5.4 Kelvin Water Dropper: Converting Gravity to Electricity
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field near the second nozzle polarizes the droplet in such a way that its detached part
is becoming positively charged. This positive charge, being brought to the second
receiver, induces a positive charge in the ring, which stimulates negatively charged
droplets from the first nozzle. Thus, subsequent cycles are enhanced by a positive
feedback loop. As a result, the potential difference between receivers reaches several
thousand volts, and one can observe periodic discharges, which bring the system
back to the initial state.
Note that the induced voltage does not depend on the distance from the nozzle to
the droplet receiver. In other words, the potential energy of the water tank, above the
generator does not play a role, explaining the low efficiency of the Kelvin’s generator.
The only important part of the work performed by gravity is the detachment of the
droplet from the rest of the water in the nozzle. As the potential of the receiver
increases with time, the electric field of the inductor also produces the work which
helps gravity to detach the droplets from the nozzle. Interestingly, measuring the
charge of the droplets produced in the natural electric field of the atmosphere, Kelvin
was able to measure the atmospheric potential with very high precision, comparable
to modern plasma-based devices.
The working principle of the device can be described in the following way. Let C 1
and C 2 be the capacitances of the receivers (1) and (2), respectively (see Fig. 5.12), b 1
and b 2 are the rates of the charge loss per unit of time due to some leakage currents,
and a 1 and a 2 are the rates of charge accumulation brought by the droplets. Let +ϕ 1
and −ϕ 2 be the time-dependent potentials of the receivers (1) and (2). Then, the
action of the device is expressed by the following system of the coupled equations:
C 1
dϕ 1
dt
= a 2 ϕ 2 − b 1 ϕ 1 ,
C 2
dϕ 2
dt
= a 1 ϕ 1 − b 2 ϕ 2 .
(5.18)
Assuming that C 1,2 , a 1,2 , and b 1,2 are constants, and that C 1 ≈ C 2 = C, a 1 ≈ a 2
= a, and b 1 ≈ b 2 = b, we get a simple solution for the potential difference between
the receivers:
ϕ =
a
a + b
exp
2(a − b)t
C
(5.19)
which shows the exponential increase of the voltage with time. If a = b, the voltage
does not change with time.
Figure 5.13 shows the droplet of water in an external electric field E 0 . The electric
field lines inside the droplet are roughly uniform, and the corresponding field E
≈
E/(0) is weakened by the value of the dielectric constant of water (0) ≈ 80. The
field lines near the poles of the droplet are getting denser due to geometric factors.
For a spherical droplet the pole field strength E p is approximately three times larger
than the strength of the external field E 0 and is determined by the formula:
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