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4.5 Local Variations of the Gain 143
from the wire, which is usually the case, we may neglect Q and obtain in first order
that
E
1
A
1
= E
2
A
2
and therefore
ρ
1
=
ρ
2
.
We have found the remarkable result that the charge density is constant along the
entire ‘flux tube’ at any distance from the wire. In case the electric field E
1
and
the ion flux
ν
are uniform around the wire surface, which is a common case, we
find a constant charge density in the entire volume that is filled by the field lines
originating from the wire.
For a wire of radius a with a surface field of E
1
, a charge deposit q at a rate of R
per unit length of the wire we therefore find a uniform space charge density of
ρ
=
qR
μ
E
1
2a
π
. (4.23)
Next we investigate the gas gain drop due to the presence of
ρ
. Setting a wire to
potential V
0
amounts to placing a certain charge
λ
on the wire. The charge density
ρ
creates an electric field E
s
(x) which is superimposed on the electric field of the wire
charge
λ
. The resulting potential difference due to
ρ
is
Δ
V =
E
s
(x)ds, where the
path of integration is taken from the wire surface to the cathode. Consequently the
wire charge
λ
is reduced such that its contribution to the potential is V
0
−
Δ
V.The
gain reduction due to the space charge
ρ
is therefore equal to a voltage reduction
of
Δ
V in the absence of space charge. This effective voltage drop of the wires in a
chamber is therefore calculated in first order by removing all wires and calculating
the potential difference between wire positions and cathode due to the charge density
ρ
. We will now discuss two examples.
First we examine a drift tube (cf. Section 3.1) exposed to a particle flux of
Φ
per
unit detector area and a total avalanche charge of q per particle. With R = 2b
Φ
and
the wire surface field of E
1
= V
0
/(aln
b
a
) we find the uniform space charge density
ρ
=
bq
Φ
ln
b
a
πμ
V
0
. (4.24)
In the absence of the wire,
ρ
produces an electric field of E
s
(r)=
ρ
r/2
ε
0
,sothe
potential difference between wire position and tube wall is
Δ
V =
b
0
E
s
(r)dr =
ρ
b
2
/4
ε
0
=
b
3
q
Φ
ln
b
a
4
πε
0
μ
V
0
. (4.25)
The voltage drop is proportional to the third power of the tube radius, so a reduction
of the tube radius is the most effective way to increase the rate capability. As an
example, Fig. 4.14 shows the gain reduction in the drift tubes of the ATLAS muon
spectrometer [ATL 00]. The measurements are well explained by the theory outlined
above.
As a second example we investigate a wire chamber schematically shown in
Fig. 4.15, where a row of wires with a pitch of s is positioned symmetrically be-
tween two grounded cathode planes at a distance h. Setting all wires to potential V
0
,