Exam 2_Practice_Problems_Part 1_Solutions.pdf

7/25/2019 Exam 2_Practice_Problems_Part 1_Solutions.pdf

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY

Department of Physics

Exam 2 Practice Problems Part 1 Solutions

Problem 1 Electric Field and Charge Distributions from Electric Potential

An electric potential ( )V z is described by the function

V(z) =

(!2V "m-1)z+ 4V ; z> 2.0 m0 ; 1.0 m


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z

dVE

dz= =!E i k

!

(i) 2.0 mz> :

( )-1 -1

( 2V m ) 4V 2V m

dV dz

dz dz =!

=! ! "

+ = "E k k k

!

(ii) 1.0 m 2.0 mz< < :

dV

dz=! =E k 0

!!

(iii) 0 m 1.0 mz< < :

( )-3 3 -3 22 2

V V m 2 V m3 3

dV dz z

dz dz

! "! "=# =# # $ = $% &% &

' (' (E k k k

!

Note that the2

z has units of 2[m ], so the value of the electric field at a point just inside

1.0 m mz !"

= " where 0! > is a very small number is given by

!

E!

= (2 V"m-3)(1m)2 k = 2

V

mk

Note that the z-component of the electric field, 2 V !m"1 , has the correct units.

(iv) 1.0 m 0 mz! < < :

!

E =!dV

dzk =!

d

dz

2

3V+

2

3V "m-3

#$%

&'(z

3#

$%&

'(k =! 2 V "m-3( )z2 k

The value of the electric field at a point just inside z+ =!1.0 m + "m is given by

!

E!

=!(2 V"m-3)(1m)2 k =!2

V

mk.

(v) 2.0 m 1.0 mz! < < ! :

dV

dz=!

=E k 0

!!

(vi) 2.0 mz


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b) Make a plot of the z-component of the electric field,z

E , as a function of z. Make

sure you label the axes to indicate the numeric magnitude of the field.

c) Qualitatively describe the distribution of charges that gives rise to this potential

landscape and hence the electric fields you calculated. That is, where are the charges,

what sign are they, what shape are they (plane, slab)?

In the region 1.0 m 1.0 mz! < < there is a non-uniform (in the z-direction) slab of

positive charge. Note that the z-component of the electric field is zero at 0 mz = ,

negative for the region 1.0 m 0 mz! < < , and positive for 0 m 1.0 mz< < as it should

for a positive slab that has zero field at the center.

In the region 1.0 m 2.0 mz< < there is a conductor where the field is zero.

On the plane 2.0 mz = , there is a positive uniform charge density ! that produces a

constant field pointing to the right in the region 2.0 mz> (hence the positive

component of the electric field).

On the plane 1.0 mz = , there is a negative uniform charge density !" .

In the region 2.0 m 1.0 mz! <


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Problem 2: Electric Field from Electric Potential

The electric potential V(x) for a planar charge distribution is given by:

V(x) =

0 forx < !d

!V0

1+x

d

"#$

%&'

2

for ! d(x < 0

!V0

1+ 2x

d

"#$

%&'

for 0(x < d

!3V0

forx > d

)

*

++++

,

++++

where !V0is the potential at the origin and dis a distance.

This function is plotted to the right, with d = 2 cm and V0 = 2 V , the x-axis with units in

cm, the y-axis in units of Volts.

(a) What is the electric field!

E(x) for this problem?

Region I: x a

Q

4#$0

1

r !

1

a

%&'

()*, r


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What can we say about the electric potential? V(r) =0 for r> c , and V(r) = constant

for r< b but the potential is very complicated defined between the two shells.


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Problem 7 N-P Semiconductors Revisited Potential Difference

When two slabs of n -type and p -type semiconductors are put in contact, the relative

affinities of the materials cause electrons to migrate out of the n -type material across thejunction to the p -type material. This leaves behind a volume in the n -type material that

is positively charged and creates a negatively charged volume in the p -type material.

Let us model this as two infinite slabs of charge, both of thickness a with the junctionlying on the plane z = 0 . The n -type material lies in the range 0 < z< a and has uniform

charge density +!0

. The adjacent p -type material lies in the range !a < z< 0 and has

uniform charge density !"0

. Thus:

!(x,y,z) = !(z) =

+!0 0 < z< a

"!0

"a< z a

#

$%

&

%

a) Find the electric field everywhere.

b) Find the potential difference between the points P1andP

2.. The point P

1.is located

on a plane parallel to the slab a distance z1> a from the center of the slab. The

pointP2.

is located on plane parallel to the slab a distance z2


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The electric field can be described by

2

1

0( )

0T

z a

a zz

z a

x d

!


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Qenc

!0

=

"#0A

cap(a +z)

!0

where the length of the Gaussian cylinder is a z+ since 0z< .

Substituting these two results into Gausss Law yields

0

2,

0

( )cap

z cap

A a zE A

!

"

# +

=

Hence the electric field in the n -type is given by

02,

0

( )x

a zE

!

"

# += .

The negative sign means that the electric field point in the z direction so the electricfield is

02

0

( ) a z!

"

# +=E k

!

.

Note when z a=! then2 =E 0

!!

and when 0z= , 02

0

a!

"

#

=E k

!

.

We make a similar calculation for the electric field in the p -type noting that the charge

density has changed sign and the expression for the length of the Gaussian cylinder is

a z! since 0z> . Also the unit normal now points in the negative z-direction. So the dot

product becomes

1 1, 1, ( )

z zda E da E da! = " ! ="E n k k

!

Thus Gausss Law becomes

0

1,

0

( )cap

z cap

A a zE A

!

"

+ #

# = .

So the electric field is

01,

0

( )z

a zE

!

"

#

=# .

The vector description is then

01

0

( ) a a!

"

# #

=E k

!


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Note when z a= then1 =E 0

!!

and when 0z= , 01

0

a!

"

#

=E k

!

.

So the resulting field is

02

0

01

0

( ) 0

( )( ) 0

T

z a

a za z

za z

z a

z a

!

"

!

"

# %'

0

E k

E

E k

0

!

!

!

!

!

.

The graph of the electric field is shown below

.

(ii) Find the potential difference between the points P1andP

2.. The point P

1.is located

on a plane parallel to the slab a distance1

z a> from the center of the slab. The

pointP2.

is located on plane parallel to the slab a distance2

z a


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2

1

0

2 1

0

( ) ( )

z zz a z z a

T T T T

z z z a z z a

V z V z d d d d

== = =!

= = = =!

" #! =! $ + $ + $ + $% &% &

' () ) ) ) E r E r E r E r

! ! ! !

! ! ! !

.

Since the fields are zero outside the slab, the only non-zero pieces are

0

2 1

0

( ) ( )z z a

T T

z a z

V z V z d d

= =!

= =

" #! =! $ + $% &

' () )E r E r

! !

! !

.

We now use our explicit result for the electric field in each region and that d dz=r k!

,

0

2 1 1, 2,

0

0

0 0

0 00

( ) ( )

( ) ( )

z z a

z a

z a z

z z a

z a z

V z V z E dz E dz

a z a z dz dz

! !

" "

= =#

= =

= =#

= =

$ %# =# & + &' (

) *$ %# # # +

=# +' () *

+ +

+ +

k k k k

We now calculate the two integrals

2 200 0

2 1 0

0 0

2 2 2 2 2

0 0 0 0

0 0 0 0

2

0

0

( / 2) ( / 2)( ) ( )

( / 2) (( ) ( ) / 2) ( / 2) ( / 2)

z z a

z a z

za z za zV z V z

a a a a a a a

a

! !

" "

! ! ! !

" " " "

!

"

= =#

= =

# +# = +

# # + #=# + =# #

=#

.

The potential difference is negative because we are moving along the direction of the

field. So the type pn -semiconductor slab established a small potential difference across

the slab.


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Problem 8: Conceptual Questions

Part (a) A parallel plate capacitor has capacitance C. It is connected to a battery of

EMFe until 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.Below are a series of questions about how other quantities changed. Although they are

related you do not need to rely on the answers to early questions in order to correctly

answer the later ones.

a) Did the potential difference increase or decrease by a factor of 4?

INCREASE DECREASE

Answer. Increase.

b) By what factor did the electric field change due to this increase in distance?

Make sure that you indicate whether the field increased or decreased.

Answer.

Since the charge cannot change (the battery is disconnected) the electric field cannot

change either. No Change!

c) By what factor did the energy stored in the electric field change?

Make sure that you indicate whether the energy increased or decreased.

Answer.

The electric field is constant but the volume in which the field exists increased, so the

energy must have increased. But by how much? The energy U = 12QV . The charge

doesnt change, the potential increased by a factor of 4, so the energy: Increased by a

factor of 4

d) A dielectric of dielectric constant k is now inserted to completely fill the volume

between the plates. Now by what factor does the energy stored in the electric field

change? Does it increase or decrease?

Answer.

Inserting a dielectric decreases the electric field by a factor of k so it decreases the

potential by a factor of k as well. So now, by using the same energy formula U = 12QV,

Energy decreases by a factor of k


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e) 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)

Answer.

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 =!0A

x

where x is the original distance between the plates. Make sure you dont use the more

typical variable dhere 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 cansolve forx:

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 (nowx+ d) to get the volume:

Volume = A x + d( ) =xC

!0

x + d( ) =dC

3!0

d

3+ d

"#$

%&' =

4d2C

9!0


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Problem 9: Spherical Capacitor and Stored Energy

A capacitor consists of two concentric spherical shells. The outer radius of the inner shell

is a = 0.1m and the inner radius of the outer shell is b = 0.2 m .

a) What is the capacitance Cof this capacitor?

b) 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?

c) What is the maximum amount of energy stored in this capacitor?

d) When E(a) = 3.0 !106 V "m-1 what is the potential difference between the shells?

Solution:

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 ).

Region I: Outside r! b :

Region III: Inside r! a :

These Gaussian surfaces contain a total charge of 0, so the electric fields in these regionsmust be 0 as well.

Region II: In between a < r< b : Choose a Gaussian sphere of radius r. The electric flux

on the surface is!E ! d

!A""" =EA =E!4#r

2


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The enclosed charge is Qenc

= +Q , and the electric field is everywhere perpendicular to

the surface. Thus Gausss Law becomes

E! 4"r2=

Q

#0

$E =Q

4"#0r2

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

Summarizing:

!

E =

Q

4!"0r2r fora < 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

$ =" Q

4%&0r2dr

b

a

$ =Q

4%&0r

b

a

=Q

4%&0

1

a"1

b

'()

*+, > 0

which is positive as we expect.

We can now calculate the capacitance using the definition

C =Q

!V=

Q

Q

4"#0

1

a $

1

b

%

&'

(

)*

=

4"#0

1

a $

1

b

%

&'

(

)*

=

4"#0ab

b $ a

C =4!"

0ab

b# a=

(0.1m)(0.2 m)

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


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that if the radii band aare very close together, the spherical capacitor begins to look verymuch like two parallel plates separated by a distance d=b ! a and area

A! 4"a + b

2

#$

%&

2

! 4"a + a

2

#$

%&

2

= 4"a2 ! 4"ab

So, in this limit, the spherical formula is the same at the plate one

C = limb!a

4"#0ab

b $ a!

#04"a

2

d=

#0A

d.

b) Suppose the maximum possible electric field at the outer surface of the inner shell

before the air starts to ionize is E(a) = 3.0 !106 V "m-1 . What is the maximum

possible charge on the inner capacitor?

Solution:

The electric field E(a) =Q

4!"0a2

. Therefore the maximum charge is

Qmax

=4!"0E

max(a)a2 =

(3.0 #106 V $m-1)(0.1m)2

(9 #109 N $m2 $C2 )=3.3#10%6 C

c) What is the maximum amount of energy stored in this capacitor?

Solution:

The energy stored is

Umax

=

Qmax

2

2C=

(3.3!10"6 C)2

(2)(2.2!10"11 F)=2.5!10"1 J

d) When E(a) = 3.0 !106 V "m-1 what is the potential difference between the shells?

Solution:

We can find the potential difference two different ways.

Using the definition of capacitance we have that


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!V =Q

C=

4"#0E(a)a2(b $ a)

4"#0ab

=

E(a)a(b $ a)

b

!V =(3.0 %106 V &m-1)(0.1m)(0.1 m)

(0.2 m)= 1.5%105 V

.

We already calculated the potential difference in part a):

!V =Q

4"#0

1

a$1

b

%&'

()*

.

Recall that E(a) =Q

4!"0a

2or

Q

4!"0

2 = E(a)a2 . Substitute this into our expression for

potential difference yielding

!V = E(a)a21

a"1

b

#$%

&'( =E(a)a2

(b" a)ab

=E(a)a(b" a)

b

in agreement with our result above.


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Problem 10: Cylindrical Capacitor

Consider two nested cylindrical conductors of height hand radii a& brespectively. Acharge +Qis evenly distributed on the outer surface of the pail (the inner cylinder), -Qon

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!7.25cm.

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 ofradius r(with a r b), height h , thickness dr, and volume 2!rhdr? Integrate your

expression to find the total energy stored in the capacitor and compare your result with

that obtained using UE

(1/ 2)C(!V)2

.


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(a) Calculate the electric field between the two cylinders (a< r< b).

For this we use Gausss Law, with a Gaussian cylinder of radius r, height l

!E ! d

!A""" = 2#rlE=

Qinside

$0

=

1

$0

Q

hl % E(r)

a


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dU = uEdV =

1

2!0

Q

2"r!0h

#

$%&

'(

2

2"rhdr =Q2

4"!0h

dr

r

Integrating we find that

U = dUa

b

! =Q2

4"#0h

dr

r=

Q2

4"#0hln(b / a)

a

b

! .

From part d) C = 2!"oh / ln(b / a) , therefore

U = dUa

b

! =Q2

4"#0h

dr

r=

Q2

4"#0hln(b /a) =

Q2

2Ca

b

! =1

2C$V2

which agrees with that obtained above.


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Problem 11Capacitance of Multiple Plates

A capacitor is made of three sets of parallel plates of area A , with the two outer plates onthe left and the right connected together by a conducting wire. The outer plates are

separated by a distanced. The distance from the middle plate to the left plate is z. The

distance from the inner plate to the right plate isd!z

. You may assume all three platesare very thin compared to the distances dand 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 theouter 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.


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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 figurebelow.

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 Gausss Law to show that the magnitude of the

electric field on the left is EL

= QL/ !

0A for parallel plates on the left. Because the

electric field is uniform the potential difference is

!V =ELz = Q

Lz/ "

0A .

Therefore the positive charge on the left satisfies


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QL =

!V"0A

z (1)

A similar argument applied to the plates on the right show that

!V =ER(d"z) = QR(d"z) / #0A .

Hence the positive charge on the right satisfies

QR

=

!V"0A

d#z (2)

The total positive charge at the higher potential is the sum QL+Q

Rand so the

capacitance of the capacitor is

C=QL

+QR

!V (3)

We can now substitute Eqs. (1) and (2) into Eq. (3) and solve for the capacitance

C=!V"

0A

!Vz+

!V"0A

!V(d#z)= "

0A

1

z+

1

d#z$

%&'

() =

"0Ad

z(d#z).

Note: Because the outer plates are at the same potential you can think of these capacitorsas connected in parallel. Therefore the equivalent capacitance is

C= CL+C

R =

QL

!V+

QR

!V

agreeing with our result above.

b) Find the total energy stored in this system.

The stored energy is just

U =1

2C!V

2=

"0Ad!V

2

2z(d#z) .


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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


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

We can calculate the electric field on both sides using Gausss Law (neglecting edgeeffects. We find

EL

= !L

/ "0, E

R = !

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

=

"Rs

2#0

. (2)

The potential difference on the two sides are the same because the upper and lower platesare held at the same potential difference

!V " !VL

= !VR

.

Therefore we can solve for a relationship between the charge densities on the two sidesby setting Eq. (1) equal to Eq. (2)


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!Ls

"0

=

!Rs

2"0

Hence

!L

=

!R

20

. (3)

The capacitance of the system is the total charge divided by the potential difference

C=QL+Q

R

!V.

Using our results for the charges and the potential difference we have that

C= !LL(L"

x)+!RLx

!Ls / #

0

(4)

We can now substitute Eq. (3) into Eq. (4) and find that the capacitance is

C=(!

R/ 2)L(L "x) +!

RLx

!Rs / 2#

0

=

#0(L(L "x) + 2Lx)

s=

#0L(L +x)

s.

b) How much energy is stored in the electric field?

The energy stored in the capacitor is then

U =Q2

2C=

Q2s

2!0L(L +x)

.

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 , thechange in potential energy is approximately !U ! (dU/ dx)!x . Will the stored

potential energy increase or decrease?

If the middle plate moves a positive distance !x to the left resulting in a larger value ofx , then the stored potential energy changes by an amount

!U =dU

dx!x = "

Q2s

2#0L(L +x)2

!x .


37/40

This change is negative hence the middle plate will move to the left decreasing the storedpotential energy of the system. Because the change in potential energy is negative, the

work done by the electric forces is positive

W =Q

2s

2!0L(L +x)2"x .

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 onthe charges on the middle plate that cause it to move.

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 platesacting on the charges on the middle plate is given by

F = Q

2

s2!

0L(L +x)2

.


38/40

Problem 13Energy 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 ofA

, pullingA

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 ofF

and relevantdimensions shown in the figure.

Solution:

a) We begin by assuming the plates are very large and use Gausss 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.


39/40

Then!E ! d

!a""" =EAcap

and

1

!0

qenc

=

1

!0

"Acap

.

Thus Gausss Law implies that

!

E =1

!0

"i , between the plates .

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

V(+)!V(!) = ! Exdx

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) =Q

V(+) !V(!)=

2"yb#0

"s=

2#0yb

s.

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

When the inserted distance is increased by


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!C =2"

0!yb

s

Because the charge on the plates is fixed, the energy stored in the capacitor is given by

U(y) =Q

2

2C(y)

When the inserted distance is increased by !y , the energy stored between the plates

decreases by

!U =dU

dC!C= "

Q2

2C2!C= "

(V(+) "V("))2

2

2#0!yb

s

c) 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

2

2#0!yb

s+F

y!y .

We can solve this equation for the y-component of the force

Fy =

(!(+) "!("))2

2

2#0b

s

or

V(+) !V(!) =sF

y

"0b

.

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Exam 2_Practice_Problems_Part 1_Solutions.pdf

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