note that there is a charge on inner surface of the surrounding sheet equal to ! yb , so the total charge on the outer sheets is Q = 2! yb . Therefore the capacitance is

C( y) =

QV (+) !V (!)

=2" yb#0

"s=

2#0 ybs

.

b) How much energy is needed to increase inserted distance by !y ? When the inserted distance is increased by


prof. anchordoqui march 17, 2015

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