“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 21

^{st}September 2012. As promised, I give below a model answer. The question also is given below for your convenience.
Two parallel flat
metal plates P

_{1}and P_{2}, each of area 0.16 m^{2}, 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 P

_{1}is +10 V, what is the potential of plate P_{2}? 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 P

_{1}and P_{2}) 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 P

_{1}is at higher potential. Since the potential of P_{1}is given as 10 V and the potential difference between P_{1}and P_{2}is given as 60 V, we have
10 –

*V*_{2}= 60 V where V_{2}is the potential of plate P_{2}.
This gives

*V*_{2}= – 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*= ε

_{0}

*A*/

*d*where ε

_{0 }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].