8.02pset1sol

MASSACHUSETTS INSTINUTE OF TECHNOLOGY ESG Physics

8.02 with Kai

Spring 2003 Problem Set 1 Solution

Problem 1: 23.10

Two identical point charges each having charge + q are fixed in space and separated by a distance d . A third point charge −Q of mass m is free to move and lies initially at rest on a perpendicular bisector of the two fixed charges a distance x distance x from the midpoint of the two fixed charges. (a) Show that if x if x is small compared with d , the motion of −Q is simple harmonic along the perpendicular bisector. Determine the period of that motion. (b) How fast will the charge −Q be moving when it is at the midpoint between the two fixed charges, if initially it is released at a distance x = a << d from the midpoint?

Solution: The top charge exerts a force on the negative charge which is directed upward and to the ke qQ d left at an angle of tan to the x-axis, x-axis, with magnitude . The bottom charge 2 2 x d   2

2 +  

exert a force with the same magnitude at an angle of − tan

d

. The resultant force would 2 have only x only x component as the y the y component of the two forces cancel out each other. And the resultant force would be

    ke qQ   x F total = 2  2   d    d 2 2    + x     + x 2  2   2 

2003 Spring 8.02 with Kai Problem Set 1 Solution

  −2 xke qQ  i i − = ( ) 3     d 2 2  2     + x    2   G

G

(1.1)

1

When x <<

d 2

, we have

Ftotal ≈

−16ke qQ d 3

As F = ma , we have a≈

−16ke qQ

x

x

md 3

(1.2)

(1.3)

Therefore the motion of the charge is simple harmonic. As the equation of simple harmonic motion is a = −ω 2 x , by comparing this with (1.3), we have 16ke qQ ω 2 = (1.4) md 3 As T =

2π ω

, we have for the period

T =

π

md 3

2

ke q

(1.5)

where m is the mass of the object with charge −Q . (c) From (1.4), we get ω = 4

ke qQ md 3

(1.6)

Therefore, the maximum velocity is vmax = aω = 4 a


ke qQ md 3

(1.7)

2

Problem 2: 23.21 Consider the electric dipole shown in Figure P23.21. Show that the electric field at a 4k qa distant point along the x axis E x ≅ e3 x

Solution: At any point x on the x-axis, the distance from that point to the two charges are

and

(

− ( − a ) ) respectively, so the electric field at any point x on the x axis is 2

2

4 axke q ( x + a ) − ( x − a ) − = = E = k q e 2 2 2 2 ( x − a ) ( x − ( − a ) ) ( x − a ) ( x + a2 ) ( x 2 − a 2 ) ke q

When

( − a)

ke q

(1.8)

>> a , we have E =

4axke q x

4

=


4ake q x3

(1.9)

3

Problem 3: 23.31 The electric field along the axis of a uniformly charged disk of radius R and total charge Q was calculated in Example 23.9. Show that the electric field at distances x that are great compared with R approaches that of a point charge Q = σπ R 2 .

x

(Hint: First show that

n (1 + δ ) ≈ 1 + nδ

2

+ R

= 2

1 1+

R

, and use the binomial expansion

2

x 2

when δ << 1

Solution: According to Example 23.9, the electric field at a distance x is



E x = 2π k eσ 1 −



  x 2 + R 2  x

(1.10)

Since −1

x + R 2

= 2

1 x +R 2

x 2

= 2

1 1+

2  R  2  1  R  = 1 + 2  ≈ 1 +  −   2  2  2  x  R  x  2

(1.11)

x2

Therefore

  1 R 2   π keσ R 2 E x = 2π k eσ  1 −  1 − = 2  2 x x2    Remember that σ =

Q π R 2

(1.12)

, substituting and we can get

E x =

keQ x 2

(1.13)

for a disk at large distance. Note that the electric field is the same as if the charge is a point charge.


4

Problem 4: 23.34

(a) Consider a uniformly charged right circular cylindrical shell having total charge Q, radius R, and height h. Determine the electric field at a point a distance d from the right side of the cylinder, as shown in Figure P23.34. (Hint: Use the result of Example 23.8 and treat the cylinder as a collection of ring charges) (b) Consider now a solid cylinder with the same dimensions and carrying the same charge, which is uniformly distributed through its volume. Use the result of Example 23.9 to find the field it creates at the same point.

Solution: (a) We define x = 0 at the point where we are to find the field. One ring, with Q thickness dx, has charge dx and produced a differential electric field at the h chosen point

dE =

ke

G

x

ke x

dq i = G

x 2 + R 2 x 2 + R 2

( x

Q 3

2

+ R2 ) 2

h

G

(1.14)

dx i

Therefore, the total charge is

E= G

∫

d +h

d



Qke x

dx

h( x + R 2

 i i= − 2 2 2   2 h d +R d h R + + ( )  

G

2

)

3 2

ke Q 

1

1

G

(b) Consider the cylinder to be a stack of disks, each with thickness dx, and charge and charge per area σ =

1 Q

π disk thus produces a field

2

h

(1.15)

Q h

dx ,

dx . With reference of Example 23.9 in the text, one such


5

 x  Q  − 1   dx i 2   π R h   x2 + R2 

dE = 2π ke  G

G

(1.16)

Therefore, by integrating, we get E= G

d +h

2π keQ 

d

π R 2 h 

∫

1 −

  dx i 2 2 x +R  x

G

(1.17)

which gives E= G

2keQ R 2 h

(

h + d 2 + R2 −


2

)

( d + h ) + R2 i

G

(1.18)

6

Problem 5: 23.61 A line of positive charge is formed into a semicircle of radius R = 60.0 cm , as shown in Figure P23.61. The charge per unit length along the semicircle is described by the expression λ = λ0 cosθ . The total charge on the semicircle is 12.0 µ C . Calculate the

total force on a charge of 3.00 µ C placed at the center of curvature. Solution: Denote Q to be the charge at the center and q to be the total charge on the semicircle, thus we have for the force that each differential dq exert to the central charge is

dF =

However, we know that

dq dl

ke Qdq

(1.19)

R 2

= λ0 cosθ , thus dq = λ0 cosθ dl = λ0 cosθ R dθ

(1.20)

Also, if we consider the whole semicircle of charge, we see that the x-component of the resultant force would be zero by symmetry, so that dF y =

ke Q R

2

( λ0 cosθ R dθ ) cosθ =

ke Qλ 0 R

cos2 θ dθ

(1.21)

Integrating, we get F y =

keQ λ 0

∫

π 2 π

−

R

cos 2 θ d θ

(1.22)

2

which gives F y =

π keQ λ 0 2

(1.23)

Now we have to solve for λ 0 . If we integrate both side of (1.20), we can get q = Rλ0

we get λ 0 =

q 2 R

∫

π 2 π

−

cosθ dθ = 2 Rλ0

(1.24)

2

and thus, substituting to (1.23),


7

F y =

π keQq 4

2

(1.25)

which gives F y = 0.707 N


(1.26)

8

Problem 6: 23.62 Two small spheres, each of mass 2.00 g, are suspended by light strings 10.0 cm in length (Fig. P23.62). A uniform electric field is applied in the x direction. The spheres have − − charges equal to −5.00 × 10 8 C and +5.00 × 10 8 C. Determine the electric field that enables the spheres to be in equilibrium at an angle of θ = 10.0° . Solution:

At equilibrium, the distance between the charges is r = 2L sinθ . If we consider the positive charge, since the net force acting on the charge is 0 when then charge is at equilibrium, we can get from x and y direction y :

T cos θ = mg

:

T sin θ = F2 − F 1

where F 2 and F 1 denote the force by the electric field and the force by the negative electric charge respectively. Eliminate T from the two equations, we have` tan θ =

F2 − F 1 mg

We can obtain the magnitude of F 1 from the coulomb law: F 1 =

kq 2 r2

=

ke q 2

(1.30)

4 L2 sin 2 θ

Therefore, according to equation (1.29), we have F2 = mg tan θ − F1 = mg tan θ −

ke q 2 4 L2 sin 2 θ

(1.31)

Since the force is proportional to the electric field, F = qE , we can obtain the electric field E =

F 2 q

=

mg tan θ q

−

ke q 4 L2 sin 2 θ

(1.32)

which gives E = 443 kN/C


(1.33)

9

Problem 7: 23.67 Three charges of equal magnitude q reside at the corners of an equilateral triangle of side length a (Fig. P23.67). (a) Find the magnitude and direction of the electric field at point P , midway between the negative charges, in terms of k e , q, and a.

(b) Where must a −4q charge be placed so that any charge located at p experiences no net electric force? In part (b), let P be the origin and let the distance between the + q charge and P be 1.00 m.

Solution: (a) Due to the symmetry of the location of the charges, the electric field contribution from each negative charge is equal and opposite to each other. Therefore, their contribution to the net field is zero. The field at point P is solely contributed by the positive charge + q which can be determined by

E =

ke q r

2

=

ke q π    a sin 3   

2

=

4 ke q 3a 2

y

x

(1.34)

in the negative y direction. (b) Since there is a force whenever there is a field, so the electric field at P must be zero if the +4q charge experiences no force at that point. In order for the field to cancel at P , the −4q charge must be above + q on the y axis. Thus, if the −4q charge is located at y above point P , then the electric charge at P would be


10

E =

ke ( − q )

(1.00 )

2

+

ke ( +4 q ) y 2

=0

(1.35)

which gives` y = 2.00

(1.36)

Note: Only the positive answer of y is acceptable since the − 4q charge must be located above + q .Therefore the charge must be located at 2.00 meters above the point P along the + y axis.


11

Problem 8: 23.68 Two identical beads each have a mass m and charge q. When placed in a hemispherical bowl of radius R with frictionless, non-conducting walls, the beads move, and at equilibrium they are a distance R apart (Fig. P23.68). Determine the charge on each bead.

Solution: Both beads experiences 3 forces: the gravitational force, the electric force exerted by the other bead, and the normal force exerted by the bowl. The normal force is exerted to the bead directed along the radius line, thus 60.0° above the horizontal. Consider the bead on the left, we have

n F e

60.0Þ

mg

y :

x :

N sin 60.0° = mg

N cos 60.0° = F e =

ke q 2 R 2

(1.37)

(1.38)

Eliminate N , we get tan 60.0° =

Knowing that tan 60.0° =

mgR 2 ke q 2

(1.39)

3 , we get

q=

mgR 2 3k e


(1.40)

12

8.02pset1sol

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