Group Problems Week 5 Feb 19

Problem 1 A particle of mass m and total energy E is in a simple

harmonic oscillator potential, V =1

2m! 2

x2 .

a) Write down the classical turning points. b) Suppose the particle is in the ground state

(

!0(x) =

m"

#h

$%&

'()

1/4

e*m" x2 /(2h)

.) What is the probability that the

particle is found outside the classically allowed region? Write your answer in terms of the error function,

erf(z) =2

!e" x2

dx

0

z

# .

Solution

a) The turning points occur when E = V =1

2m! 2

xc

2 , so xc= ±

2E

m! 2.

b) The probability that the particle is outside the classically

allowed region is P = 1! |"0|2dx

! xc

xc

# . First consider the integral

|!0|2dx

" xc

xc

# =m$

%h

&'(

)*+

1

2

e"m$ x2 /h

dx

" xc

xc

# = 2m$

%h

&'(

)*+

1

2

e"m$ x2 /h

dx

0

xc

# , which

can be expressed after a change of variable

u =m!

hx to

2m!

"he#m! x2 /h

dx

0

xc

$ = 2m!

"h

h

m!e#u2

du

0

uc = 2E h!

$ =2

"e#u2

du

0

uc = 2E h!

$ = erf( 2E h! )

(Note that this quantity is the probability that the particle is found within the classically allowed region.) Therefore, the probability that the particle is found outside the classically allowed region is given by

P = 1! erf( 2E h" ) .

Problem 2 Four point charges of equal magnitudes are arranged at the corners of a square of side L as shown in the figure. Find the magnitude and direction of the force exerted on the charge in the lower left corner by the other 3 charges.

Solution

1 2

3 4

Problem 3 A dipole of moment 0.50 e ·cm is composed of two charges of 2.0 pC. a) What is the electric field at a point of the axis of symmetry of the dipole at a distance y = 2 cm from the middle point between the two charges? The dipole is placed in a uniform electric field that has a magnitude of 4.0 x 104 N/C. What is the magnitude of the torque on the dipole when: b) the dipole is aligned with the electric field? d) Defining the potential energy to be zero when the dipole is transverse to the electric field, find the potential energy of the dipole when the direction of the dipole makes an angle of 30o with the direction of the electric field.

a) Since p = qs = 0.50 e cm, s = 0.50 e cm/2.0 pC = 0.50 x 1.6 x 10-

19/2.0 x 10-12 = 4.0 x 10-8 cm

The point is at a distance y along the axis of symmetry of the dipole and the resultant E-field is directed along –x.

!

r E P " #k

p

y3

ˆ i = #9 $109$

0.5 $10#2$1.6 $10

#19

2 $10#2( )

3= #9.0 $10

#7N /C

b)

!

r " = lF = sF sin# = qsE sin# = pE sin#

When the torque is aligned to E, θ=0 and sinθ = 0 => the torque is null. c)

!

U = "pE cos# = "0.5 $10"2$1.6 $10

"19$ 4.0 $10

4$ cos30 = 2.8 $10

"17J

y

θ

r