Electromagnetic Fields and Energy - Chapter 5 ... - MIT

Sec. 5.6 Solutions to Poisson’s Equation 23
Fig. 5.6.2 Equipotentials and field lines for configuration of Fig. 5.6.1
showing graphically the superposition of particular and homogeneous
parts that gives the required potential.
The variable x in (5) has been replaced by x − vt. With this moving charge distribution,
the field also moves. Thus, (10) becomes
ρ o cosh βy
Φ = 1 − cos β(x − vt) (12)
o β 2 cosh βa
Note that the homogeneous solution is now a linear combination of the first and
third solutions in the middle column of Table 5.4.1.
As the space charge wave moves by, the charges induced on the perfectly
conducting walls follow along in synchronism. The current that accompanies the
redistribution of surface charges is detected if a section of the wall is insulated from
the rest and connected to ground through a resistor, as shown in Fig. 5.6.1. Under
the assumption that the resistance is small enough so that the segment remains at
essentially zero potential, what is the output voltage v o ?
The current through the resistor is found by invoking charge conservation for
the segment to find the current that is the time rate of change of the net charge on
the segment. The latter follows from Gauss’ integral law and (12) as
l/2
q = w o E y dx
−l/2 y=−a
wρ o
= tanh βa sin β l
− − vt (13)
β
2
2
+ sin β l
+ vt
2
It follows that the dynamics of the traveling wave of space charge is reflected in a
measured voltage of
dq 2Rwρ ov βl
v o = −R = − tanh βa sin sin βvt (14)
dt β
2