Solutions to Miscellaneous Physics Problems

Homer Reid

January 15, 2003

D. Yoshioka,

The Quantum Hall Effect, Problem 1.2

An electron on the surface of liquid He is bound 10 nm above the surface by the
polarization of the liquid He. Determine the electrostatic energy of the electron
relative to infinity.

We’ll put the electron on the positive z axis a distance d above the xy plane.

Then the only charge in the region z > 0 is the single electron, so we may write
the electrostatic potential in that region as the electrostatic potential of the
electron plus a term that satisfies the Laplace equation. In the region z < 0
there are no charges, so we may write the potential in that region simply as
a term satisfying the Laplace equation. Using the most general azimuthally
symmetric solution to the Laplace equation that is regular at the origin, we
have

Φ(ρ, z) =















Z

dk A(k)e

−

kz

J

0

(kρ) −



e

4π

0



1

pρ

2

+ (z − d)

2

,

z > 0

Z

dk B(k)e

kz

J

0

(kρ),

z < 0.

(1)

The boundary conditions at the helium surface are






(z)

∂φ

∂z






z

+

z

−

= 0,






∂φ

∂ρ






z

+

z

−

= 0.

Applying these to (1) we obtain

Z

dk

h

A(k) + 

r

B(k)

i

J

0

(kρ) = −



e

4π

0



d

(ρ

2

+ d

2

)

3/2

Z

dk

h

A(k) − B(k)

i

J

1

(kρ) =



e

4π

0



ρ

(ρ

2

+ d

2

)

3/2

where 

r

is the relative dielectric constant of liquid helium.

1

Homer Reid’s Solutions to Miscellaneous Physics Problems

2

Next we multiply the first of these by ρJ

0

(k

0

ρ) and the second by ρJ

1

(k

0

ρ),

integrate from ρ = 0 to ∞, and use the identity

Z

∞

0

ρJ

ν

(kρ)J

ν

(k

0

ρ) dρ =

1
k

δ(k − k

0

)

to obtain

A(k) + 

r

B(k) = −

 ed

4π

 Z

∞

0

dρ

ρJ

0

(kρ)

(ρ

2

+ d

2

)

3/2

A(k) − B(k) =



e

4π



Z

∞

0

dρ

ρ

2

J

1

(kρ)

(ρ

2

+ d

2

)

3/2

.

With help from Mathematica, we find that the first integral equals e

−

kd

/d, while

the second integral equals e

−

kd

, and we then solve the system to obtain

A(k) =



r

−

1



r

+ 1



e

4π

0



e

−

kd

B(k) =

−

2



r

+ 1



e

4π

0



e

−

kd

.

Then (1) becomes

Φ(ρ, z) =





















e

4π

0



"



r

−

1



r

+ 1

Z

dk e

−

k(z+d)

J

0

(kρ) −

1

pρ

2

+ (z − d)

2

#

z > 0

−



e

4π

0



2



r

+ 1

Z

dk e

k(z−d)

J

0

(kρ),

z < 0.

(2)

Again using Mathematica to evaluate the integrals, we may simplify this to

Φ(ρ, z) =

























e

4π

0



"



r

−

1



r

+ 1

1

p(ρ

2

+ (z + d)

2

−

1

pρ

2

+ (z − d)

2

#

z > 0

−



e

4π

0



2



r

+ 1

1

pρ

2

+ (z − d)

2

z < 0.

(3)

We check that the potential in both regions simply reduces to the potential of
the point charge when 

r

= 1.

To find the electrostatic energy of the point charge, we observe that the first

term in the top line of equation (3) represents the potential arising from the
surface charge at the helium surface. The electrostatic energy of the electron is
just e times this term evaluated at the position of the electron, or

U =



e

2

4π

0

  

r

−

1



r

+ 1



1

2d

.

Putting in d = 10 nm and 

r

= 1.057 for liquid helium at 1 K, we find U ≈ 2.0

meV.