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 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 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].

AP Physics B Electrostatics – A Free Response Practice Question

“This time, like all times, is a very good one if we but know what to do with it”

– Ralph Waldo Emerson

Today I give you a free response practice question meant for testing your knowledge, understanding and capacity for application of basic principles in electrostatics:

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.

The above question carries 15 points. You may take up to 17 minutes for answering it. Try to answer it yourself within the stipulated time. I’ll come back shortly with a model answer for your benefit.

AP Physics C - Multiple Choice Practice Questions on Electrostatics

Multiple choice as well as free response practice questions on electrostatics were discussed earlier on this blog. You can access them by clicking on the label ‘electrostatics’ below this post or by trying a search using the search box provided on this page. Today we will discuss a few more multiple choice practice questions in this section:

(1) Ten identical capacitors, each of value 2 μF are connected in series and this series combination is connected across a regulated power supply of output 10 V. The energy stored in any one of the 2 μF capacitors is

(a) 1 J

(b) 2 J

(c) 5 J

(d) 10 J

(e) 100 J

The effective capacitance of the series combination is 2 μF/10 = 0.2 μF.

[If n capacitors each of value C are in series, the effective capacitance is C/n].

The energy of the series combination on charging with the 10 volt supply is ½ CV² = ½×0.2×10² = 10 J.

Therefore, the energy of one capacitor = 10 J/10 = 1 J.

(2) Two equal positive charges of value 8 μC are placed in a region of space where there are no external fields. When a third charge q is placed at the mid point of the line joining the other two charges, the system is found to be in equilibrium. The third charge q must be

(a) 8 μC

(b) – 8 μC

(c) 4 μC

(d) – 4 μC

(e) – 2 μC

The third charge q must be negative so that the mutual repulsive force between the positive charges (Q each, let us say) is counteracted by the attractive force exerted by the negative charge.

Of course the above fact will come out from the expression for the net force on Q, which we will put equal to zero. Therefore we have

(1/4πε₀)Q²/r² + (1/4πε₀)Qq /(r/2)² = 0

Or, Q = – 4q so that q = – Q/4 = – 8 μC/4 = – 2 μC

(3) The figure shows 4 identical parallel metallic plates [(1), (2), (3) and (4)] arranged with equal separation d between neighbouring plates. The surface area of one side of each plate is A and the medium between the plates is air. Alternate plates are joined to terminals T₁ and T₂ so that the system makes a parallel plate capacitor. Suppose there are n plates (instead of 4) in the system where n > 1 and may be odd or even. What is the capacitance of the system made of these n plates?

(a) 2n ε₀A/d

(b) n ε₀A/d

(c) (n – 1)ε₀A/d

(d) (n + 1)ε₀A/d

(e) n ε₀A/2d

The lower surface of plate (1) and the upper surface of plate (2) makes a capacitor of capacitance ε₀A/d. The lower surface of plate (2) and the upper surface of plate (3) makes another capacitor of capacitance ε₀A/d. Similarly the lower surface of plate (3) and the upper surface of plate (4) makes a third capacitor of capacitance ε₀A/d. These three capacitors are connected in parallel and the system give a total capacitance of 3ε₀A/d.

If there are n plates, the effective capacitance C will be given by

C = (n – 1)ε₀A/d.

(4) In the combination of capacitors shown in the adjoining figure, what is the effective capacitance between the terminals A and B?

(a) 20 C

(b) 11 C

(c) 8 C

(d) 6 C

(e) 3 C

On connecting a voltage source between the terminals A and B, the potential at the junction of the two capacitors of value 2C is the same as the potential at the junction of the two capacitors of value 4C.

[The capacitors need not necessarily be equal. It is enough that the ratios of capacitance are equal to balance the Wheatstone bridge]

The capacitors of value 8C is therefore connected between equi-potential points and it can be ignored (since it does not get charged). The network thus reduces to four capacitors with the series combination of 2C and 2C connected in parallel with the series combination of 4C and 4C. Therefore, the effective value is C + 2C = 3C.

(5) Point charges +2q, +2q and –q are placed at the vertices A, B and C respectively of an equilateral triangle ABC of side 2a. How much external work is to be done to move these charges so that the side of the equilateral triangle becomes a?

(a) (1/4πε₀)(4q²/a)

(b) – (1/4πε₀)(4q²/a)

(c) (1/4πε₀)(2q²/a)

(d) – (1/4πε₀)(2q²/a)

(e) Zero

The external work W to be done is given by

W = U₂ – U₁ where U₂ and U₁ are respectively the final and initial electrostatic potential energies of the system.

Now, U₂ = (1/4πε₀)[(2q×2q)/a + 2q(–q)/a + (–q)2q/a] = 0

U₁ = (1/4πε₀)[(2q×2q)/2a + 2q(–q)/2a + (–q)2q/2a] = 0

Therefore, W = 0.

(6) A conducting sphere of radius R is arranged concentrically inside a thin conducting spherical shell of radius 2R. The sphere carries a charge +q and the spherical shell carries a charge –Q. The potential difference between the sphere and the shell is

(a) (1/4πε₀)(q/2R)

(b) (1/4πε₀)(q/R – Q/2R)

(c) (1/4πε₀)(Q/2R)

(d) (1/4πε₀) (q– Q)/2R

(e) (1/4πε₀) (q– Q)/R

The potential V₁ of the inner sphere is equal to the sum of the potentials due to its own charge q and the charge –Q on the shell:

V₁ = (1/4πε₀)(q/R) +(1/4πε₀)(–Q/2R)

[Note that the potential due to the shell is constant everywhere inside it and is equal to (1/4πε₀)(–Q/2R)].

The potential V₂ of the outer shell is equal to the sum of the potentials due to its own charge –Q and the charge q on the sphere inside it. Since the total charge contributing to the potential of the shell is q – Q, its net potential is given by

V₂ = (1/4πε₀)(q–Q)/2R

The potential difference between the sphere and the shell is given by

V₁ – V₂ = (1/4πε₀)[(q/R) –(Q/2R) –(q/2R) +(Q/2R)]

Or, V₁ – V₂ = (1/4πε₀) [(q/R) –(q/2R)] = (1/4πε₀)(q/2R)

* * * * * * * * * * * * * * * *

You can easily work out this problem if you note that the potential due to the charge on the shell appears in the net potential of the sphere inside it. Therefore, when you find the potential difference between the sphere and the shell, the contribution by the shell gets canceled and it is enough to find the potentials due to the charge on the sphere (sphere alone) at its surface and at distance 2R and find the difference:

Therefore the answer is simply (1/4πε₀) (q/R) – (1/4πε₀) (q/2R), which is equal to (1/4πε₀)(q/2R).

Now suppose we have a conducting sphere of radius R₁ carrying a charge Q₁ arranged concentrically inside a thin conducting spherical shell of radius R₂ carrying a charge Q₂. The potential difference between the sphere and the shell is(1/4πε₀) (Q₁/R₁) – (1/4πε₀) (Q₁/R₂), which is equal to (1/4πε₀)[(Q₁/R₁) – (Q₁/R₂)].

AP Physics C – Electrostatics – Multiple Choice Practice Questions on Capacitors

Let us discuss some practice questions on capacitors meant for AP Physics C aspirants. You should try to find the answers yourself and then only go through the solution part. Here are the questions:

(1) A parallel plate air capacitor with separation d between the plates negligibly small compared to the length and breadth of the plates, is fully charged by connecting it across a battery of emf V volt. If the capacitance of the capacitor is C and the area of each plate is A, the electric field at a point P in between the plates, at distance d/4 from the positive plate (Fig.) is

(a) CV/ε₀A

(b) 4CV/ε₀Ad

(c) CV/ε₀Ad

(d) CV/ 4ε₀Ad

(e) 3CV/4ε₀A

The point P being sufficiently away from the edge of the plate, the electric field at P is constant and is independent of the distance from the plate. The electric field due to a plane sheet of charge is σ/2ε₀ where σ is the surface charge density (charge per unit area). [This can be easily obtained from Gauss’s law].

In the region between the plates, the fields due to the positive plate and the negative plate act in the same direction ( from positive plate to negative plate) and hence they add up to produce a net field of magnitude 2×σ/2ε₀ = σ/ε₀.

Since σ = Q/A and Q = CV, the field at P is CV/ε₀A.

[Note also that at a point outside, the electric field is zero since the positive and negative charges will produce equal and opposite fields].

(2) Two capacitors C₁ and C₂ are identical in all respects except for the dielectric media between their plates. C₁ has air as dielectric where as C₂ has a medium of dielectric constant K in place of air. The capacitor C₁ is charged to V₁ volt and the charging battery is disconnected. Then the uncharged capacitor C₂ is connected across C₁. If the common voltage across the capacitors is V₂, the value of the dielectric constant K is

(a) V₁/ V₂

(b) V₂/ V₁

(c) (V₁ + V₂)/ (V₁ – V₂)

(d) (V₁ – V₂)/ V₁

(e) (V₁ – V₂)/ V₂

If the capacitance of C₁ is C the capacitance of C₂ is KC.

Since the charge is conserved, the initial charge (Q) on C₁ must be equal to the sum of the final charges on C₁ and C₂. Therefore we have

Q = CV₁ = CV₂ + KCV₂

This gives K =(V₁ – V₂)/ V₂

(3) Two thin metal plates of the same area are given positive charges Q₁ and Q₂ and kept parallel to each other to form a capacitor of capacitance C. If Q₂ > Q₁ what will be the potential difference between the plates?

(a) (Q₂ – Q₁) /2C

(b) (Q₂ + Q₁) /2C

(c) (Q₂ + Q₁) /C

(d) (Q₂ – Q₁) /C

(e) (Q₂ + Q₁) /4C

This question may mislead many of you to a wrong answer…

The potential difference (V) between the plates is related to the electric field (E) between the plates and the distance (d) between the plates as

E = V/d

The plates produce opposing fields of magnitudes E₁ (due to Q₁) and E₂ (due to Q₂) in the region between the plates and the resultant field is directed from the plate carrying greater positive charge Q₂ to that carrying smaller positive charge Q₁.

Therefore, E = E₂ – E₁ = V/d so that

V = (E₂ – E₁)d

But E₁ = σ₁ /2ε₀ and E₂ = σ₂ /2ε₀ where σ₁ and σ₂ are the surface densities of charges on the plates given by σ₁ = Q₁/A and σ₂ = Q₂/A

Substituting these values, V = [(Q₂/2ε₀A) – (Q₁/2ε₀A)] /d

Since ε₀A/d = C, we obtain V = (Q₂ – Q₁) /2C.

(4) A parallel plate capacitor with air as dielectric has plates of area A and separation d. It is charged to a potential difference V. The charging battery is then disconnected and the plates are pulled apart so that the separation becomes 3d. What is the work done for pulling the plates?

(a) ε₀AV²/2d

(b) ε₀AV²/3d

(c) ε₀AV²/d

(d) 3ε₀AV²/d

(e) 2ε₀AV²/3d

It is enough to find the difference between the energies of the capacitor. The potential difference between the plates will increase on pulling the plates apart; but the charge will remain unchanged. The energy (U₁) before pulling is given by

U₁ = Q²/2C where C is the initial value of capacitance given by C = ε₀A/d.

When the separation between the plates is increased to 3d, the capacitance is reduced to C/3 and the energy (U₂) is given by

U₂ = Q²/(2C/3) = 3Q²/2C

The work done (W) for pulling the plates apart is given by

W = U₂ – U₁ = 3Q²/2C – Q²/2C = Q²/C

Since Q = CV where C = ε₀A/d, we have

W = CV² = ε₀AV²/d

You will find some useful multiple choice questions with solution at physicsplus.

AP Physics B & C – Electrostatics – Multiple Choice Questions on Capacitors

Here are some typical multiple choice questions (for practice) on capacitors:

(1) Capacitors C₁, C₂ and C₃ of values 15 μF, 10 μF, 3 μF are connected in series and the series combination is connected across a battery of emf 10 V. When the capacitors are fully charged, the charge on one plate of the 3 μF capacitor will be of magnitude

(a) 10 μC

(b) 15 μC

(c) 3 μC

(d) 20 μC

(e) 280 μC

The series combined value (C) of the three capacitors is given by

1/C = 1/C₁ + 1/C₂ + 1/C₃ = 1/15 + 1/10 + 1/3 so that C = 2 μF

The charge (Q) on this equivalent capacitance is given by

Q=CV = 2 μF ×10 V = 20 μC

When capacitors are connected in series, the charges on the plates of all capacitors are of the same magnitude so that the correct option is 20 μC [Option (d)].

(2) A 2 μF capacitor connected in a circuit has one plate at + 6 V and the other plate at – 6 V. The charge on the negative plate of the capacitor is

(a) + 12 μC

(b) – 12 μC

(c) + 24 μC

(d) – 24 μC

(e) zero

The magnitude of charge on either plate is given by

Q = CV

Note that V is the potential difference between the plates and is equal to 6 – (– 6) = 12 V.

Therefore, Q = 2 μF × 12 V = 24 μC

Since the charge on the negative plate must be negative, the answer is – 24 μC.

(3) A parallel plate capacitor with air as dielectric remains connected across a battery of emf 6 V. The charge on the capacitor in this condition is Q. If the separation between the plates is decreased by 10% in this condition and sufficient time is allowed to attain steady state, the charge on the capacitor will be

(a) unchanged

(b) increased by 9%, approximately

(c) decreased by 9%, approximately

(d) decreased by 11%, approximately

(e) increased by 11%, approximately

The correct option is (e) since the capacitance will increase by approximately 11% in accordance with the expression,

C = ε₀A/d when the separation d changes to 0.9d.

[Remember that Q = CV]

Now, note the following in connection with the above question:

If the battery is disconnected after charging the capacitor to have an initial charge Q,

(i) the charge on the plates will be unchanged if the plate separation is decreased or increased (since there is no battery to control the charge). The law of conservation of charge is very strictly obeyed.

(ii) The potential difference between the plates of the capacitor will be decreased on decreasing the separation between the plates since V = Q/C and Q is unchanged where as C is increased.

(iii) The potential difference between the plates of the capacitor will be increased on increasing the separation between the plates since V = Q/C and Q is unchanged where as C is decreased.

(iv) If a dielectric slab is inserted in to the gap between the plates, the potential difference between the plates will be decreased since the capacitance is increased (with the charge on the plates unchanged).

* * * * * * * * * * * * * *

(4) A parallel plate capacitor with air as dielectric remains connected across a battery. After the capacitor is fully charged in this condition, a slab of dielectric constant 5 is slowly introduced into the gap between the plates. Which one of the following statements is true during the introduction of the slab?

(a) The charge on the capacitor will gradually decrease

(b) The capacitance will gradually decrease

(c) A current will flow through the leads connecting the capacitor to the battery

(d) The potential difference between the plates of the capacitor will gradually increase

(e) The potential difference between the plates of the capacitor will gradually decrease

When the slab is introduced, the capacitance gradually increases and the battery supplies more charges to the capacitor since the charge Q has to increase in accordance with the equation, Q = CV. The potential difference across the capacitor will be unchanged since it is connected across the battery (whose emf is fixed).

The only correct option therefore is (c).

(5) Half of the space between the plates of a parallel plate air capacitor of capacitance C is filled as shown with a material of dielectric constant K. The new capacitance will be

(a) KC

(b) KC/2

(c) (K + 1)C/2

(d) 2C/K

(e) (K – 1)C

Initially the capacitance C is given by

C = ε₀A/d where A and d are the plate area and plate separation respectively.

On introducing the dielectric material, you can treat the new capacitor to be made of two capacitors, one with air as dielectric and the other with the introduced material as the dielectric. But the area of each capacitor is half that of the full capacitor. Further, the two capacitors are in parallel and hence the capacitance (C’) of the new capacitor is given by

C’ = ε₀A/2d + Kε₀A/2d

Putting C = ε₀A/d, we obtain C’ = (K + 1)C/2.

We will discuss more questions on capacitors in due course. Meanwhile, find some useful and interesting multiple choice questions (with solution) at physicsplus.