Homework # 2 Ch 26

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Homework # 2 Ch 26

Due: 11:59pm on Friday, October 23, 2015

T o u n d e r s t a n d h o w p o i n t s a r e a w a r d e d , r e a d t h e G r a d i n g P o l i c y f o r t h i s a s s i g n m e n t .

Electric Field due to Multiple Point Charges

Two point charges are placed on the x axis. The f irst charge,

= 8. 00 , is placed a dist ance 16. 0 f rom t he origin

along t he posit ive x axis; t he second charge, = 6. 00 ,

is placed a dist ance 9.00 from the origin along the

negat ive x a x i s .

Part A

Calculat e t he elect ric f ield at point A, locat ed at coordinat es (0 , 12. 0 ).

Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per

coulomb to three significant figures.

Hint 1. How to approach the problem

Find t he cont ribut ions t o t he elect ric f ield at point A separat ely f or and , t hen add t hem t oget her (using

vect or addit ion) t o f ind t he t ot al elect ric f ield at t hat point . You will need t o use t he Pyt hagorean t heorem t o

f ind t he dist ance of each charge f rom point A.

Hint 2. Calculate the distance from each charge to point A

Calculate the distance from each charge to point A.

Enter the two distances, separated by a comma, in meters to three significant figures.

ANSWER:

Hint 3. Determine the directions of the electric fields

Which of t he f ollowing describes t he direct ions of t he elect ric f ields and creat ed by charges

and at point A?

ANSWER:

1

n C m

2

n C

m

m m

1

2

, = 2 0. 0 , 15 . 0 A 1

A 2

m

E

⃗

A 1

E

⃗

A 2

1

2

T y p e s e t t i n g m a t h : 2 3 %

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Hint 4. Calculate the components of

Calculate the x andy component s of t he elect ric f ield at point A due t o charge .

Express your answers in newtons per coulomb, separated by a comma, to three significant figures.

Hint 1. Calculate the magnitude of the total field

Calculate the magnitude of the field at point A due to charge only.

Express your answer in newtons per coulomb to three significant figures.

ANSWER:

Hint 2. How to find the components of the total field

Once you have found the magnitude of the field, use trigonometry to determine the x andy

component s of t he f ield. The elect ric f ield of a posit ive point charge point s direct ly away f rom t he

charge, so t he direct ion of t he elect ric f ield at point A due t o charge will be along t he line joining

the two. Use the posit ion coordinates of and point A to find the angle that the line joining the twomakes wit h t he x ory axis. Then use t his angle t o resolve t he elect ric f ield vect or int o component s.

ANSWER:

Hint 5. Calculate the components of

Calculate the x andy component s of t he elect ric f ield at point A due t o charge .

Express your answers in newtons per coulomb, separated by a comma, to three significant figures.

Hint 1. Calculate the magnitude of the total field

Calculate the magnitude of the field at point A due to charge only.

Express your answer in newtons per coulomb to three significant figures.

ANSWER:

points up and left and points up and right.

points up and left and points down and left .

points down and right and points up and right.

points down and right and points down and left .

E

⃗

A 1

E

⃗

A 2

E

⃗

A 1

E

⃗

A 2

E

⃗

A 1

E

⃗

A 2

E

⃗

A 1

E

⃗

A 2

E

⃗

A 1

E

⃗

A 1

1

E

A 1

1

= 0. 180E A 1

N / C

1

1

= -0.144,0.108 , E A 1

E

A 1

N / C

E

⃗

A 2

2

E

A 2

2

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Hint 2. How to find the components of the total field

Once you have found the magnitude of the field, use trigonometry to determine the x andy

component s of t he f ield. The elect ric f ield of a posit ive point charge point s direct ly away f rom t he

charge, so t he direct ion of t he elect ric f ield at point A due t o charge will be along t he line joining

the two. Use the posit ion coordinates of and point A to find the angle that the line joining the twomakes wit h t he x ory axis. Then use t his angle t o resolve t he elect ric f ield vect or int o component s.

ANSWER:

ANSWER:

Correct

Part B

An unknown addit ional charge is now placed at point B, locat ed at coordinat es (0 , 15. 0 {\rm m} ).

Find the magnitude and sign of \texttip{q_{\rm 3}}{q_3} needed t o make t he t ot al elect ric f ield at point A equal t o

zero.

Express your answer in nanocoulombs to three significant figures.


You have already calculat ed t he elect ric f ield at point A due t o \texttip{q_{\rm 1}}{q_1} and \ t ext t ip{q_{\ rm 2}}

{q_2}. Now find the charge \texttip{q_{\rm 3}}{q_3} needed to make an opposite field at point A, so when the

two are added together the total field is zero.

Hint 2. Determine the sign of the charge

Which sign of charge \texttip{q_{\rm 3}}{q_3} is needed t o creat e an elect ric f ield \ t ext t ip{\ vec{E}_{\ rm A3}}

{E_A3_vec} t hat point s in t he opposit e direct ion of t he t ot al f ield due t o t he ot her t wo charges,

\texttip{q_{\rm 1}}{q_1} and\texttip{q_{\rm 2}}{q_2}?

ANSWER:

= 0. 240E A 2

N / C

2

2

= 0.144,0.192 , E A 2

E

A 2

N / C

= 0, 0. 300 , E A

E

A

N / C

3

m

posit ive

negat ive

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Hint 3. Calculating the magnitude of the new charge

Keep in mind that the magnitude of the field due to \texttip{q_{\rm 3}}{q_3} is E_{\rm A3}=kq_3/r_{\rm A3}^2,

and the field must be equal in magnitude to the field due to charges \texttip{q_{\rm 1}}{q_1} and

\texttip{q_{\rm 2}}{q_2}.

ANSWER:

Correct

PhET Tutorial: Charges and Electric Fields

Learning Goal:

To underst and t he spat ial dist ribut ion of t he elect ric f ield f or a variet y of simple charge conf igurat ions.

For t his problem, use t he PhET simulat ion Charges and Fields. This simulat ion allows you t o place mult iple posit ive

and negative point-charges in any configuration and look at the resulting electric field.

Start the simulation. You can click and drag positive charges (red) or negative charges (blue) into the main screen. If

you select Show E-field in the green menu, red arrows will appear, showing the direction of the electric field. Faint re

arrows indicate that the electric field is weaker than at locations where the arrows are brighter (this simulation does not

use arrow length as a measure of field magnitude).

Feel free to play around with the simulation. When you are done, click Clear All before beginning Part A.

Part A

Select Show E-field andgrid in the green menu. Drag one positive charge and place it near the middle of the

screen, right on t op of t wo int ersect ing bold grid lines. You should see somet hing similar t o t he f igure below.

\texttip{q_{\rm 3}}{q_3} = 0 .3 00 \rm{nC}

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ANSWER:

Correct

This means that another positive charge, if placed near the original charge, would experience a force directed

radially away from the original charge.

Part B

Now, let 's look at how t he dist ance f rom t he charge af f ect s t he magnit ude of t he elect ric f ield. Select Show

numbers on the green menu, and then click and drag one of the orange E-Field Sensors. You will see the

magnit ude of t he elect ric f ield given in unit s of V/ m (volt s per met er, which is t he same as newt ons per coulomb).

Place t he E-Field Sensor 1 \ rm m away from the positive charge (1 \ rm m is two bold grid lines away if going in a

horizont al or vert ical direct ion), and look at t he result ing f ield st rengt h.

Consider t he locat ions t o t he right , lef t , above, and below t he posit ive charge, all 1 \ rm m away.

ANSWER:

The elect ric f ield produced by t he posit ive charge

is directed radially away from the charge at all locations near the charge.

is directed radially toward the charge at all locations near the charge.

wraps circularly around the positive charge.

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Correct

This result implies t hat t he st rengt h of t he elect ric f ield due t o one point charge depends solely on t he

dist ance away f rom t he charge. Mat hemat ically, we say t he elect ric f ield is spherically symmet ric.

Part C

What is the magnitude of the electric field 1 m away from the positive charge compared to the magnitude of the

elect ric f ield 2 m away?


Use an E-Field Sensor to determine the field strength both at 1 \ rm m away and at 2 \ rm m away from the

charge. Then, t ake t he rat io of t he t wo f ield st rengt hs.

ANSWER:

Correct

The magnit ude of t he f ield decreases more quickly t han t he inverse of t he dist ance f rom t he charge. The

magnit ude of t he elect ric f ield is proport ional t o t he inverse of t he dist ance squared ( E \propto1/r^2, where\ t ext t ip{r}{r} is t he dist ance f rom t he charge). You should verif y t his by looking at t he f ield st rengt h 3 or 4

met ers away. This is consist ent wit h Coulombâs law, which st at es t hat t he magnit ude of t he f orce bet ween

t wo charged part icles is F = k Q_1Q_2/ r^2.

Part D

I f t he f ield st rengt h is \ t ext t ip{E}{E} = 9 \ rm V/ m a dist ance of 1 \ rm m from the charge, what is the field strength

\ t ext t ip{E}{E} a dist ance of 3 \ rm m from the charge?

For t hese f our locat ions, t he magnit ude of t he elect ric f ield is

greatest below the charge.

great est t o t he lef t of t he charge.

the same.

greatest above the charge.

greatest to the right of the charge.

The magnit ude of t he elect ric f ield 1 \ rm m away f rom t he posit ive charge is

equal to

t wo t imes

f our t imes

one-half

one-quarter

t he magnit ude of t he elect ric f ield 2 \ rm m away.

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The magnit ude of t he elect ric f ield is inversely proport ional t o dist ance squared (E \propto 1/r^2). So if t he

dist ance is increased by a f act or of t hree, t he f ield st rengt h must decrease by a f act or of t hree squared.

You could use the simulation to make a measurement (you might have to drag the charge away from the

center so you have enough room to get 3 \ rm m away).

ANSWER:

Correct

Correct . Since E \propto 1/r^2, if t he dist ance is increased by a f act or of t hree, t he elect ric f ield is decreased

by a f act or of nine.

Part E

Remove the positive charge by dragging it back to the basket, and drag a negative charge (blue) toward the midd

of t he screen. Det ermine how t he elect ric f ield is dif f erent f rom t hat of t he posit ive charge.

Which st at ement best describes t he dif f erences in t he elect ric f ield due t o a negat ive charge as compared t o a

posit ive charge?

ANSWER:

Correct

The elect ric f ield is now direct ed t oward t he negat ive charge, but t he f ield st rengt h doesnât change. The

elect ric f ield of a point charge is given by \ vec{E}=(kQ/ r^2)\ hat {r}. Because of t he sign of t he charge, t he f ield

produced by a negat ive charge is direct ed opposit e t o t hat of a posit ive charge but t he magnit ude of t he f ield

is t he same.

Part F

Now, remove the negative charge, and drag two positive charges, placing them 1 \ rm m apart, as shown below.

\ t ext t ip{E}{E} = 1 \ rm V/ m

The elect ric f ield changes direct ion (now point s radially inward), and t he magnit ude of t he elect ric f ield

decreases at all locat ions.

The elect ric f ield changes direct ion (now point s radially inward), but t he elect ric f ield st rengt h does not

change.

Not hing changes; t he elect ric f ield remains direct ed radially out ward, and t he elect ric f ield st rengt h

doesnât change.

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Let âs look at t he result ing elect ric f ield due t o bot h charges. Recall t hat t he elect ric f ield is a vect or, so t he net

elect ric f ield is t he vect or sum of t he elect ric f ields due t o each of t he t wo charges.

Where is the magnitude of the electric field roughly equal to zero (other than very far away from the charges)?

ANSWER:

Correct

Directly between the two charges, the electric fields produced by each charge are equal in magnitude and

point in opposit e direct ions, so t he t wo vect ors add up t o zero.

Part G

Consider a point 0.5 \ rm m above t he midpoint of t he t wo charges. As you can verif y by removing one of t he

posit ive charges, t he elect ric f ield due t o only one of t he posit ive charges is about 18 \ rm V/ m. What is t he

magnit ude of t he t ot al elect ric f ield due t o bot h charges at t his locat ion?

ANSWER:

The elect ric f ield is zero at any locat ion along a vert ical line going t hrough t he point direct ly bet ween t he

two charges.

The elect ric f ield is nonzero everywhere on t he screen.

The electric field is roughly zero near the midpoint of the two charges.

zero

25 \ rm V/ m

36 \ rm V/ m

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Correct

Not ice t hat t his number is less t han t wice t he magnit ude of t he f ield due t o each charge. This occurs because

t he horizont al component s of t he elect ric f ield due t o each charge exact ly cancel out (add t o zero). Only t he

vert ical component s of t he f ields add t oget her.

Part H

Make an elect ric dipole by replacing one of t he posit ive charges wit h a negat ive charge, so t he f inal conf igurat ion

looks like the figure shown below.

ANSWER:

Correct

The elect ric f ield due t o t he posit ive charge is direct ed t o t he right , as is t he elect ric f ield due t o t he negat ive

charge. So t he net elect ric f ield, which is t he sum of t hese t wo f ields, is also t o t he right .

Part I

Make a small dipole by bringing the two charges very close to each other, where they are barely touching. The

midpoint of the two charges should still be on one of the grid point intersections (see figure below).

The elect ric f ield at t he midpoint is

direct ed t o t he lef t .

zero.

directed to the right.

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Measure t he st rengt h of t he elect ric f ield 0. 5 \ rm m direct ly above t he midpoint as well as 1 \ rm m direct ly above

Does t he st rengt h of t he elect ric f ield decrease as 1 over dist ance squared ( 1/r^2)?


I f t he st rengt h of t he f ield is decreasing as 1/r^2, t hen t he rat io of t he magnit udes of t he elect ric f ield

measured at t wo dist ances, say 0. 5 \ rm m away and 1 \ rm m away, would be

E_{r=0.5}/E_{r=1} =(1/0.5)^2/(1/1)^2=4.

Compare this value to the value you measure with an E-Field Sensor.

ANSWER:

Correct

I n f act , it t urns out t hat t he st rengt h of t he elect ric f ield decreases roughly as 1/r^3! So t he f ield 1 \ rm m above

t he midpoint is roughly eight t imes weaker t han at 0. 5 \ rm m above the midpoint. The important lesson here is

t hat , in general, a dist ribut ion of charges produces an elect ric f ield t hat is very dif f erent f rom t hat of a single

charge.

Part J

Make a long line of posit ive charges, similar t o t hat shown in t he f igure below. Try t o place all of t he charges

cent ered along a horizont al grid line. Feel f ree t o look at t he elect ric f ield, as it is int erest ing.

No, it decreases more quickly wit h dist ance.

Yes, it does.

No, it decreases less quickly wit h dist ance.

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Measure t he st rengt h of t he elect ric f ield 1 \ rm m direct ly above t he middle as well as 2 \ rm m direct ly above. Do

t he st rengt h of t he elect ric f ield decrease as 1 over dist ance squared ( 1/r^2)?

ANSWER:

Correct

I n f act , it t urns out t hat t he st rengt h of t he elect ric f ield decreases roughly as 1/ r . So t he f ield 1 m above t he

midpoint is roughly half t he st rengt h at 0. 5 \ rm m. This is anot her example showing t hat a dist ribut ion of

charges produces an elect ric f ield t hat is very different from that of a single charge.

PhET Interactive SimulationsUniversity of Colorado

http://phet.colorado.edu

Charged Ring

Consider a uniformly charged ring in the xy plane, centered at the origin. The ring has radius \ t ext t ip{a}{a} and posit ive

charge \ t ext t ip{q}{q} dist ribut ed evenly along it s circumf erence.

No, it decreases less quickly wit h dist ance.

Yes, it does.

No, it decreases more quickly wit h dist ance.

http://phet.colorado.edu/

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Part A

What is t he direct ion of t he elect ric f ield at any point on t he z a x i s ?


Approach 1

In what direction is the field due to a point on the ring? Add to this the field from a point on the opposite side

of t he ring. I n what direct ion is t he net f ield? What if you did t his f or every pair of point s on opposit e sides of

the ring?

Approach 2

Consider a general electric field at a point on the z axis, i. e. , one t hat has a z component as well as a

component in the xy plane. Now imagine that you make a copy of the ring and rotate this copy about itsaxis. As a result of t he rot at ion, t he component of t he elect ric f ield in t he xy plane will rotate also. Now you

ask a friend to look at both rings. Your friend wouldn't be able to tell them apart, because the ring that is

rot at ed looks just like t he one t hat isn't . However, t hey have t he component of t he elect ric f ield in t he xy

plane point ing in dif f erent direct ions! This apparent cont radict ion can be resolved if t his component of t he

f ield has a part icular value. What is t his value?

Does a similar argument hold for the z component of t he f ield?

ANSWER:

Correct

parallel to the x a x i s

parallel to the y a x i s

parallel to the z a x i s

in a circle parallel t o t he xy plane

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Part B

What is t he magnit ude of t he elect ric f ield along t he posit ive z a x i s ?

Use \texttip{k}{k} in your answer, where \large{k = \frac{1}{4 \pi \epsilon_0}}.

Hint 1. Formula for the electric field

You can always use Coulomb's law, \large{F = k\frac{q_1q_2}{r^2}}, t o f ind t he elect ric f ield (t he Coulomb

f orce per unit charge) due t o a point charge. Given t he f orce, t he elect ric f ield say at \texttip{q_{\rm 2}}{q_2}

due to \texttip{q_{\rm 1}}{q_1} is \large{E = \frac{F}{q_2} = k\frac{q_1}{r^2}} .

In the situation below, you should use Coulomb's law to find the contribution \ t ext t ip{dE}{dE} t o t he elect ric

f ield at t he point (0, 0, z) from a piece of charge \texttip{dq}{dq} on the ring at a distance \ t ext t ip{r}{r} away.

Then, you can integrate over the ring to find the value of \ t ext t ip{E}{E}. Consider an inf init esimal piece of t he

ring with charge \ t ext t ip{dq}{dq}. Use Coulomb's law to write the magnitude of the infinitesimal \ t ext t ip{dE}

{dE} at a point on t he posit ive z axis due to the charge \ t ext t ip{dq}{dq} shown in the figure.

Use \texttip{k}{k} in your answer, where \large{k = \frac{1}{4 \pi \epsilon_0}}. You may also use some

or all of the variables \texttip{dq}{dq}, \texttip{z}{z}, and \texttip{a}{a}.

ANSWER:

Hint 2. Simplifying with symmetry

By symmet ry, t he net f ield must point along t he z axis, away f rom t he ring, because t he horizont al

component of each contribution of magnitude \ t ext t ip{dE}{dE} is exact ly canceled by t he horizont al

component of a similar cont ribut ion of magnit ude \ t ext t ip{dE}{dE} from the other side of the ring. Therefore,

all we care about is the z component of each such cont ribut ion. What is t he component \ t ext t ip{dE_{\ mit z}}

{dE_z} of t he elect ric f ield caused by t he charge on an inf init esimally small port ion of t he ring \texttip{dq}{dq}

in t he z direct ion?

\ t ext t ip{dE}{dE} = \large{\frac{k dq}{z^{2}+a^{2}}}

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Express your answer in terms of \texttip{dE}{dE}, the infinitesimally small contribution to the electric

field; \texttip{z}{z}, the coordinate of the point on the z axis; and \texttip{a}{a}, the radius of the ring.

ANSWER:

Hint 3. Integrating around the ring

I f you combine your result s f rom t he f irst t wo hint s, you will have an expression f or \ t ext t ip{dE_{\ mit z}}

{dE_z}, t he vert ical component of t he f ield due t o t he inf init esimal charge \texttip{dq}{dq} . The t ot al f ield is

\ vec{E}=E_z\ hat {k}=\ hat {k}\ displayst yle{\ oint }_{\ rm ring}dE_z .

If you are not comfortable integrating \ t ext t ip{dq}{dq} over t he ring, change t o a spat ial variable. Since t he

total charge \ t ext t ip{q}{q} is dist ribut ed evenly about t he ring, convince yourself t hat

\large{\displaystyle{\oint}_{\rm ring}dq=\int_{0}^{2\pi}\frac{q}{2\pi}d\theta}.

ANSWER:

\ t ext t ip{dE_{\ mit z}}{dE_z} = \large{\frac{dE z}{\sqrt{z^{2}+a^{2}}}}

\ t ext t ip{E\ lef t (z\ right )}{E(z)} = \large{\frac{kqz}{\left(z^{2}+a^{2}\right)^{\frac{3}{2}}}}

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Correct

Not ice t hat t his expression is valid f or bot h posit ive and negat ive charges as well as f or point s locat ed on t he

posit ive and negat ive z axis. I f t he charge is posit ive, t he elect ric f ield should point out ward. For point s on t he

posit ive z axis, t he f ield point s in t he posit ivez direction, which is outward from the origin. For points on the

negat ive z axis, t he f ield point s in t he negat ivez direct ion, which is also out ward f rom t he origin. I f t he charge

is negat ive, t he elect ric f ield should point t oward t he origin. For point s on t he posit ive z axis, t he negat ive sign

f rom t he charge causes t he elect ric f ield t o point in t he negat ive z direction, which points toward the origin. Fo

point s on t he negat ive z axis, t he negat ive sign f rom t he z coordinate and the negative sign from the chargecancel, and t he f ield point s in t he posit ive z direction, which also points toward the origin. Therefore, even

though we obtained the above result for postive \ t ext t ip{q}{q} and\ t ext t ip{z}{z} , t he algebraic expression is

valid f or any signs of t he paramet ers. As a check, it is good t o see t hat if \ t ext t ip{|z|}{|z|} is much greater than

\ t ext t ip{a}{a} the magnitude of \ t ext t ip{E\ lef t (z\ right )}{E(z)} is approximat ely \large{k \frac{q}{z^2}}, independent

of t he size of t he ring: The f ield due t o t he ring is almost t he same as t hat due t o a point charge \ t ext t ip{q}{q}

at the origin.

Part C

I magine a small met al ball of mass \ t ext t ip{m}{m} and negative charge -q_0. The ball is released f rom rest at t he

point (0, 0, d) and constrained to move along thez axis, wit h no damping. I f 0 < d \ l l a , what will be the ball's

subsequent t raject ory?

ANSWER:

Correct

Part D

The ball will oscillat e along t he z axis between z=d and z = - d in simple harmonic motion. What will be the angular

frequency \texttip{\omega }{omega} of t hese oscillat ions? Use t he approximat ion d \ ll a t o simplif y your calculat io

t hat is, assume t hat d^2+a^2 \approx a^2.

Express your answer in terms of given charges, dimensions, and constants.

Hint 1. Simple harmonic motion

Recall the nature of simple harmonic motion of an object attached to a spring. Newton's second law for the

s y s t e m s t a t e s t h a t

\large{F_x= m\frac{d^2 x}{dt^2} = - {k'} x}, leading t o oscillat ion at a f requency of

\ large{\ omega=\ dis playst yl e{\ sqrt {\ f rac{k'}{m}}}}

(here, t he prime on t he symbol represent ing t he spring const ant is t o dist inguish it f rom \ large{k=\ f rac{1}

{4\pi\epsilon_0}} ). The solut ion t o t his dif f erent ial equat ion is a sinusoidal f unct ion of t ime wit h angular

frequency \texttip{\omega }{omega}. Write an analogous equation for the ball near the charged ring in order

t o f ind t he \texttip{\omega }{omega} t erm.

repelled from the origin

attracted toward the origin and coming to rest

oscillat ing along t he z axis between z = d and z = -d

circling around the z a x i s a t z = d

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Hint 2. Find the force on the charge

What is \ t ext t ip{F_{\ mit z}}{F_z}, t he z component of the force on the ball on the ball at the point (0,0,d)?

Use the approximation d^2 + a^2 \approx a^2.

Express your answer in terms of \texttip{q_{\rm 0}}{q_0}, \texttip{k}{k}, \texttip{q}{q}, \texttip{d}{d},

and \texttip{a}{a}.

Hint 1. A formula for the force on a charge in an electric field

The formula for the force \ t ext t ip{\ vec{F}}{F_vec} on a charge \ t ext t ip{q}{q} in an elect ric f ield

\ t ext t ip{\ vec{E}}{E_vec} is

\ vec{F} = q\ vec{E}.

Therfore, in particular,

F_z = qE_z .

You have already found \ t ext t ip{E_{\ mit z}\ lef t (z\ right )}{E_z(z)} in Part B. Use t hat expression in t he

equation above to find an expression for the z component of the force \ t ext t ip{F_{\ mit z}}{F_z} on the

ball at t he point (0,0,d). Don't forget to use the approximation given.

ANSWER:

ANSWER:

Correct

PSS 26.1: The Electric Field of Multiple Point Charges

Learning Goal:

To practice Problem-Solving Strategy 26.1 for point charge problems.

Three posit ively charged part icles, wit h charges q_1=q, q_2=2q, and q_3=q (where q > 0), are located at the corners o

a square wit h sides of lengt h \ t ext t ip{d}{d}. The charge q_2 is located diagonally from the remaining (empty) corner.

Find t he magnit ude of t he result ant elect ric f ield \ t ext t ip{\ vec{E}_{\ rm net }}{E_net _vec} in the empty corner of the

square.

PROBLEM-SOLVING STRATEGY 26.1 The elect ric f ield of mult iple point charges

MODEL: Model charged objects as point charges.

VISUALIZE: For the pictorial representation:

Est ablish a coordinat e syst em, and show t he locat ion of t he charges.

I dent if y t he point P at which you want t o calculat e t he elect ric f ield.

\ t ext t ip{F_{\ mit z}}{F_z} = \large{\frac{-k q q_{0} d}{a^{3}}}

\texttip{\omega }{omega} = \large{\sqrt{\frac{kqq_{0}}{a^{3}m}}}

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Correct

Part B

Now, consider t he result ant elect ric f ield \ t ext t ip{\ vec{E}_{\ rm net }}{E_net _vec} at P. Wit h ref erence t o t he

coordinat e syst em shown in t he previous part , which component of \ t ext t ip{\ vec{E}_{\ rm net }}{E_net _vec}, i f a n y ,

zero in this problem?

ANSWER:

Correct

Alt hough neit her of t he elect ric f ield component s is zero in t his sit uat ion, t here is st il l symmet ry t hat you can

exploit . I n part icular, t he charge conf igurat ion is symmet ric about t he line y=x . This means t hat t he x andy

component s of t he net elect ric f ield must be equal. Take advant age of t his f act t o reduce t he amount of work

you need t o do in t he Solve st ep.

Solve

Part C

Determine the magnitude \ t ext t ip{E_{\ rm net }}{E_net } of t he net elect ric f ield at point P. Use \ t ext t ip{K}{K} for the

elect rost at ic const ant .

Express your answer in terms of \texttip{q}{q}, \texttip{d}{d}, and \texttip{K}{K}.

Hint 1. Exploit symmetry

At t he end of t he Visualize st ep, it was det ermined t hat t he x andy component s of \ t ext t ip{\ vec{E}_{\ rm

net}}{E_net_vec} are equal. So you only need to compute one of them (say, the x component ) t o solve t he

problem. Note that \ t ext t ip{\ vec{E}_{\ rm 1}}{E_1_vec} has no x component, as you can see from the diagram

you drew in Part B.

Hint 2. Find the x component of the electric field due to charge 3

Find (E_3)_x, t he x component of t he elect ric f ield due t o \texttip{q_{\rm 3}}{q_3} at point P. Use \ t e x t t i p { K }

{K} f or t he elect rost at ic const ant .


ANSWER:

only t he x component

only t he y component

both the x andy components

neither the x nor they component

(E_3)_x = \large{\frac{K q}{d^{2}}}

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Hint 3. Find the x component of the electric field due to charge 2

Find (E_2)_x, t he x component of t he elect ric f ield due t o \texttip{q_{\rm 2}}{q_2} at point P. Use \ t e x t t i p { K }

{K} f or t he elect rost at ic const ant .


Hint 1. Find the magnitude of the electric field due to charge 2

What is the magnitude of \ t ext t ip{\ vec{E}_{\ rm 2}}{E_2_vec}, t he elect ric f ield due t o \ t ext t ip{q_{\ rm

2}}{q_2} at point P? Use \ t ext t ip{K}{K} f or t he elect rost at ic const ant .


Hint 1. Find the distance between charge 2 and point P

Find t he dist ance \texttip{d_{\rm 2}}{d_2} between q_2 and point P.

Express your answer in terms of \texttip{d}{d}.

ANSWER:

ANSWER:

Hint 2. How to calculate the x component

Recall t hat (E_2)_x = |\ vec{E}_2|\ cos(\ t het a), where \ t ext t ip{\ t het a }{t het a} is the angle between the x

axis and \ t ext t ip{\ vec{E}_{\ rm 2}}{E_2_vec}. What is \ t ext t ip{\ t het a }{t het a} in t his case?

ANSWER:

Hint 4. Find the x component of the net electric field

Find the x component of t he net elect ric f ield \ t ext t ip{\ vec{E}_{\ rm net }}{E_net _vec} at point P. Use

\ t ext t ip{K}{K} f or t he elect rost at ic const ant .


Hint 1. The net electric field

Recall t hat \ vec{E}_{\ rm net }=\ sum \ vec{E}_i . I n component f orm, t his relat ion becomes (\ vec E_{\ rm

net })_x=\ sum (E_i)_x and (\ vec{E}_{\ rm net })_y=\ sum (E_i)_y . So t o f ind (\vec{E}_{\rm net})_x ,

simply add (E_1)_x, (E_2)_x, and (E_3)_x.

\texttip{d_{\rm 2}}{d_2} = d \ sqrt {2}

|\ vec E_2| = \large{\frac{K q}{d^{2}}}

(E_2)_x = \large{\frac{K q \sqrt{2}}{2 d^{2}}}

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ANSWER:

ANSWER:

Correct

Assess

Part D

I nt uit ively, which of t he f ollowing would happen t o \ t ext t ip{\ vec{E}_{\ rm net }}{E_net _vec} if \ t ext t ip{d}{d} became

very large?

A. \ t ext t ip{\ vec{E}_{\ rm net }}{E_net _vec} should reduce to the field of a point charge of magnitude

\ t ext t ip{q}{q}.

B. \ t ext t ip{\ vec{E}_{\ rm net }}{E_net _vec} should reduce to the field of a point charge of magnitude 4q.

C. The larger \ t ext t ip{d}{d} becomes, t he smaller t he magnit ude of \ t ext t ip{\ vec{E}_{\ rm net }}{E_net _vec}

will be.

D. The larger \ t ext t ip{d}{d} becomes, the greater the magnitude of \ t ext t ip{\ vec{E}_{\ rm net }}{E_net _vec}

will be.

Enter the letters of all the correct answers in alphabetical order. Do not use commas. For instance, if you

think that A and D are correct, enter AD.

ANSWER:

Correct

Looking at your answer derived in Part C, you should see that the magnitude \ t ext t ip{E_{\ rm net }}{E_net } will

decrease as \ t ext t ip{d}{d} get s larger, just as you would expect . Your result s do make sense! I t is int erest ing

t o not ice t hat t he f ield will not reduce t o t hat of a point charge (of magnit ude 4q) unless t he point P is moved

farther away from the three charges while they themselves remain at their original positions.

PSS 26.2 The Electric Field of a Continuous Distribution of Charge

Learning Goal:

To pract ice Problem-Solving St rat egy 26. 2 f or cont inuous charge dist ribut ion problems.

A st raight wire of lengt h \ t ext t ip{L}{L} has a posit ive charge \ t ext t ip{Q}{Q} dist ribut ed along it s lengt h. Find t he

magnit ude of t he elect ric f ield due t o t he wire at a point locat ed a dist ance \ t ext t ip{d}{d} from one end of the wire along

(\vec E_{\rm net})_x = \large{\frac{K q \left(2+\sqrt{2}\right)}{2 d^{2}}}

\ t ext t ip{E_{\ rm net }}{E_net } = \large{{\frac{Kq}{d^{2}}}\left(1+\sqrt{2}\right)}

C

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the line extending from the wire.

PROBLEM-SOLVING STRATEGY 26.2 The elect ric f ield of a cont inuous dist ribut ion of charge

MODEL: Model t he dist ribut ion as a simple shape, such as a line of charge or a disk of charge. Assume t he charge isunif ormly dist ribut ed.

VISUALIZE: For the pictorial representation:

1. Draw a picture and establish a coordinate system.

2. I dent if y t he point P at which you want t o calculat e t he elect ric f ield.

3. Divide t he t ot al charge \ t ext t ip{Q}{Q} int o small pieces of charge \ t ext t ip{\ Delt a Q}{Delt aQ} using shapes

for which you already know how to determine \ t ext t ip{\ vec{E}}{E_vec}. This is of t en, but not always, a

division int o point charges.

4. Draw t he elect ric f ield vect or at P f or one or t wo small pieces of charge. This will help you ident if y

dist ances and angles t hat need t o be calculat ed.

5. Look f or symmet ries in t he charge dist ribut ion t hat simplif y t he f ield. You may conclude t hat some

component s of \ t ext t ip{\ vec{E}}{E_vec} are zero.

SOLVE: The mat hemat ical represent at ion is \ vec E_{\ rm net }=\ sum \ vec E_i .

Use superposition to form an algebraic expression for each of the three components of \ t ext t ip{\ vec{E}}

{E_vec} at point P. (Note that one or more components may be zero.)

Let t he ( x , y , z ) coordinat es of t he point remain as variables.

Replace the small charge \ t ext t ip{\ Delt a Q}{Delt aQ} with an equivalent expression involving a charge

densit y and a coordinat e, such as \ t ext t ip{dx}{dx}, t hat describes t he shape of charge \ t ext t ip{\ Delt a Q}

{DeltaQ}. This is the critical step in making the transition from a sum to an integral because you need a

coordinate to serve as the integration variable.

Express all angles and dist ances in t erms of t he coordinat es.

Let the sum become an integral. The integration will be over the coordinate variable that is related to\ t ext t ip{\ Delt a Q}{Delt aQ}. The int egrat ion limit s f or t his variable will depend on your choice of coordinat e

syst em. Carry out t he int egrat ion, and simplif y t he result as much as possible.

ASSESS: Check t hat your result is consist ent wit h any limit s f or which you know what t he f ield should be.

Model

No inf ormat ion is given on t he cross sect ion of t he wire. For simplicit y, assume t hat it s diamet er is much smaller t han

the wire's length, and model the wire as a line of charge. Also, assume t hat t he t ot al charge is unif ormly dist ribut ed

along t he wire. Not e t hat t he point at which you want t o calculat e t he elect ric f ield is close t o one of t he ends of t he

wire, so you cannot make use of any symmet ry t o simplif y t he problem. I nst ead, you will need t o divide t he t ot al charg

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distributed along the wire into many small segments and model each segment as a point-like charge.

Visualize

Part A

The diagram below is an incomplet e pict orial represent at ion of t he sit uat ion described in t his problem: I t shows t he

posit ively charged wire, a suit able coordinat e syst em wit h t he origin at t he lef t end of t he wire, and t he point P at

which you need t o calculat e t he elect ric f ield. The t ot al charge dist ribut ed along t he wire has been divided int o

small segment s, each of lengt h \ t ext t ip{\ Delt a x}{Delt ax} and charge \ t ext t ip{\ Delt a Q}{Delt aQ}. For clarit y, only t h

i -t h segment at point \ t ext t ip{x_{\ it i}}{x_i} is shown in the diagram.

Complet e t his pict orial represent at ion by drawing t he elect ric f ield \ t ext t ip{\ vec{E}_{\ rm i }}{E_i _vec} at P due t o

segment i .

Draw the vector starting at P.

ANSWER:

Correct

Each small segment will cont ribut e t o t he net elect ric f ield at P. By modeling each segment as a point charge

and using t he principle of superposit ion, you can make a quick est imat e of t he direct ion of t he net f ield at P.

To do that, it is helpful to draw the electric field due to one or two more charge segments.

Part B

Wit h ref erence t o t he coordinat e syst em used in t he previous part , which component , if any, of t he elect ric f ield

due t o t he t ot al charge is zero at P?

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ANSWER:

Correct

Since t he point at which you want t o calculat e \ t ext t ip{\ vec{E}}{E_vec} lies on a st raight line ext ending f rom

the wire, \ t ext t ip{\ vec{E}}{E_vec} must be direct ed along t he same line. Theref ore, by choosing t he x a x i s t o

be parallel to the wire, as we did in Part A, we simplified the problem: Now the y component of t he f ield at P is

zero and the magnitude of \ t ext t ip{\ vec{E}}{E_vec} is simply given by t he magnit ude of it s x component.

Solve

Part C

Find \ t ext t ip{E_{\ rm net }}{E_net }, t he magnit ude of t he elect ric f ield at point P due t o t he t ot al charge.

Express your answer in terms of some or all of the variables \texttip{Q}{Q}, \texttip{L}{L}, and \texttip{d}{d

and the constants \texttip{\pi }{pi} and \texttip{\epsilon _{\rm 0}}{epsilon_0}.


As explained in t he st rat egy above, t he mat hemat ical expression f or t he net f ield is \ vec E_{\ rm net }=\ sum

\ v e c E _ i . Because of the nature of the problem, which deals with a continuous distribution of charge rather

than a point-like distribution, the sum must be treated as an integral. To set up the summation and find whatt o int egrat e, f ollow t he st eps list ed in t he st rat egy. Af t er dividing t he t ot al charge int o small segment s, and

modeling each segment as a point charge, as we did in Part A, const ruct a mat hemat ical expression f or

\ t ext t ip{\ vec{E}_{\ rm i}}{E_i_vec}, t he elect ric f ield due t o t he i -t h segment . Treat \ t ext t ip{x_{\ it i}}{x_i}, t he x

coordinate of the i -t h segment , as a variable. Express t he charge on t he i -t h segment in t erms of

\ t ext t ip{x_{\ it i}}{x_i}. Then, let \ s u m \ v e c E _ i become an integral and apply your knowledge of calculus to

evaluate the integral.

Hint 2. Find an expression for the electric field due to segment i

Find t he magnit ude of t he elect ric f ield \ t ext t ip{E_{\ rm i}}{E_i} at point P due to a segment of charge

\ t ext t ip{\ Delt a Q}{Delt aQ} locat ed at \ t ext t ip{x_{\ it i}}{x_i} on the x a x i s .

Express your answer in terms of some or all of the variables \texttip{x_{\it i}}{x_i}, \texttip{\Delta Q}

{DeltaQ}, \texttip{L}{L}, and \texttip{d}{d}, and the constants \texttip{\pi }{pi} and \texttip{\epsilon

_{\rm 0}}{epsilon_0}.

Hint 1. Find the distance between point P and a segment of charge at \texttip{x_{\it i}}{x_i}

Assume t he origin of t he x axis is at t he lef t end of t he wire and point P is locat ed at a dist ance

\ t ext t ip{d}{d} from the other end of the wire. \ t ext t ip{L}{L} is t he lengt h of t he wire. What is t he

dist ance \ t ext t ip{r_{\ it i}}{r_i} between point P and a segment of charge located at \ t ext t ip{x_{\ it i}}{x_i}

on the x a x i s ?

neither the x nor they component

t he x component

t he y component

both the x andy component

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Express your answer in terms of \texttip{x_{\it i}}{x_i}, \texttip{L}{L}, and \texttip{d}{d}.

ANSWER:

ANSWER:

Hint 3. Find an expression for the charge of a small segment

Since you will int egrat e wit h respect t o a coordinat e, it is import ant t o express t he charge on a short

segment of wire, \ t ext t ip{\ Delt a Q}{Delt aQ}, in t erms of it s lengt h \ t ext t ip{\ Delt a x}{Delt ax}.

Find a mat hemat ical expression f or \ t ext t ip{\ Delt a Q}{Delt aQ} assuming t he charge is unif ormly dist ribut ed

along the wire. Recall that the total charge on the wire is \ t ext t ip{Q}{Q} and the length of the wire is

\ t ext t ip{L}{L}.

Express your answer in terms of \texttip{\Delta x}{Deltax} and some or all of the variables \texttip{L}

{L}, \texttip{Q}{Q}, and \texttip{d}{d}.

Hint 1. Find the linear charge density

Assuming t he wire, whose lengt h is \ t ext t ip{L}{L}, has a t ot al charge \ t ext t ip{Q}{Q} unif ormly

dist ribut ed along it s lengt h, what is t he linear charge densit y \ t ext t ip{\ lambda }{lambda} of the wire?

Express your answer in terms of \texttip{L}{L} and \texttip{Q}{Q}.

ANSWER:

ANSWER:

Hint 4. How to make the transition from a sum to an integral

At t his point in t he problem, you should have f ound an expressi on f or t he magnit ude of t he elect ric f ield

\ t ext t ip{E_{\ rm i}}{E_i} due to a small charge segment of length \ t ext t ip{\ Delt a x}{Delt ax} locat ed at

\ t ext t ip{x_{\ it i}}{x_i} along the x axis. Make sure t hat your expression f or \ t ext t ip{E_{\ rm i}}{E_i} cont ains

only t he f ollowing quant it ies, besides t he elect rost at ic const ant 1/(4\pi\epsilon_0) : \ t ext t ip{d}{d}, \ t ext t ip{L}

{L}, \ t ext t ip{Q}{Q}, \ t ext t ip{x_{\ it i}}{x_i}, and \ t ext t ip{\ Delt a x}{Delt ax}. I f t his is not t he case, go back and

make t he necessary subst it ut ions (t he previous hint s may help you do t hat ).

Assuming each charge segment has t he same lengt h \ t ext t ip{\ Delt a x}{Delt ax}, you can const ruct an

algebraic expression f or t he magnit ude of t he net f ield using t he principle of superposit ion, t hat is, E = \ s u m

E_i, where t he sum cont ains as many t erms as t he number of charge segment s. I f you assume t hat t here is

an infinite number of segments in the wire, \ t ext t ip{\ Delt a x}{Delt ax} will become inf init esimally small and

\ t ext t ip{r_{\ it i}}{r_i} = L+d-x_{i}

\ t ext t ip{E}{E} = \large{\frac{\Delta{Q}}{4 {\pi} {\epsilon}_{0} \left(L+d-x_{i}\right)^{2}}}

\ t ext t ip{\ lambda }{lambda} = \large{\frac{Q}{L}}

\ t ext t ip{\ Delt a Q}{Delt aQ} = \large{\frac{{\Delta{x}} Q}{L}}

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can be replaced with \ t ext t ip{dx}{dx}. At this point, you can replace the sum with an integral and drop the

subscript i from the variable \ t ext t ip{x_{\ it i}}{x_i}, which will then become the integration variable \ t e x t t i p { x }

{x}. Note that the remaining quantities \ t ext t ip{d}{d}, \ t ext t ip{L}{L}, and \ t ext t ip{Q}{Q} are const ant s: They are

known quantities given in the problem statement.

Hint 5. Find the limits of integration

When you change the sum over charge segments into an integral, the variable of integration will be the

coordinate \ t ext t ip{x}{x} . What should t he limit s of int egrat ion be?

ANSWER:

Hint 6. How to simplify the integral

You can simplif y t he int egral by changing t he variable of int egrat ion f rom \ t ext t ip{x}{x} t o x'=L+d-x. Don't

forget to adjust the limits of integration appropriately when you make this change. The following formula will

also be usef ul:

\large{\int \frac{dx}{x^2} = -\frac{1}{x}}.

ANSWER:

Correct

Assess

Part D

I magine t hat dist ance \ t ext t ip{d}{d} is much great er t han t he lengt h of t he wire. I nt uit ively, what should t he

magnit ude of t he elect ric f ield at point P be in t his case?

Express your answer in terms of some or all of the variables \texttip{Q}{Q} and \texttip{d}{d}, and the

constants \texttip{\pi }{pi} and \texttip{\epsilon _{\rm 0}}{epsilon_0}. The variable \texttip{L}{L} should not

appear in your answer.

Hint 1. Envisioning the limiting case

I n t he limit ing case d\gg L, the length of the wire is not relevant and the wire appears to be a point charge in

t he dist ance. What is t he magnit ude of t he elect ric f ield due t o a point charge?

0 t o \ i n f t y

0 t o \ t ext t ip{L}{L}

\ t ext t ip{d}{d} t o \ t ext t ip{L}{L}

\ t ext t ip{d}{d} t o L+d

0 t o \ t ext t ip{d}{d}

\ t ext t ip{E_{\ rm net }}{E_net } = \large{{\frac{1}{4{\pi}{\epsilon}_{0}}} {\frac{Q}{d\left(L+d\right)}}}

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ANSWER:

Correct

A short wire f ar away f rom point P creat es a f ield very similar t o t he one creat ed by a point charge. Not

surprisingly, in t he limit ing case d \gg L, t he mat hemat ical expression you derived f or \ t ext t ip{E_{\ rm net }}

{E_net} reduces t o t hat of t he elect ric f ield due t o a point charge.

To see t his, not e t hat one way of writ ing t he answer you obt ained in t he Solve st ep is:

\large{E_{\rm net}=\frac{Q}{4\pi\epsilon_0 d (d+L)}}.

When d \gg L, \ t ext t ip{L}{L} becomes negligible compared to \ t ext t ip{d}{d} andd + L \approx d . Thus, t he

denominator reduces to (4\pi\epsilon_0 d^2), as we expect ed.

Problem 26.4

What are t he st rengt h and direct ion of t he elect ric f ield at t he posit ion indicat ed by t he dot in t he f igure ?

Part A

Specif y t he st rengt h of t he elect ric f ield. Let \ t ext t ip{r}{r} = 6. 3 {\ rm cm} .

Express your answer using two significant figures.

ANSWER:

All attempts used; correct answer displayed

\ t ext t ip{E_{\ rm net }}{E_net } = \large{{\frac{Q}{4{\pi}{\epsilon}_{0}}} {\frac{1}{d^{2}}}}

E = 4 800 \rm N/C

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Part B

Specif y t he direct ion.

Express your answer using two significant figures.

ANSWER:

Correct

Problem 26.11

Two 10-\ r m c m-diameter charged rings face each other, 23 {\ rm cm} apart. The left ring is charged to -18 {\rm nC} and

the right ring is charged to +18 {\rm nC} .

Part A

What is t he magnit ude of t he elect ric f ield \ t ext t ip{\ vec{E}}{E_vec} at the midpoint between the two rings?

Express your answer to two significant figures and include the appropriate units.

ANSWER:

Correct

Part B

What is t he direct ion of t he elect ric f ield \ t ext t ip{\ vec{E}}{E_vec} at the midpoint between the two rings?

ANSWER:

Correct

Part C

What is the magnitude of the force \ t ext t ip{\ vec{F}}{F_vec} on a -1.0 \rm nC charge placed at the midpoint?


\ t het a = 90 ^\ circ below horizontal

E = 1. 9×104 \large{{\rm \frac{N}{C}}}

To t he lef t ring.

Parallel to the plane of the rings.

To the right ring.

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ANSWER:

Correct

Part D

What is t he direct ion of t he f orce \ t ext t ip{\ vec{F}}{F_vec} on a -1.0 \rm nC charge placed at the midpoint?

ANSWER:

Correct

Problem 26.22

An elect ron moving parallel t o a unif orm elect ric f ield increases it s speed f rom 2. 0×107 {\ rm m/ s} t o 3. 6×107 {\ rm m/ s}

over a dist ance of 1. 8 {\ rm cm} .

Part AWhat is t he elect ric f ield st rengt h?


ANSWER:

Correct

Problem 26.36

The f igure is a cross sect ion of t wo inf init e lines of charge t hat ext end out of t he page. Bot h have linear charge densit y

\ lambda.

F = 1. 9×10−5 {\rm N}

To t he lef t ring.

Parallel to the plane of the rings.

To the right ring.

E = 1. 4×105 \large{{\rm \frac{N}{C}}}

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Part A

Find an expression f or t he elect ric f ield st rengt h E at height y above the midpoint between the lines.

Express your answer in terms of the variables \texttip{\lambda ,}{lambda,} \texttip{y}{y} , \texttip{d}{d} and

appropriate constants.

ANSWER:

Correct

Problem 26.40

The figure shows a thin rod of length L with total charge Q.

\large{{\frac{1}{4{\pi}{\epsilon}_{0}}} {\frac{y4{\lambda}}{y^{2}+\left(\frac{d}{2}\right)^{2}}}}

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Part A

Find an expression f or t he elect ric f ield st rengt h on t he axis of t he rod at dist ance r f rom t he cent er.

Express your answer in terms of the variables \texttip{L}{L}, \texttip{Q}{Q}, \texttip{r}{r}, and appropriate

constants.

ANSWER:

Correct

Part B

Verif y t hat your expression has t he expect ed behavior if r \gg L.

Express your answer in terms of variables Q, r and constants \pi, \varepsilon _0.

ANSWER:

Correct

Part C

Evaluat e E at \ t ext t ip{r}{vr} = 4. 3 {\ rm cm} if L = 5. 0 {\ rm cm} and \ t ext t ip{Q}{vQ} = 5. 0 {\rm nC} .


ANSWER:

Correct

Problem 26.44

Charge Q is unif ormly dist ribut ed along a t hin, f lexible rod of lengt h L. The rod is t hen bent int o t he semicircle shown i

the figure .

E = \large{{\frac{1}{4{\pi}{\epsilon}_{0}}} {\frac{Q}{r^{2}-{\frac{L^{2}}{4}}}}}

E = \large{{\frac{1}{4{\pi}{\epsilon}_{0}}} {\frac{Q}{r^{2}}}}

E = 3. 7×104 \large{{\rm \frac{N}{C}}}

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Part A

Find an expression f or t he elect ric f ield \ v e c E at t he cent er of t he semicircle.Hint: A small piece of arc lengt h \ Delt a s spans a small angle \ Delt a \ t het a = \ Delt a s / R, where R is t he radius.

Express your answer in terms of the variables Q, L, unit vectors \texttip{\hat{i}}{i_unit}, \texttip{\hat{j}}

{j_unit}, and appropriate constants.

ANSWER:

Correct

Part B

Evaluat e t he f ield st rengt h if \ t ext t ip{L}{vL} = 14 {\ rm cm} and \ t ext t ip{Q}{vQ} = 36 {\rm nC} .

Express your answer with the appropriate units.

ANSWER:

Correct

Problem 26.50

An elect ron is launched at a 45^\ circ angle and a speed of 5. 0 \ t imes \ ; \ ; \ ; \ ; 10^{6}\ ; {\ rm m}/ {\ rm s} f rom t he posit ive

plate of the parallel-plate capacitor shown in the figure . The electron lands 4.0 cm away.

\ v e c E = \large{{\frac{Q}{2\left(L^{2}\right)\left({\epsilon}_{0}\right)}}\hat{i}}

E = 1. 0×105 \large{{\rm \frac{N}{C}}}

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Part A

What is t he elect ric f ield st rengt h inside t he capacit or?


ANSWER:

Correct

Part B

What is t he smallest possible spacing bet ween t he plat es?


ANSWER:

Correct

Problem 26.55

In a classical model of the hydrogen atom, the electron orbits the proton in a circular orbit of radius 0.053 \ rm nm.

Part A

What is t he orbit al f requency? The prot on is so much more massive t han t he elect ron t hat you can assume t he

prot on is at rest .


ANSWER:

3550 \large{{\rm \frac{N}{C}}}

1.00 {\ rm cm}

f = 6. 6×1015 {\rm Hz}

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Correct

Problem 26.57

An elect ric f ield can induce an electric dipole in a neutral atom or molecule by pushing the positive and negative charg

in opposit e direct ions. The dipole moment of an induced dipole is direct ly proport ional t o t he elect ric f ield. That is, \ v e cp = \ alpha \ vec E , where \alpha is called t he polarizability of t he molecule. A bigger f ield st ret ches t he molecule f art he

and causes a larger dipole moment.

Part A

What are the units of \ alpha?

ANSWER:

Correct

Part B

An ion wit h charge q is dist ance r f rom a molecule wit h polarizabilit y \alpha. Find an expression for the magnitudeof t he f orce \vec F_{{\rm{ion\; on\; dipole}}}.

Express your answer in terms of the variables \texttip{q}{q}, \texttip{r}{r}, \texttip{\alpha }{alpha} and

appropriate constants.

ANSWER:

Correct

Part C

What is t he direct ion of t his f orce?

ANSWER:

\ rm C^2/ kg

\ rm C^2s^2 m/ kg\ rm C^2s^2 m^2/ kg

\ rm C^2s^2/ kg

F_{{\rm{ion\; on\; dipole}}} = \large{\left(\frac{1}{4{\pi}{\epsilon}_{0}}\right)^{2}{\frac{2q^{2}{\alpha}}{r^{5}}}}

Toward ion

Away f rom ion

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Correct

Score Summary:

Your score on t his assignment is 94. 4%.

You received 132. 17 out of a possible t ot al of 140 point s.

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Homework # 2 Ch 26