“There is no democracy in physics. We can't
say that some second-rate guy has as much right to an opinion as Fermi.”
– Luis Walter Alvarez
A free response
practice question meant for testing your knowledge, understanding and capacity
for application of basic principles in electrostatics was given to you in the
post dated 21st September 2012. As promised, I give below a model
answer. The question also is given below for your convenience.
Two parallel flat
metal plates P1 and P2 , each of area 0.16 m2,
are arranged horizontally (Fig.) in air with a separation of 1.5×10–2 m. [Separation is shown exaggerated in the
figure for convenience]. A constant potential
difference of 60 V is established between the plates so that the electric field
in the region between the plates is directed vertically downwards as indicated in the figure. Now, answer the following
questions:
(a) If the potential
of plate P1 is +10 V, what is the potential of plate P2?
Justify your answer.
(b) Determine the
magnitude of the electric field between the plates.
(c) Determine the
capacitance of the parallel plate capacitor formed by the plates.
(d) Calculate the
increase in the energy stored in the capacitor (formed by the plates P1
and P2) when additional charges equal to ten electronic charges.are
added to it.
(e) A charged oil
droplet of effective weight 3.2×10–14
newton is found to remain stationary in the space between the plates. Determine
the charge on the oil droplet.
(a) The electric
field is pointed vertically downwards. This means that the upper plate P1
is at higher potential. Since the potential of P1 is given as 10 V
and the potential difference between P1 and P2 is given
as 60 V, we have
10 –
V2 = 60 V where V2
is the potential of plate P2.
This gives V2
= – 50 V
(b) The magnitude (E)
of the electric field between the plates is given by
E = V/d
where V is the potential difference
and d is the separation between the
plates.
Therefore, E = 60/(1.5×10–2) = 4000 NC–1
(c) The capacitance
(C) of a parallel plate capacitor
with air (or free space) as dielectric is given by
C
= ε0A/d where ε0 is the permittivity of free space (or air,
very nearly), A is the area of the
plates and d is the separation
between the plates.
Therefore, C.= (8.85×10–12×0.16)/(1.5) = 9.44×10–11
F
(d) Ten electronic charges is a very small amount of
charge which is equal to 10×1.6×10–19 coulomb.
The potential difference between the plates can therefore be assumed to be
unchanged. Therefore, the increase in the energy of the capacitor is V(∆q) = 60×1.6×10–18 joule = 9.6×10–17
joule.
(e) Since the
charged droplet is stationery, the effective weight of the droplet acting vertically downwards is balanced by the electric force on it. The electric
force is therefore acting vertically upwards. This means that the charge on the
droplet is negative.
Equating the
magnitudes of the electric force qE and
the effective weight W, we have
q×4000
= 3.2×10–14
Therefore q = (3.2×10–14)/4000 = 8×10–18
coulomb.
Since the charge is
negative, the correct answer is –8×10–18 coulomb.
[If you were asked
to find the number of electrons gained (or lost) by the droplet, you will
answer that the droplet has gained 50
electrons since (8×10–18)/(1.6×10–19) = 50].
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