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Sunday, September 23, 2012

AP Physics B Electrostatics – Answer to Free Response Practice Question

“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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