charges, forces, fields, capacitance, and energy

8/13/2019 Charges, Forces, Fields, Capacitance, and Energy

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Version 001 HW#3 - Charges, Forces, Fields, Capacitance, and Energy arts (00224)

This print-out should have 10 questions.Multiple-choice questions may continue onthe next column or page find all choicesbefore answering.

Charging Two Metal Balls

001 10.0 pointsTwo uncharged metal balls, Z and X, standon insulating glass rods. A third ball, carryinga positive charge, is brought near the ball Xas shown in the figure. A conducting wire isthen run betweenZandXand then removed.Finally the third ball is removed.

X Z+

conducting wire

When all this is finished

1.ballZ is negative and ballX is neutral.

2.ballZ is neutral and ballX is positive.

3. ball Z is positive and ball X is negative.correct

4.ballZ is positive and ballX is neutral.

5.ballZis negative and ball Xis positive.

6.ballsZ and X are both negative, but ballXcarries more charge than ball Z.

7.ballsZ and X are both positive.

8.ballZ is neutral and ballX is negative.

9.ballsZ and X are still uncharged.

10.ballsZ and X are both negative, but ballZcarries more charge than ball X.

Explanation:When the conducting wire is run between

Z andX, some positive charge flows fromXtoZunder the influence of the positive charge

of the third ball.Therefore, after the wire is removed, Z i

charged positive andX is charged negative.

Three Point Charges 14002 10.0 points

Three equal charges of 4 C are in the x-plane. One is placed at the origin, another iplaced at (0, 28 cm), and the last is placed a(80 cm, 0).

Calculate the magnitude of the force on thcharge at the origin. The Coulomb constanis 9109 N m2/C2 .

Correct answer: 1.85046 N.

Explanation:

Let : q= 4C = 4106 C ,(x1, y1) = (0, 28 cm) = (0, 0.28 m) ,

(x2, y2) = (80 cm, 0) = (0.8 m, 0) and

(x0, y0) = (0, 0) .

The electric fields are

Ey= ke q

x21

+y21

=9

109 N m2/C2 4106 C

0 + (0.28 m)2

= 4.59184105 N/C , andEx=

ke q

x22

+y22

=

9109 N m2/C2 4106 C

(0.8 m)2 + 0

= 56250 N/C , so

E2 =E2x+E2y

= (4.59184 105 N/C)2 + (56250 N/C)2= 2.14014 1011 N2/C2 and

F =q E=q

E2x+E2y

=

4106 C

2.140141011 N2/C2

= 1.85046 N .


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Four Point Charges 02 S003 10.0 points

Four point charges are placed at the four cor-ners of a square. Each side of the square hasa lengthL.

q1=q

q3= q

q2= q

q4= q

P

L

L

Find the magnitude of the electric force onq2 due to all three charges q1, q3 and q4 forL= 1 m and q= 1.46C.

Correct answer: 0.0287368 N.

Explanation:

F1

F4y F4

F3

F4xq2

From the figure,

F1x=k q2

L2

= 8.98755109 Nm2/C2(1.4610

6 C)2

(1 m)2

=0.0191579 N ,

F3y =k q2

L2

= 8.98755 109 Nm2/C2

(1.46106 C)2

(1 m)2

=0.0191579 N ,

F4x= k q2

2 L21

2

=8.98755109 Nm2/C2

2(1 m)2

(1.46106 C)2

2= 0.00677333 N , and

F4y =k q2

2 L21

2

=8.98755109 Nm2/C2

2(1 m)2

(1.46106 C)2

2=0.00677333 N .

Fx= F1x+F4x

=0.0191579 N + 0.00677333 N=0.0123845 N and

Fy =F3y+F4y

=

0.0191579 N

+ (0.00677333 N)=0.0259312 N , so

F2 =F2x +F2y= (0.0123845 N)2

+ (0.0259312 N)2= 0.000825803 N2

F= 0.000825803 N

2

= 0.0287368 N .

Electric Potential and Field 01004 (part 1 of 4) 10.0 points

Two charges are located in the (x, y) planas shown in the figure below. The fields produced by these charges are observed at a poin

pwith coordinates (0, 0).


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7.6 C8.1 C

p

1.6 m1.7 m

3 m2.8 m

Use Coulombs law to find thex-componentof the electric field atp.

Correct answer: 1.10131010 N/C.Explanation:

Let : ke= 8.98755109 Nm2/C2 ,(xp, yp) = (0, 0) ,

r=

x2 +y2 ,

(x1, y1) = (3 m,1.6 m)Q1=7.6 C ,

(x2, y2) = (2.8 m,1.7 m) , andQ2= 8.1 C .

q1q2

p

y1y2

x1x2

The figure below shows the electric fieldvectors.

E1

E2

7.6 C8.1 C

Let : 1= 331.928

2= 31.2637

In thex-direction, the contributions from the

two charges are

Ex1 =keQ1r21

cos(180 1) (1

=keQ1r21

(0x1)r1

(2

=(8.98755109 Nm2/C2)7.6 C

(3.4 m)23 m

3.4 m

= 5.21362 109 N/C .

Ex2 =keQ2r22

cos(180 2) (1

=keQ2r22

(0x2)r2

(2

=(8.98755109 Nm2/C2) 8.1 C

(3.27567 m)22.8 m

3.27567 m

= 5.79942 109 N/C .

Ex= Ex1+Ex2= 5.21362109 N/C

+ 5.79942109 N/C

= 1.10131010

N/C .

005 (part 2 of 4) 10.0 pointsLet: V= 0 at infinity.

Find the electric potential atpwith coordinates (0, 0) .

Correct answer: 2.1344109 V.Explanation:

The potential for a point chargeQ is

V =keQ

r .

r1=

x21

+y21

=

(3 m)2 + (1.6 m)2= 3.4 m .


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r2=

x22

+y22

=

(2.8 m)2 + (1.7 m)2= 3.27567 m .

For the two charges in this problem, wehave

V1 = keQ1r1

= (8.98755 109 Nm2/C2)7.6 C

3.4 m=2.008981010 V .

V2 = keQ2r2

= (8.98755 109 Nm2/C2) 8.1 C

3.27567 m= 2.22242 1010 V .

Vp= V1+V2

=2.008981010

V+ 2.22242 1010 V

= 2.1344109 V .

006 (part 3 of 4) 10.0 pointsFind the y-component of the electric field at

pby taking the appropriate partial derivativeof the potential function from part 2.

Correct answer: 7.40481 108

N/C.Explanation:

r= (y2 +x2)1/2

so r

y =

1

2

y2 +x2

1/2 2 y V

r =k Q

r2.

To calculate the electric field from the potential

Ey= Vy

= Vr

r

y

=k Q

r2

1

2 y2 +x2

1/2 2 y

=kQ

r2 y

r =k

Q y

r3 ,

Ey1 =keQ1r21

sin(180 1) (3

=keQ1r21

(0y1)r1

(4

=(8.98755109 Nm2/C2)

7.6 C

(3.4 m)

2

1.6 m

3.4 m=2.7806109 N/C .

Ey2 =keQ2r22

sin(180 2) (3

=keQ2r22

(0y2)r2

(4

=(8.98755109 Nm2/C2) 8.1 C

(3.27567 m)21.7 m

3.27567 m

= 3.52108109 N/C .

Ey =Ey1+Ey2

=2.7806109 N/C+ 3.52108 109 N/C

= 7.40481 108 N/C .

007 (part 4 of 4) 0.0 pointsLet: Points be at position (0, 1).

FindV s V pby integratingE dsover straight line path between the two points.

Correct answer:6.24542108 V.Explanation:

In general, the potential difference betweenany two pointsAand B is

VBVA= BA

Eds


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LetAcorrespond to the originpandB to thepoints on the positivey-axis

VsVp= sp

Eds

=

s

p

Eydy .

and from Part 3

VsVp =k Q sp

y

r3dy .

Notice that

d

dy

1

r

=1

r2 y

r =y

r3,

using the chain rule and ry

from Part 3. So

VsVp= k Qr

sp

=k Q

rsk Q

rp.

But, of course,

Vs=k Q

rs

andVp=

k Qrp

,

so all we need to do is evaluate the potentialat points and at point p and subtract them.

From above, Vp = 2.1344109 V. To findVs, we need the distance from each charge topoints, that isr1 ,sand r2 ,s.

r1 ,s= x21

+ (

y1+ 1)2

=

(3 m)2 + (2.6 m)2

= 3.96989 m .

r2 ,s=

x22

+ (y2+ 1)2

=

(2.8 m)2 + (2.7 m)2= 3.88973 m .

For the two charges in this problem, whave

V1 ,s= keQ1r1 ,s

= (8.98755 109 Nm2/C2)

7.6 C3.96989 m=1.720591010 V .

V2 ,s= keQ2r2 ,s

= (8.98755 109 Nm2/C2) 8.1 C

3.88973 m= 1.871571010 V .

Vs= V1 ,s+V2 ,s

=1.720591010 V+ 1.87157 1010 V

= 1.50986109 V .The potential difference Vis given by

V =VsVp= 1.50986

109 V

2.1344109 V=6.24542108 V .

Equipotential Surfaces 02008 (part 1 of 2) 10.0 points

Consider the figure+Q

#1+++++++

++++

Q

#2

A

B

C Dx

y

Of the following elements, identify all thacorrespond to an equipotential line or surface

1.neitherAB nor CD


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2.lineCDonly

3.bothAB andCD

4.lineAB onlycorrect

Explanation:Consider the electric field

+Q

#1+++++++++++

Q

#2

A

B

C D x

y

An equipotential line or surface (AB) isnormal to the electric field lines.

009 (part 2 of 2) 10.0 pointsConsider the figure

Aq +B

+q

C

DOf the following elements, identify all that

correspond to an equipotential line or surface.

1.lineAB only

2.bothAB andCD

3.lineCDonlycorrect

4.neitherAB norC D

Explanation:Consider the electric field:

A

+

B

C

D

An equipotential line or surface (CD) inormal to the electric field lines.

Capacitor Circuit 09010 10.0 points

Consider the group of capacitors shown in thfigure.

EB

C

C

C 2 Cda b c

Find the equivalent capacitance Cad between pointsa and d.

1.Cad =1

2Ccorrect

2.Cad = 4 C

3.Cad =3

5C

4.Cad =1

3C

5.Cad = 5 C

6.Cad =25

C

7.Cad =2

3C

8.Cad =3

4C

9.Cad = 3 C

10.Cad = 2 C


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Explanation:For capacitors in series,

1

Cseries= 1

Ci

Vseries = Vi,

and the individual charges are the same.For parallel capacitors,

Cparallel =

Ci

Qparallel =

Qi,

and the individual voltages are the same.CandCare connected parallel with equiv-

alent capacitance

Cbc= C+C

= 2 C .

EB

2 CC 2 Cda b c

C, 2 C,and 2 Care connected in series withequivalent capacitance

Cad =

1

C+

1

2 C+

1

2 C

1

=

4

2 C

1

=1

2C .

charges, forces, fields, capacitance, and energy

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