Physics 112 Homework 2 (solutions) (2004 Fall)

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Solutions to Homework Questions 2

Chapt16, Problem-1: A proton moves 2.00 cm parallel to a uniform electric field with E = 200N/C. (a) How much work is done by the field on the proton? (b) What change occurs in the potential energyof the proton? (c) Through what potential difference did the proton move?

Solut ion:

(a) The work done is W = F !s cos" = qE( ) !s cos" , or

W = 1.60 !10

"19 C( ) 200 N C( ) 2.00 !10

"2 m( )cos 0° =

6.40 !10"19

J

(b) The change in the electrical potential energy is

!PEe = "W =

! 6.40" 10!19

J

(c) The change in the electrical potential is

!V =

!PEe

q="6.40 #10

"19 J

1.60# 10-19

C=

! 4.00 V

Chapt16, Problem-3: A potential difference of 90 mV exists between the inner and outer surfacesof a cell membrane. The inner surface is negative relative to the outer surface. How much work is requiredto eject a positive sodium ion (Na

+) from the interior of the cell?

Solut ion:

The work done by the agent moving the charge out of the cell is

Winput = !Wfield = ! !"PEe( ) = +q "V( )

= 1.60 !10

"19 C( ) + 90! 10

"3

J

C

# $ %

& ' ( =

1.4!10"20

J

Chapt16, Problem-7: A pair of oppositely charged parallel plates are separated by 5.33 mm. Apotential difference of 600 V exists between the plates. (a) What is the magnitude of the electric fieldbetween the plates? (b) What is the magnitude of the force on an electron between the plates? (c) How muchwork must be done on the electron to move it to the negative plate if it is initially positioned 2.90 mm fromthe positive plate?

Solut ion:

(a) E =

!V

d=

600 J C

5.33 "10#3

m=

1.13! 10

5 N C

(b) F = q E= 1.60 !10

"19 C( ) 1.13 !10

5 N C( )=

1.80! 10"14

N

(c) W = F !s cos"

= 1.80 !10

"14 N( ) 5.33 " 2.90( )!10

"3 m[ ]cos 0° =

4.38 !10"17

J

Physics 112 Homework 2 (solutions) (2004 Fall)

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Chapt16, Problem-8: Suppose an electron is released from rest in a uniform electric field whosestrength is 5.90x10

3 V/m. (a) Through what potential difference will it have passed after moving 1.00 cm?

(b) How fast will the electron be moving after it has traveled 1.00 cm?

Solut ion:

(a) !V = Ed = 5.90 "10

3 V m( ) 1.00" 10

#2 m( )=

59.0 V

(b)

1

2me v2

! 0 = "KE =W = !"PEe = q "V( ) ,

so

v=2 q !V( )

me

=2 1.60 "10#19 C( ) 59.0 J C( )

9.11"10#31 kg=

4.55! 10

6 m s

Chapt16, Problem-15: Two point charges Q1 = +5.00 nC and Q2 = –3.00 nC are separated by35.0 cm. (a) What is the electric potential at a point midway between the charges? (b) What is the potentialenergy of the pair of charges? What is the significance of the algebraic sign of your answer?

Solut ion:

(a) V =

ke qi

rii

!

= 8.99! 109 N "m 2

C2

#

$ %

&

' (

5.00 !10)9

C

0.175 m)

3.00 !10)9

C

0.175 m

#

$ %

&

' ( =

103 V

(b) PE=

ke qiq2

r12

= 8.99! 109 N "m 2

C2

#

$ %

&

' (

5.00! 10)9

C( ) ) 3.00! 10)9

C( )0.350 m

= ! 3.85 "10

!7 J

The negative sign means that positive work must be done to separate the charges (i.e., bring them up to

a state of zero potential energy).

Chapt16, Problem-19: In Rutherford’s famous scattering experiments (which led to theplanetary model of the atom), alpha particles (having charges of +2e and masses of 6.64x10

–27 kg) were

fired toward a gold nucleus with charge +79e. An alpha particle, initially very far from the gold nucleus, isfired at 2.00x10

7 m/s directly toward the gold nucleus as in Figure P16.19. How close does the alpha particle

get to the gold nucleus before turning around? Assume the gold nucleus remains stationary.

Solut ion:

From conservation of energy, KE +PEe( )

f= KE+ PEe( )

i, which gives

0+keQq

rf

=1

2m! vi

2+ 0 or

rf =

2 keQq

m! vi2=

2 ke 79e( ) 2e( )m! vi

2

rf =

2 8.99! 109 N "m2

C2

#

$ %

&

' ( 158( ) 1.60! 10)19 C( )

2

6.64 !10)27 kg( ) 2 .00 !107 m s( )2 =

2 .74 !10

"14 m

Physics 112 Homework 2 (solutions) (2004 Fall)

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Chapt16, Problem-20: Starting with the definition of work, prove that at every point on anequipotential surface the surface must be perpendicular to the local electric field

Solut ion:

By definition, the work required to move a charge from one point to any other point on an equipotential

surface is zero. From the definition of work, W = Fcos!( )" s , the work is zero only if s= 0 or Fcos! = 0 .

The displacement s cannot be assumed to be zero in all cases. Thus, one must require that Fcos! = 0 . The

force F is given by F = qE and neither the charge q nor the field strength E can be assumed to be zero in

all cases. Therefore, the only way the work can be zero in all cases is if cos! = 0 . But if cos! = 0 , then

! = 90° or the force (and hence the electric field) must be perpendicular to the displacement s (which is

tangent to the surface). That is, the field must be perpendicular to the equipotential surface at all points

on that surface.

Chapt16, Problem-25: An air-filled capacitor consists of two parallel plates, each with an area of7.60 cm

2, separated by a distance of 1.80 mm. If a 20.0-V potential difference is applied to these plates,

calculate (a) the electric field between the plates, (b) the capacitance, and (c) the charge on each plate.

Solut ion:

(a) E =

!V

d=

20.0 V

1.80" 10-3

m= 1.11" 10

4 V m =

11.1 kV m directed toward the negative plate

(b) C =

!0

A

d=

8.85" 10#12

C2

N $m2( ) 7.60" 10

#4 m

2( )1.80"10

-3 m

= 3.74 !10"12

F = 3.74 pF

(c) Q = C !V( ) = 3.74" 10

#12 F( ) 20.0 V( ) = 7.47 "10

#11 C = 74.7 pC on one plate and ! 74.7 pC on the

other plate.

Chapt16, Problem-28: A small object with a mass of 350 mg carries a charge of 30.0 nC and issuspended by a thread between the vertical plates of a parallel-plate capacitor. The plates are separated by4.00 cm. If the thread makes an angle of 15.0° with the vertical, what is the potential difference between theplates?

Solut ion:

!Fy = 0 " T cos15.0° =mg or T =

mg

cos15.0°

!Fx = 0 " qE= T sin15.0° =mg tan15.0°

or

E =

mg tan15.0°

q

!V =Ed =

mgd tan15.0°

q

!V =

350"10#6 kg( ) 9.80 m s2( ) 0.0400 m( ) tan15.0°

30.0" 10#9 C= 1.23 "103 V =

1.23 kV

15.0°T

F = qE

mg

Physics 112 Homework 2 (solutions) (2004 Fall)

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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 212.0 V

3.00 µF6.00 µF

a c

Figure 312.0 V

2.00 µF

a b c

4.00 µF

2.00 µF

3.00 µF

Figure 112.0 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

Physics 112 Homework 2 (solutions) (2004 Fall)

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Chapt16, Problem-51: A model of a red blood cell portrays the cell as a spherical capacitor—apositively charged liquid sphere of surface area A, separated by a membrane of thickness t from thesurrounding negatively charged fluid. Tiny electrodes introduced into the interior of the cell show apotential difference of 100 mV across the membrane. The membrane’s thickness is estimated to be 100 nmand its dielectric constant to be 5.00. (a) If an average red blood cell has a mass of 1.00x10

–12 kg, estimate

the volume of the cell and thus find its surface area. The density of blood is 1100 kg/m3. (b) Estimate the

capacitance of the cell. (c) Calculate the charge on the surface of the membrane. How many electroniccharges does this represent?

Solut ion:

(a)

V =m

!=

1.00 "10#12 kg

1100 kg m3 = 9.09! 10

"16 m

3

Since

V =4!r

3

3, the radius is

r =3V

4!

"

# $

%

& '

1 3

, and the surface area is

A = 4!r2 = 4!

3V

4!

"

# $

%

& '

2 3

= 4!3 9.09( 10

)16 m

3( )4!

"

#

$ $

%

&

' '

2 3

= 4.54! 10

"10 m

2

(b) C =

! "0 A

d

=5.00( ) 8.85 !10

"12 C

2N #m

2( ) 4.54! 10"10

m2( )

100! 10"9

m=

2 .01! 10

"13 F

(c) Q = C !V( ) = 2 .01"10

#13 F( ) 100 "10

-3 V( )=

2 .01! 10

"14 C ,

and the number of electronic charges is

n =

Q

e=

2.01!10"14

C

1.60! 10-19

C=

1.26! 10

5

Chapt16, Problem-54: Charges of equal magnitude 1.00x10–15

C and opposite sign aredistributed over the inner and outer surfaces of the cell wall in Figure P16.54. Find the force on thepotassium on (K

+) if the ion is (a) 2.70 µm from the center of the cell, (b) 2.92 µm from the center, and (c)

4.00 µm from the center.

Solut ion:

(a) At r = 2.70 µm from the cell center, the ion is located inside the spherically symmetric charge

distributions on the surfaces of the cell wall. From Gauss’s law, the field is zero at such a location, so

F = qE = q 0( ) = 0 .

Physics 112 Homework 2 (solutions) (2004 Fall)

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(b) With spherically symmetric charge distributions, consider a spherical gaussian surface of radius

r = 2.92 µm centered on the cell center. From Gauss’s law, the electric field at points on the gaussian

surface is the same as if the total charge inside the surface, Q = !1.00" 10!15

C , was located at the

center. Thus,

F = qE = qkeQ

r2

! " #

$ % &

= +1.60' 10(19

C( ) 8.99 '109

N )m 2

C2

!

" #

$

% & (1.00' 10

(15 C( )

2.92 '10-6

m( )2

*

+

, , ,

-

.

/ / /

F = !1.69" 10!13

N

or F = 1.69! 10"13 N directed radially inward

(c) At r = 4.00 µm from the cell center, the ion is located outside a spherically symmetric charge

distribution having zero net charge. From Gauss’s law, the field is zero at such a location, so

F = qE = q 0( ) = 0 .

Chapt16, Problem-55: A virus rests on the bottom plate of oppositely charged parallel plates inthe vacuum chamber of an electron microscope. The electric field strength between the plates is 2.00x10

5

N/C, and the bottom plate is negative. If the virus has a mass of 1.00x10–15

kg and suddenly acquires acharge of –1.60x10

–19 C, what are its velocity and position 75.0 ms later? Do not disregard gravity.

Solut ion:

The electric field is directed vertically downward with Ey = !2.00" 10

5 N C . Thus,

ay =

!Fy

m=

qEy "mg

m=

"1.60 #10"19 C( ) "2 .00 #105 N C( )1.00# 10-15 kg

" 9.80 m s2 ,

or ay = +22.2 m s

2

At t = 75.0 ms = 0.0750 s ,

vy = vyi + ay t = 0+ 22.2 m s

2( ) 0.0750 s( ) = 1.67 m s upward , and

y = vyi t+

1

2ay t2 = 0+

1

222.2 m s

2( ) 0.0750 s( )2

= 0.0624 m = 6.24 cm above bottom plate

Chapt16, Conceptual-5: Suppose you are sitting in a car and a 20 kV power line drops acrossthe car. Should you stay in the car or get out? The power line potential is 20 kV compared to the potentialof the ground.

Solut ion:

[Adapted from the solution in the text] The rubber (a good insulator) tires on the car means that there is

not good ‘electrical contact’ between the rest of the car (including occupants) and the Earth (ground). Thus

if the power line makes electrical contact with the metal of the car, it will raise the potential of the car to

20kV. It will also most-likely raise the potential of your body to 20kV, because you are undoubtedly in good

electrical contact with the car. In itself, this is not a problem – electrostatic equilibrium will be

established, and charge will quickly cease to flow. However if you step out of the car, your body (with a

potential of 20 kV) will make contact with the ground, which is at a potential of zero Volts (by definition)

As a result, a current will pass through your body, and you are likely to be injured. Thus it is best to stay in

the car until help arrives.

Physics 112 Homework 2 (solutions) (2004 Fall)

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Chapt16, Conceptual-6: Why is it important to avoid sharp edges or points on conductors usedin high-voltage equipment?

Solut ion:

A sharp point on a charged conductor would produce a large electric field in the region near the point (see

Section 15.6). Therefore an unintended electric discharge could take place at such a location.

Chapt16, Conceptual-9: [Sorry, I did not spot the typo in the textbook –here is a rewording of the question] Why is it dangerous to touch the terminals of a high-voltage capacitor even after the voltage source that charged the battery capacitor is disconnected? from thecapacitor? What can be done to make the capacitor safe to handle after the voltage source has beenremoved?

Solut ion:

The capacitor often remains charged long after the voltage source is disconnected. This residual charge

can be lethal. The capacitor can be safely handled after discharging the plates by short-circuiting the

device with a conductor, such as a screwdriver with an insulating handle.

Chapt16, Conceptual-15: If you were asked to design a capacitor where small size and largecapacitance were required. What factors would be important in your design?

Solut ion:

You should use a dielectric-filled capacitor whose dielectric constant is very large. Furthermore, you should

make the dielectric as thin as possible, keeping in mind that it cannot be too thin or else dielectric

breakdown will occur.