Physics7B Sp04 Mt2 Smoot Soln

8/11/2019 Physics7B Sp04 Mt2 Smoot Soln

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M i d t e r m 2 - S o l u t i o n s

1 . 2 5 p o i n t s ] S h o r t Q u e s t i o n s

( a ) 5 p o i n t s ] C i r c l e ( b l a c k i n ) T o r F f o r T r u e o r F a l s e

T ( i ) U n l i k e e l e c t r i c c h a r g e s a t t r a c t a n d l i k e c h a r g e s r e p e l .

T ( i i ) T h e n e t a m o u n t o f c h a r g e p r o d u c e d i n a n y p r o c e s s i s z e r o .

F ( i i i ) T h e e l e c t r i c e l d i n t e n s i t y a n y w h e r e i s t h e s u m o f t h e e l e c t r i c e l d i n t e n s i t y f r o m a l l

t h e c o n t r i b u t i n g c h a r g e s .

a n s w e r : T h e e l d s t r e n g t h I S t h e s u m , b u t t h e e l d I N T E N S I T Y i s n o t .

T ( i v ) E l e c t r i c c h a r g e i s q u a n t i z e d .

T F ( v ) T h e e l e c t r i c p o t e n t i a l o f a s y s t e m o f t w o u n l i k e c h a r g e s i s p o s i t i v e .

N O T E : T h i s q u e s t i o n i s a m b i g u o u s s o i t w a s n o t g r a d e d .

F ( v i ) I f t h e r e i s n o c h a r g e i n a r e g i o n o f s p a c e , t h e e l e c t r i c e l d o n a s u r f a c e s u r r o u n d i n g

t h e r e g i o n m u s t b e z e r o e v e r y w h e r e .

a n s w e r : t h e F L U X m u s t b e z e r o e v e r y w h e r e , b u t n o t t h e F I E L D .

T ( v i i ) I n e l e c t r o s t a t i c e q u i l i b r i u m , t h e e l e c t r i c e l d i n s i d e a c o n d u c t o r i s z e r o .

F ( v i i i ) I f t h e n e t c h a r g e o n a c o n d u c t o r i s z e r o , t h e c h a r g e d e n s i t y m u s t b e z e r o a t e v e r y

p o i n t o n t h e s u r f a c e o f t h e c o n d u c t o r .

a n s w e r : i m a g i n e a c o n d u c t i n g s h e l l w i t h + Q o n i t s o u t e r s u r f a c e a n d - Q o n i t s i n n e r s u r f a c e .

T ( i x ) T h e r e s u l t t h a t E = 0 i n s i d e a c o n d u c t o r c a n b e d e r i v e d f r o m G a u s s ' s L a w .

F ( x ) O n e c a n i n d u c e c h a r g e o n a c o n d u c t o r w i t h o u t t o u c h i n g .

F ( x i ) R e s i s t a n c e i n c r e a s e s a s t e m p e r a t u r e i n c r e a s e s .

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( b ) 5 p o i n t s ] C i r c l e ( b l a c k i n l e t t e r o f ) c o r r e c t a n s w e r

( i ) M e t a l s a r e i n g e n e r a l e l e c t r i c a l

( A ) n o n c o n d u c t o r s .

( B ) c o n d u c t o r s .

( C ) s e m i c o n d u c t o r s

( D ) i n s u l a t o r s . .

( E ) c o n d u c t a n c e d e p e n d s u p o n c o n g u r a t i o n .

A n s w e r : ( B )

( i i ) W h a t i s t h e m a x i m u m v o l t a g e t h a t a s p h e r e o f r a d i u s 5 c m c a n h o l d i n a i r ?

( A ) 3 1 0

6

V .

( B ) 1 : 5 1 0

6

V .

( C ) 3 1 0

5

V .

( D ) 1 : 5 1 0

5

V .

( E ) 3 1 0

4

V .

A n s w e r : ( D ) 1 : 5 1 0

5

V .

R e a s o n : A i r b r e a k s d o w n a t a n e l e c t r i c e l d o f 3 1 0

6

V / m . T h e e l e c t r i c e l d i s E =

Q = ( 4

0

r

2

) = 3 1 0

6

V / m o r Q = 4

0

r

2

E = 4 8 : 8 5 1 0

; 1 2

0 : 0 5

2

3 1 0

6

= 8 : 3 4 1 0

; 7

c o u l o m b s . V o l t a g e i s g i v e n b y V = E r = 3 1 0

6

0 : 0 5 = 1 : 5 1 0

5

V .

( i i i ) H o w m u c h v o l t a g e i s n e c e s s a r y t o a c c e l e r a t e a p r o t o n s o t h a t i t h a s j u s t s u c i e n t e n e r g y

t o t o u c h t h e s u r f a c e o f a n i r o n n u c l e u s ? A n i r o n n u c l e u s h a s a c h a r g e o f 2 6 t i m e s t h a t o f a

p r o t o n ( e ) a n d i t s r a d i u s i s a b o u t 4 : 0 1 0

; 1 5

m . W h e r e a s t h e p r o t o n h a s a r a d i u s o f a b o u t

1 : 2 1 0

; 1 5

m . A s s u m e t h e n u c l e u s i s s p h e r i c a l a n d u n i f o r m l y c h a r g e d .

( A ) 1 3 . 6 V .

( B ) 3 5 4 V

( C ) 7 0 0 0 V .

( D ) 5 0 0 , 0 0 0 V .

( E ) 7 , 0 0 0 , 0 0 0 V .

A n s w e r : ( E ) 7 , 0 0 0 , 0 0 0 V .

R e a s o n : P o t e n t i a l E n e r g y i s e V =

e 2 6 e

4

0

d

2

= 2 6 1 0

; 1 9

e = ( 4 8 : 8 5 1 0

; 1 2

) ( 1 : 2 + 4 : 0 )

1 0

; 1 5

e V = 7 : 2 3 ( 9 : 4 i f u s e i r o n r a d i u s ) 1 0

6

e V

( i v ) A c u r r e n t o f 1 0 a m p s i n a 2 - m m d i a m e t e r c o p p e r w i r e i s t h e r e s u l t o f a n e l e c t r o n d r i f t

v e l o c i t y o f a b o u t

( A ) 2 : 5 1 0

; 6

m / s

( B ) 2 : 5 1 0

; 4

m / s

( C ) 2 : 5 1 0

; 2

m / s

( D ) 2 : 5 1 0

0

m / s

( E ) 2 : 5 1 0

2

m / s

A n s w e r : ( B ) 2 : 5 1 0

; 4

m / s

R e a s o n : I = v r

2

s o v =

I

r

2

=

1 0 A

9 1 0

2 8

m

; 3

1 6 1 0

; 1 9

( 0 0 0 1 )

2

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( v ) T h e e l e c t r i c u x f r o m a c u b i c a l b o x 2 8 c m o n a s i d e i m m e r s e d i n w a t e r ( d i e l e c t r i c c o n s t a n t

= 8 0 ) i s 1 . 4 5 1 0

3

N m

2

= C f o r a G a u s s i a n s u r f a c e i n t h e w a t e r . W h a t i s t h e n e t c h a r g e

e n c l o s e d i n t h e b o x ?

( A ) 1 p C

( B ) 1 n C

( C ) 1 C

( D ) 1 m C

( E ) 1 C .

A n s w e r : ( C ) 1 C

R e a s o n i n g : U s e G a u s s ' s L a w

E

= q

n e t

= = q

n e t

= ( K

0

) q

n e t

= 1 : 4 5 1 0

3

8 : 8 5 1 0

; 1 2

8 0 = 1 : 0 2 7 1 0

; 6

C

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( d ) 5 p o i n t s ] C i r c l e c o r r e c t a n s w e r

( i ) A t a p o i n t h i g h i n t h e a t m o s p h e r e , H e

+ +

i o n s i n a c o n c e n t r a t i o n o f 2 : 8 1 0

1 2

= m

3

m o v e

n o r t h w i t h a s p e e d o f 2 1 0

6

m = s . T h e c u r r e n t d e n s i t y a t t h a t p o i n t i s

( A ) 5 : 6 1 0

1 8

A = m

2

n o r t h

( B ) 1 1 : 2 1 0

1 8

A = m

2

n o r t h

( C ) 1 : 8 A = m

2

n o r t h

( D ) 3 : 6 A = m

2

n o r t h

( E ) n o n e o f t h e a b o v e a s i t i s g o i n g s o u t h

A n s w e r : ( C ) 1 : 8 A = m

2

n o r t h

R e a s o n :

~

j = n q ~ v = 2 : 8 1 0

1 2

= m

3

2 1 : 6 1 0

; 1 9

C 2 1 0

6

m = s e c = 1 : 7 9 C = m

2

s e c

( i i ) A p r o x i m i t y ( d o n ' t h a v e t o t o u c h o n l y g e t v e r y n e a r ) b u t t o n m a k e s u s e o f w h a t e e c t ?

( A ) E a s e o f s e n s i n g t h e r e s i d u a l e l e c t r i c e l d o n h u m a n b o d y .

( B ) H e a t s e n s i n g d u e t o t h e b l o o d i n t h e n g e r t i p s . .

( C ) S e n s i n g o f t h e e l e c t r i c a l c u r r e n t s i n n g e r t i p s .

( D ) S e n s i n g t h e p o t e n t i a l d i e r e n c e b e t w e e n t h e b o d y a n d t h e b u t t o n .

( E ) C a p a c i t a n c e , w h i c h c h a n g e s a s a c o n d u c t i n g n g e r i s b r o u g h t n e a r t h e b u t t o n .

A n s w e r : ( E )

( i i i ) T h e w i r i n g i n a h o u s e m u s t b e t h i c k e n o u g h s o i t d o e s n o t b e c o m e s o h o t a s t o s t a r t a

r e . W h a t d i a m e t e r m u s t a c o p p e r w i r e , = 1 : 6 8 1 0

; 8

m , i f i t i s t o c a r r y a m a x i m u m

c u r r e n t o f 2 0 A a n d p r o d u c e n o m o r e t h a n 2 W o f h e a t p e r m e t e r ?

( A ) 0 . 5 m m .

( B ) 1 m m .

( C ) 1 . 5 m m .

( D ) 2 m m .

( E ) 2 . 5 m m .

A n s w e r : ( D ) 2 m m .

R e a s o n : R = L = A = 1 : 6 8 1 0

; 8

m 1 m = ( d

2

= 4 ) P = I

2

R = 2 0

2

1 : 6 8 1 0

; 8

m

2

= ( d

2

= 4 ) = 2 w d =

p

4 2 0

2

1 : 6 8 1 0

; 8

m

2

= 2 w = 2 : 0 6 8 1 0

; 4

m = 2 : 0 6 8 m m

( i v ) A t w h a t c u r r e n t d o h u m a n s e x p e r i e n c e s e v e r e h e a r t b r i l l a t i o n m o s t l e t h a l l y ?

( A ) 1 m A .

( B ) 1 0 m A .

( C ) 1 0 0 m A .

( D ) 1 A .

( E ) 1 A .

A n s w e r : ( C ) 1 0 0 m A

( v ) A b i r d p e r c h e s o n a D C e l e c t r i c t r a n s m i s s i o n l i n e c a r r y i n g 2 5 0 0 A . T h e l i n e h a s a

r e s i s t a n c e o f 2 : 5 1 0

; 5

p e r m e t e r a n d t h e b i r d ' s f e e t a r e 4 c m a p a r t . W h a t p o t e n t i a l

d i e r e n c e d o e s t h e b i r d f e e l ?

( A ) 2 . 5 V .

( B ) 2 . 5 m V .

( C ) 2 . 5 V .

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( D ) 2 5 V .

( E ) 2 5 0 V .

A n s w e r : ( B ) 2 . 5 m V .

V = R I = 0 : 0 4 2 : 5 1 0

; 5

2 5 0 0 A = 2 : 5 m V

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( e ) 5 p o i n t s ] C i r c l e c o r r e c t a n s w e r

( i ) F o u r 1 0 0 r e s i s t o r s a r e c o n n e c t e d i n p a r a l l e l . W h a t i s t h e e q u i v a l e n t r e s i s t a n c e ?

( A ) 2 5 .

( B ) 1 0 0 .

( C ) 4 0 0 .

( D ) 1 0 .

( E ) n o n e o f t h e a b o v e .

A n s w e r : ( A ) 2 5 .

R e a s o n : R

e q u

= (

4

1 0 0

)

; 1

( i i ) F o u r 1 0 0 F c a p a c i t o r s a r e c o n n e c t e d i n p a r a l l e l . W h a t i s t h e e q u i v a l e n t c a p a c i t a n c e ?

( A ) 2 5 F .

( B ) 1 0 0 F .

( C ) 4 0 0 F .

( D ) 1 0 F .


A n s w e r : ( C ) 4 0 0 F .

R e a s o n : C

e q u

= 4 1 0 0 F

( i i i ) A 1 2 0 0 w a t t h a i r d r y e r r u n s o n 1 2 0 V A C p o w e r i n y o u r h o u s e . W h a t i s t h e p e a k c u r r e n t

t h r o u g h t h e h a i r d r y e r ?

( A ) 1 A .

( B ) 1 . 4 A .

( C ) 1 0 A .

( D ) 1 4 A .

( E ) n o n e o f t h e a b o v e

A n s w e r : ( D ) 1 4 A

R e a s o n : P = ( 1 = 2 ) I

0

V

0

s o I

0

=

2 P

v 0

=

2 1 2 0 0 W

p

2 1 2 0 V

( i v ) T h e E V - 1 e l e c t r i c c a r i s p o w e r e d b y 2 6 b a t t e r i e s , e a c h 1 2 V a n d 5 2 A - h r . A s s u m e t h e

c a r i s o n t h e l e v e l m o v i n g a t 4 0 k m / h r s o t h a t h a s a n a v e r a g e r e t a r d i n g f o r c e o f 2 4 0 N . A f t e r

h o w m a n y k i l o m e t e r s m u s t t h e b a t t e r i e s b e r e c h a r g e d ?

( A ) 2 . 5 k m

( B ) 2 5 k m .

( C ) 2 5 0 k m .

( D ) 2 5 0 0 k m .


A n s w e r : ( C ) 2 5 0 k m .

P = F v = 2 4 0 N 4 0 k m = h r = 2 6 6 6 : 7 w W o r k = F d = 2 4 0 N d = 2 6 1 2 V 5 2 A ; h r =

1 6 2 2 4 W ; h r = 5 : 8 4 1 0

7

j o u l e s d = W = F = 5 : 8 4 1 0

7

j o u l e s = 2 4 0 N = 2 4 3 1 0

3

m = 2 4 3 k m

( v ) A c a p a c i t o r i s o f t e n u s e d i n e l e c t r o n i c s t o k e e p e n e r g y p o w e r i n g a c i r c u i t e v e n i f t h e r e i s

a m o m e n t a r y l o s s o f p o w e r f r o m t h e e l e c t r i c c o m p a n y . W h a t c a p a c i t a n c e w o u l d b e r e q u i r e d

f o r a T V , o p e r a t i n g a t a n i n t e r n a l v o l t a g e o f 1 2 0 V a t 1 5 0 W t o p r o v i d e s u c i e n t e n e r g y

d u r i n g a 0 . 1 s e c o n d l a p s e i n p o w e r ?

( A ) 2 p F = 2 1 0

; 1 2

F

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( B ) 2 n F = 2 1 0

; 9

F

( C ) 2 F = 2 1 0

; 6

F

( D ) 2 m F = 2 0 0 0 F = 2 1 0

; 3

F

( E ) 1 2 F

( F ) 2 0 0 0 F

A n s w e r : ( D ) 2 m F = 2 0 0 0 F = 2 1 0

; 3

F

R e a s o n i n g : E = P T = 1 5 0 W 0 : 1 s e c = 1 5 J =

1

2

C V

2

=

1

2

1 2 0

2

C

C = 2 E = V

2

= 2 1 5 J = 1 2 0

2

= 2 : 0 8 3 1 0

; 3

F

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4. [20 points] A thin rod of length 2Lis centered on and lying on thexaxis. The rod carries

a uniformly distributed chargeQ.

LL

2L

y

x

(a) [5 points] Determine the potentialV(y) as a function ofy (perpendicular direction torod) for points on the y-axis. LetV= 0 at infinity.Answer: dV = dQ/40r where dQ = Qdx/(2L) and r

2 = x2 + y2 or r =

x2 +y2

V(y) = Q402L

LL

dxx2+y2

= Q402L

arcsinh(x/y)|LL= Q402L(arsinh(L/y) arcsinh(L/y))

which also equals = Q402L

log(

L2+y2+L

L2

+y2L

)

(b) [5 points] Determine the potentialV(x) for points along the x axis outside the rod.Answer: dV =dQ/40rwheredQ= Qdx/(2L) andr

2 = (xx)2+y2 orr=

(x x)2 +y2and y = 0 V(y) = Q

402L

LL

dxxx =

Q

402Lln(x x)|Ll= Q402L ln(x+L)/(x L))

(c) [5 points] What is the electric field along the y axis?Answer: We can derive this either of two ways:

(1) E= V(y) = LL Qy402L(x2+y2)3/2 dxy(2) Integrate d E = dQr

40r2 = dQr

40(x2+y2) = Q

402L

dxx2+y2

r = Q402L

ydx

(x2+y2)3/2y since only

the haty direction adds coherently the x has canceling components and there is no hatzcomponent along they direction by symmetry.

E= Q

402Lyx

y2x2+y2 |LL= Q

402L L

yy2+L2 L

yy2+L2 = Q40 1yy2+L2Note that this has the correct dependence: it goes down as 1/rnear (d < L) the rod and

as 1/r2 far from the rod.(d) [5 points] Place a sphere of radius R at y = d, x = 0 with d < R < L. What is the

electric flux through the sphere?Answer: The easiest way to work this problem is to use Gausss law E = Qenclosed/0 Wecan find the enclosed charge by multiplying the linear charge density times the length of therod enclosed by the sphere. By simple trigonometry =

R2 d2 is half the length in the

sphere. Qenclosed= (2) = Q

2L2

R2 d2 =QR2 d2/LThus the flux isE=Qenclosed/0 =

Q

0

R2d2L

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Problem 5 solutions

a. This is a simple circuit consisting of the 110 V source and the persons resistance.

I =V /R = 110/1100 = 0.1A

b. We have two resistors in parallel now, but each still sees the same total voltage, thus the body

receives the same current, 0.1A. The alternate path receives 110/40 = 2.75A, and so the voltagesource is supplying a total of2.85A. However, this is irrelevant; the person sees only 0.1A(100mA).

c. Now if the voltage source can supply at most 1.5A, then we have to worry about the otherresistor. The equivalent single resistor corresponding to the person and alternate path in parallel

must have a resistance of38.6. The voltage source therefore has an effective maximum voltageof1.5 38.6 = 57.9V. Thus the person sees a current of57.9/1100 = 52.6mA. Alternatively, wemay realize that the total current in the case of a perfect EMF was 2.85A, and so we must scale allthe currents we actually see down by a factor of1.5/2.85 = 0.526, which turns 100mA into

52.6mA.

d. Heart fibrillation requires 100mA. In part b, we will see this amount and so will be in danger.

In part c, we see only 52.6mA and so were safe.

e. The bodys resistance is 1100. If we are to supply 1A across a person, the voltage source

must be1100V. Recall the energy stored in a capacitor is given by U =CV2/2. Since we knowU = 200J andV = 1100V we can calculateC, givingC = 331F.


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R

R

R

Q

(QQ)

Q

1

2

3


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R

R

R

1

2

3

dr

r


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Documents

Physics7B Sp04 Mt2 Smoot Soln