SOLUTIONS MANUAL
PLASMA PHYSICS: AN INTRODUCTION
2ND EDITION
CHAPTER 1: PLASMA PARAMETERS
Solutions to Exercises
1.1 (a) Consider a one-dimensional slab of plasma whose whose bounding surfaces are nor-
mal to the x-axis. Suppose that the electrons (whose mass, charge, and number density
are me , −e, and ne , respectively) displace a distance δxe parallel to the x-axis, whereas
the ions (whose mass, charge, and number density are mi , +Z e, and ni = ne /Z, re-
spectively) displace a distance δxi . The resulting charge density that develops on the
leading edge of the slab is
σ = −e ne δxe + Z e ni δxi = e ne (δxi − δxe ). (1)
An equal and opposite charge density develops on the opposite face of the slab. The
x-directed electric field generated inside the slab is
σ e ne
Ex = − =− (δxi − δxe ). (2)
ǫ0 ǫ0
The equation of motion of an individual electron inside the slab is thus
.. e2 n e
me δ xe = −e E x = (δxi − δxe ). (3)
ǫ0
Likewise, the equation of motion of an individual ion is
.. Z 2 e2 n i
mi δ x i = Z e E x = − (δxi − δxe ). (4)
ǫ0
Let us search for simultaneous solutions of Equations (3) and (4) of the form
δxe (t) = δ x̂e cos(ω t), (5)
δxi (t) = δ x̂i cos(ω t). (6)
It follows that
(ω2 − Πe2 ) δ x̂e + ω p2 e δ x̂i = 0, (7)
2 2
Πi δ x̂e + (ω − ω p2 i ) δ x̂i = 0, (8)
where Πe = (e2 ne /ǫ0 me )1/2 and Πi = (Z 2 e2 ni /ǫ0 mi )1/2 . The solutions are ω = 0
with δ x̂e = δ x̂i , and ω2 = Πe2 + Πi2 with Πi2 δ x̂e + Πe2 δ x̂i = 0. The former mode
corresponds to a uniform translation of the slab. The latter mode is a plasma oscillation
whose frequency, Π, satisfies
1/2
Π = Πe2 + Πi2 , (9)
, and whose characteristic ratio of ion to electron displacement amplitudes is
δ x̂i Π2 me
= − i2 = −Z . (10)
δ x̂e Πe mi
(b) Suppose that the electrons, whose temperature is T e , are distributed according to the
Maxwell-Boltzmann law,
ne + δne = ne exp(+e δΦ/T e), (11)
where ne is the equilibrium number density, and δne is the number density perturbation
due to the perturbing potential δΦ. Likewise, the ions, whose temperature is T i , are
distributed according to
ni + δni = ni exp(−Z e δΦ/T i ). (12)
Thus, in the limit that δΦ is small, we obtain
e ne
δne = δΦ, (13)
Te
Z e ni
δni = − δΦ. (14)
Ti
If δΦ is a consequence of a small perturbing charge density, δρext , then the total charge
density is
e2 n e Z 2 e2 n i
!
δρ = δρext + Z e δni − e δne = δρext − + . (15)
Te Ti
Thus, Poisson’s equation,
δρ
∇2 δΦ = − , (16)
ǫ0
yields
2 δρ
∇ − 2 δΦ = − ext ,
2
(17)
λD ǫ0
where !2 !2 !2
1 1 1 1
= + , (18)
λD 2 λD e λD i
with λD e = (ǫ0 T e /ne e2 )1/2 and λD i = (ǫ0 T i /ni Z 2 e2 )1/2 . Comparison of Equation (17)
with Eq. (1.14) in the book reveals that λD is the effective Debye length.
1.2 It is reasonable to assume, by symmetry, that the perturbed potential is a function only of the
radial spherical coordinate r. In other words, δΦ = δΦ(r). Thus, the governing differential
equation becomes !
1 d 2 dδΦ 2
2
r − 2 δΦ = 0 (19)
r dr dr λD
for r , 0. However, in the limit r → 0 we expect the perturbed potential to approach the
Coulomb potential: i.e.,
q
δΦ → (20)
4π ǫ0 r
, as r → 0. We also expect the potential to be well behaved in the limit r → ∞. Let δΦ =
V(r)/r. Equation (19) transforms to give
d2 V 2
− 2 V = 0. (21)
dr2 λD
The solution that is consistent with the boundary conditions at r = 0 and r = ∞ is
√
q 2 r
V(r) = exp − . (22)
4π ǫ0 λD
Thus, √
q 2 r
δΦ(r) = exp − . (23)
4π ǫ0 r λD
According to Poisson’s equation, the charge density of the shielding cloud is
δρ(r) = −ǫ0 ∇2 δΦ. (24)
However, according to the governing differential equation,
2
∇2 δΦ = δΦ (25)
λD2
for r , 0. Hence, √
2q 2 r
δρ(r) = − 2
exp − . (26)
4π r λD λD
The net shielding charge contained within a sphere of radius r, centered on the origin, is
Z r √ ′
2q r ′
Z
2 r ′
δρ(r′ ) r′ 2 dr′ = − 2
Q(r) = 4π r exp − dr . (27)
0 λ D 0
λ D
Thus, √ √
Z λD r/ 2 √ Z λD r/ 2
−x λD r/ 2
−x
−x
Q(r) = −q x e dx = −q −x e 0
+ e dx , (28)
0 0
which reduces to √ √
2 r 2 r
Q(r) = −q 1 − 1 + exp − . (29)
λD λD
1.3 Consider a one-dimensional slab of plasma whose bounding surfaces are normal to the x-
axis. Suppose that the electrons (whose mass, charge, and number density are me , −e, and
ne , respectively) displace a distance δxe parallel to the x-axis, whereas the ions remain sta-
tionary. The resulting charge density that develops on the leading edge of the slab is
σ = −e ne δxe . (30)
An equal and opposite charge density develops on the opposite face of the slab. The x-
directed electric field generated inside the slab is
σ e ne
Ex = − = δxe . (31)
ǫ0 ǫ0
, If E0 (t) = Ê0 cos(ω t) is the externally generated x-directed electric field then the total
electric field inside the slab is
e ne
E1 (t) = δxe (t) + Ê0 cos(ω t). (32)
ǫ0
The equation of motion of an individual electron inside the slab is
..
me δ xe = −e E1 (33)
or
..
δ xe + Π 2 δxe = −
e
Ê0 cos(ω t), (34)
me
where Π = (e2 ne /ǫ0 me )1/2 . Let us search for a solution of the form δxe (t) = δ x̂e cos(ω t).
We obtain
(e/me ) Ê0
δ x̂e = 2 . (35)
ω −Π2
Thus, writing E1 (t) = Ê1 cos(ω t), it follows from Equation (32) that
Π2
! !
1
Ê1 = 2 + 1 Ê0 = Ê0 . (36)
ω −Π2 1 − Π 2 /ω2
However, if a dielectric slab is placed in a uniform external field then we expect the internal
field to be reduced by a factor ǫ, where ǫ is the relative dielectric constant. Thus, it follows
that
Π2
ǫ =1− 2. (37)
ω
1.4 Let x measure perpendicular distance between the plates. Suppose that one plate lies at
x = −d/2, and the other at x = d/2. Because the spacing between the plates is relatively
small, we can assume that the potential, V, is only a function of x. Suppose that the plate
at x = −d/2 is held at the potential −V0 /2, whereas that at x = d/2 is held at the potential
V0 /2. According to Eq. (1.14) in the book, the potential between the plates satisfies
d2 V 2
− V = 0. (38)
dx2 λD
Moreover, V(−d/2) = −V0 /2 and V(d/2) = V0 /2. It follows that
√
V0 sinh( 2 x/λD )
V(x) = √ . (39)
2 sinh( 2 d/2 λD )
Hence, the electric field between the plates is
√
dV V0 cosh( 2 x/λD )
E(x) = − =−√ √ . (40)
dx 2 λD sinh( 2 d/2 λD )
Now, by Gauss’ law, the charge density on the plate at d = x/2 is
V0 ǫ0 1
σ = −ǫ0 E(d/2) = √ √ . (41)
2 λD tanh(d/ 2 λD )
Hence, the charge on the plate is
V0 ǫ0 A 1
Q = Aσ = √ √ . (42)
2 λD tanh(d/ 2 λD )
There is an equal and opposite charge on the other plate. Thus, the capacitance is
√
Q ǫ0 A d/ 2 λD
C= = √ . (43)
V0 d tanh(d/ 2 λD )
PLASMA PHYSICS: AN INTRODUCTION
2ND EDITION
CHAPTER 1: PLASMA PARAMETERS
Solutions to Exercises
1.1 (a) Consider a one-dimensional slab of plasma whose whose bounding surfaces are nor-
mal to the x-axis. Suppose that the electrons (whose mass, charge, and number density
are me , −e, and ne , respectively) displace a distance δxe parallel to the x-axis, whereas
the ions (whose mass, charge, and number density are mi , +Z e, and ni = ne /Z, re-
spectively) displace a distance δxi . The resulting charge density that develops on the
leading edge of the slab is
σ = −e ne δxe + Z e ni δxi = e ne (δxi − δxe ). (1)
An equal and opposite charge density develops on the opposite face of the slab. The
x-directed electric field generated inside the slab is
σ e ne
Ex = − =− (δxi − δxe ). (2)
ǫ0 ǫ0
The equation of motion of an individual electron inside the slab is thus
.. e2 n e
me δ xe = −e E x = (δxi − δxe ). (3)
ǫ0
Likewise, the equation of motion of an individual ion is
.. Z 2 e2 n i
mi δ x i = Z e E x = − (δxi − δxe ). (4)
ǫ0
Let us search for simultaneous solutions of Equations (3) and (4) of the form
δxe (t) = δ x̂e cos(ω t), (5)
δxi (t) = δ x̂i cos(ω t). (6)
It follows that
(ω2 − Πe2 ) δ x̂e + ω p2 e δ x̂i = 0, (7)
2 2
Πi δ x̂e + (ω − ω p2 i ) δ x̂i = 0, (8)
where Πe = (e2 ne /ǫ0 me )1/2 and Πi = (Z 2 e2 ni /ǫ0 mi )1/2 . The solutions are ω = 0
with δ x̂e = δ x̂i , and ω2 = Πe2 + Πi2 with Πi2 δ x̂e + Πe2 δ x̂i = 0. The former mode
corresponds to a uniform translation of the slab. The latter mode is a plasma oscillation
whose frequency, Π, satisfies
1/2
Π = Πe2 + Πi2 , (9)
, and whose characteristic ratio of ion to electron displacement amplitudes is
δ x̂i Π2 me
= − i2 = −Z . (10)
δ x̂e Πe mi
(b) Suppose that the electrons, whose temperature is T e , are distributed according to the
Maxwell-Boltzmann law,
ne + δne = ne exp(+e δΦ/T e), (11)
where ne is the equilibrium number density, and δne is the number density perturbation
due to the perturbing potential δΦ. Likewise, the ions, whose temperature is T i , are
distributed according to
ni + δni = ni exp(−Z e δΦ/T i ). (12)
Thus, in the limit that δΦ is small, we obtain
e ne
δne = δΦ, (13)
Te
Z e ni
δni = − δΦ. (14)
Ti
If δΦ is a consequence of a small perturbing charge density, δρext , then the total charge
density is
e2 n e Z 2 e2 n i
!
δρ = δρext + Z e δni − e δne = δρext − + . (15)
Te Ti
Thus, Poisson’s equation,
δρ
∇2 δΦ = − , (16)
ǫ0
yields
2 δρ
∇ − 2 δΦ = − ext ,
2
(17)
λD ǫ0
where !2 !2 !2
1 1 1 1
= + , (18)
λD 2 λD e λD i
with λD e = (ǫ0 T e /ne e2 )1/2 and λD i = (ǫ0 T i /ni Z 2 e2 )1/2 . Comparison of Equation (17)
with Eq. (1.14) in the book reveals that λD is the effective Debye length.
1.2 It is reasonable to assume, by symmetry, that the perturbed potential is a function only of the
radial spherical coordinate r. In other words, δΦ = δΦ(r). Thus, the governing differential
equation becomes !
1 d 2 dδΦ 2
2
r − 2 δΦ = 0 (19)
r dr dr λD
for r , 0. However, in the limit r → 0 we expect the perturbed potential to approach the
Coulomb potential: i.e.,
q
δΦ → (20)
4π ǫ0 r
, as r → 0. We also expect the potential to be well behaved in the limit r → ∞. Let δΦ =
V(r)/r. Equation (19) transforms to give
d2 V 2
− 2 V = 0. (21)
dr2 λD
The solution that is consistent with the boundary conditions at r = 0 and r = ∞ is
√
q 2 r
V(r) = exp − . (22)
4π ǫ0 λD
Thus, √
q 2 r
δΦ(r) = exp − . (23)
4π ǫ0 r λD
According to Poisson’s equation, the charge density of the shielding cloud is
δρ(r) = −ǫ0 ∇2 δΦ. (24)
However, according to the governing differential equation,
2
∇2 δΦ = δΦ (25)
λD2
for r , 0. Hence, √
2q 2 r
δρ(r) = − 2
exp − . (26)
4π r λD λD
The net shielding charge contained within a sphere of radius r, centered on the origin, is
Z r √ ′
2q r ′
Z
2 r ′
δρ(r′ ) r′ 2 dr′ = − 2
Q(r) = 4π r exp − dr . (27)
0 λ D 0
λ D
Thus, √ √
Z λD r/ 2 √ Z λD r/ 2
−x λD r/ 2
−x
−x
Q(r) = −q x e dx = −q −x e 0
+ e dx , (28)
0 0
which reduces to √ √
2 r 2 r
Q(r) = −q 1 − 1 + exp − . (29)
λD λD
1.3 Consider a one-dimensional slab of plasma whose bounding surfaces are normal to the x-
axis. Suppose that the electrons (whose mass, charge, and number density are me , −e, and
ne , respectively) displace a distance δxe parallel to the x-axis, whereas the ions remain sta-
tionary. The resulting charge density that develops on the leading edge of the slab is
σ = −e ne δxe . (30)
An equal and opposite charge density develops on the opposite face of the slab. The x-
directed electric field generated inside the slab is
σ e ne
Ex = − = δxe . (31)
ǫ0 ǫ0
, If E0 (t) = Ê0 cos(ω t) is the externally generated x-directed electric field then the total
electric field inside the slab is
e ne
E1 (t) = δxe (t) + Ê0 cos(ω t). (32)
ǫ0
The equation of motion of an individual electron inside the slab is
..
me δ xe = −e E1 (33)
or
..
δ xe + Π 2 δxe = −
e
Ê0 cos(ω t), (34)
me
where Π = (e2 ne /ǫ0 me )1/2 . Let us search for a solution of the form δxe (t) = δ x̂e cos(ω t).
We obtain
(e/me ) Ê0
δ x̂e = 2 . (35)
ω −Π2
Thus, writing E1 (t) = Ê1 cos(ω t), it follows from Equation (32) that
Π2
! !
1
Ê1 = 2 + 1 Ê0 = Ê0 . (36)
ω −Π2 1 − Π 2 /ω2
However, if a dielectric slab is placed in a uniform external field then we expect the internal
field to be reduced by a factor ǫ, where ǫ is the relative dielectric constant. Thus, it follows
that
Π2
ǫ =1− 2. (37)
ω
1.4 Let x measure perpendicular distance between the plates. Suppose that one plate lies at
x = −d/2, and the other at x = d/2. Because the spacing between the plates is relatively
small, we can assume that the potential, V, is only a function of x. Suppose that the plate
at x = −d/2 is held at the potential −V0 /2, whereas that at x = d/2 is held at the potential
V0 /2. According to Eq. (1.14) in the book, the potential between the plates satisfies
d2 V 2
− V = 0. (38)
dx2 λD
Moreover, V(−d/2) = −V0 /2 and V(d/2) = V0 /2. It follows that
√
V0 sinh( 2 x/λD )
V(x) = √ . (39)
2 sinh( 2 d/2 λD )
Hence, the electric field between the plates is
√
dV V0 cosh( 2 x/λD )
E(x) = − =−√ √ . (40)
dx 2 λD sinh( 2 d/2 λD )
Now, by Gauss’ law, the charge density on the plate at d = x/2 is
V0 ǫ0 1
σ = −ǫ0 E(d/2) = √ √ . (41)
2 λD tanh(d/ 2 λD )
Hence, the charge on the plate is
V0 ǫ0 A 1
Q = Aσ = √ √ . (42)
2 λD tanh(d/ 2 λD )
There is an equal and opposite charge on the other plate. Thus, the capacitance is
√
Q ǫ0 A d/ 2 λD
C= = √ . (43)
V0 d tanh(d/ 2 λD )
