AP Physics B & C - Electrostatics - Multiple Choice Practice Questions

“Iron rusts from disuse, stagnant water loses its purity and in cold weather becomes frozen; so does inaction sap the vigors of the mind.”

– Leonardo da Vinci

 

Questions on electrostatics were discussed on many occasions on this site. You can access them by clicking on the label ‘electrostatics’ below this post or by trying a search for ‘electrostatics’ using the search box provided on this page. Today we shall discuss a few more questions in this section.

(1) Right angled triangle ABC is located in a uniform electric field E. Sides AC and BC have lengths 0.5 m and 0.3 m respectively and BC is at right angles to the electric field lines (Fig.). If the electric potential difference between A and C is 80 V, what is the magnitude of the electric field E?

(a) 80 V/m

(b) 160 V/m

(c) 200 V/m

(d) 100 V/m

(e) 16 V/m

The electric potential at the point B is the same as that at the point C (since the straight line BC is at rjght angles to the direction of the uniform electric field E). Therefore the potential difference between points A and B is 80 V. The electric field E is directed along AB and hence the magnitude of E is (80 V)/(0.4 m) = 200 V/m.

[AB = √(AC² – BC²) = √(0.5² – 0.3²)  = 0.4 m]

(2) In the above question, what is the component of electric field along the direction AC?

(a) 80 V/m

(b) 160 V/m

(c) 200 V/m

(d) 100 V/m

(e) 40 V/m

The potential difference between the points A and C is 80 V and the distance between these points is 0.5 m. Therefore, the component of electric field along the direction AC is (80 V)/(0.5 m) = 160 V/m.

(3) A cube of side a has a charge Q at its centre (Fig.). What is the electric flux through one face of the cube?

(a) Q/ε₀a

(b) Qa/ε₀

(c) Q/6ε₀

(d) Q/8ε₀

(e) Qε₀/6

The electric flux over a closed surface, according to Gauss theorem, is Q/ε₀ where Q is the net charge enclosed by the surface and ε₀ is the permittivity of free space.

[You can understand the above even without knowing Gauss law. You know (from inverse square law) that the electric field at distance r from a point charge Q is Q/4πε₀r². The electric field at any point is the electric flux through unit area held with the plane of the area perpendicular to the electric field lines. Imagine a spherical surface of radius r such that the charge Q is at the centre. Since the area of the spherical surface is 4πr² the electric field at any point on the surface must be Ф/4πr² where Ф is the total electric flux produced by the charge Q. Therefore we have

            Q/4πε₀r² = Ф/4πr²

This gives Ф = Q/ε₀ as stated in Gauss law.]

Since the charge Q is at the centre of the cube, all the six faces of the cube receive equal electric flux so that the flux through one face of the cube is Ф/6 = Q/6ε₀

(4) The effective capacitance between terminals A and B in the network shown in the adjoining figure is

(a) 16 μF

(b) 8 μF

(c) 6 μF

(d) 16/3 μF

(e) 8/3 μF

Since the capacitors C₁, C₂, C₃ and C₄ are of values satisfying the condition C₁/C₂ = C₃/C₄, the junction of C₁ and C₂ is at the same potential as the junction of C₃ and C₄ (as in the case of a balanced Wheatstone brige). Therefore the capacitors C₅ and C₆ connected across the diagonal of the bridge have no effect and can be ignored.

The network therefore simplifies to the series combination of C₁ and C₂ connected in parallel with the series combination of C₃ and C₄.

Series combined value of C₁ and C₂ = (3×6)/(3+6) = 2 μF

Series combined value of C₃ and C₄ = (1×2)/(1+2) = 2/3 μF

Therefore the effective capacitance between terminals A and B = 2 μF + (2/3) μF = 8/3 μF

The capacitors C₁ and C₄ in the above question are short circuited and the circuit then gets modified as shown in the figure. What is the effective capacitance between the terminals A and B in this situation?   

(a) 11 μF

(b) 9 μF

(c) 8 μF

(d) 7.5 μF

(e) 6.5 μF

Have a careful look at the circuit. You will find that one plate of C₂, C₃ and C₅ is connected to terminal A. The other plate of C₂ as well as C₃ is connected to terminal B while that of C₅ is connected 6through C₆ to terminal B. The series combination of C₅ and C₆ gives an effective capacitance of 1μF. Thus we have three capacitances 1μF, 1μF and 6μF connected in parallel across the terminals A and B.

Therefore the effective capacitance between terminals A and B in this case is 1μF + 1μF + 6μF = 8μF.

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 B and C- Electrostatics: Answers to Free Response Practice Questions on Capacitors

In the post dated 5^th August 2008, two free response practice questions on capacitors were given to you. As promised, I give below model answers along with the questions:

Question No1 (For AP Physics B & C):

Two metal plates A and B, each of area 0.2 m² are arranged parallel to each other as shown, with a separation of 1 cm in air. A regulated direct voltage source of output V volt remains connected between the plates so that there is an electric field of 2000 N/C in the region between the plates.

(a) Determine the output voltage V of the regulated voltage source.

(b) Determine the capacitance of the parallel plate capacitor formed by the plates A and B

(c) If the separation between the plates is gradually reduced (with the voltage source remaining connected between the plates), what quantities among the following will change? Put tick (√) marks against the changing quantities:

(i) Capacitance____ (ii) Charge on the plates____ (iii) Potential difference between the plates___

Give reasons for change (if any) and for no change (if any).

(d) During the decrease in the separation between the plates, will there be a current through the wires connecting the plates to the voltage source? Justify your answer.

(e) Determine the energy of the capacitor when the separation between the plates is reduced to half the initial value.

(a) The potential difference between the plates is equal to the otput voltage V of the voltage source. Since the electric field (E) between the plates is given by E = V/d where d is the separation between the plates, we have

2000 = V/10^–2 from which V = 20 volt.

(b) Capacitance (C) of the parallel plate capacitor is given by

C = ε₀A/d where A is the area of a plate.

Therefore, C = 8.85×10^–12×0.2/10^–2 = 1.77×10^–10 farad.

[The value of ε₀ will be given in your question paper].

(c) (i) Capacitance__√__ (ii) Charge on the plates__√__ (iii) Voltage between the plates____

The capacitance is given by C = ε₀A/d where A is the area of a plate and d is the seaparation between the plates. Since the separation is decreased, the capacitance is increased.

The charge on the plates is given by Q = CV. The capacitance C is increased but the voltage V between the plates is unchanged (since the plates are connected across the fixed voltage source). The charge Q on the plates is therefore increased.

(d) During the decrease in the separation between the plates, the capacitance goes on increasing and hence more charges flow from the voltage source to the capacitor, satisfying the equation, Q = CV. So there is a charging current in the wires connecting the plates to the voltage source.

(e) The energy (U) of a capacitor of capacitance C charged to a potential difference V between its plates is given by

U = (½) CV²

Initially the capacitance of the capacitor made by the two plates is 1.77×10^–10 farad as shown in part (a) above. When the separation (d) between the plates is reduced to half the initial value, the capacitance is doubled in accordance with the expression for capacitance, C = ε₀A/d. Therefore, the energy is given by

U =(½) (2×1.77×10^–10)×20² = 7.08×10^–8 joule.

Question No.2 (For AP Physics C):

A spherical capacitor is made of two conducting spherical shells of radii r₁ and r₂ (r₂ > r₁). The inner shell has a total charge +Q and the outer shell has a total charge –Q. Now, answer the following questions:

(a) The space between the shells is filled with air of dielectric constant very nearly equal to 1. Calculate the potential difference between the shells.

(b) Determine the electric field at a point in between the shells, at distance a from the common centre of the shells.

(c) What are the values of the electric field and potential at a point outside both shells, at distance b (b > r₂) from the common centre of the shells?

Justify your answer.

(d) Determine the potential at the common centre of the shells.

(e) The air in between the shells is replaced by a material whose dielectric constant has value K₁ from r₁ to R and value K₂ from R to r₂ ((r₂ > R > r₁). Show that the capacitance of this spherical capacitor is given by

C = 4πε₀/[(1/K₂R – 1/K₁R) + (1/K₁r₁ – 1/K₂r₂)]

(a) The potential at all points inside the larger shell is the same as the potential at its surface. Therefore, the potential difference (between the shells) produced by the charge –Q on the outer shell is zero and we need consider the contribution (to the potential difference) by the charge +Q on the inner shell alone.

The potential (V₁) of the inner shell due to its charge +Q is given by

V₁ = Q/4πε₀r₁

The potential (V₂) of the outer shell due to the charge +Q on the inner shell is given by

V₂ = Q/4πε₀r₂

Therefore, potential difference V₁ – V₂ = (Q/4πε₀)[(1/ r₁) – (1/ r₂)]

(b) The electric field at points inside the outer shell due to the charge on it is zero (since there is no potential gradient because of its charge). The field (E) in between the shells at distance a is therefore due to the charge +Q on the inner shell alone and is given by

E = Q/4πε₀a²

This field is directed radially outwards.

(c) The electric potential and field at all points outside both shells (at distance b > r₂) is zero since the potential and field at points outside any charged spherical shell is produced as though the entire charge is concentrated at the centre. We have therefore positive and negative charges of equal magnitude and the net charge imagined to be concentrated at the centre is zero.

[The electric field can be proved to be zero by applying Gauss’s law, considering a Gaussian spherical surface of radius b passing through the point. The net charge enclosed by the gaussian surface is zero so that the field is zero].

(d) The potential (V) at the common centre of the shells is the sum of the potentials due to the charges on the shells and is given by

V = (1/4πε₀)[(Q/r₁) – (Q/r₂)] = (Q/4πε₀)[(1/ r₁) – (1/ r₂)]

(e) The capacitance (C) is given by

C = Q/V where V is the potential difference between the shells.

But V = V₁ – V₂ where V₁ and V₂ are respectively potentials on the inner and outer shells with the composite dielectric in between them.

If E is the electric field at a point in between the shells, at distance r from the centre, we have

V = – _r2∫^r1 Edr = _r1∫^r2 (Q/4πε₀K r²)dr where K is the dielectric constant which varies with r (K has value K₁ from r₁ to R and value K₂ from R to r₂).

Therefore, V = (Q/4πε₀) _r1 ∫^r2 dr/ Kr²

= (Q/4πε₀) [{– 1/ K₁r}^R _r1 + {– 1/ K₂r}^r2_R]

= (Q/4πε₀) [(1/K₂R – 1/K₁R) + (1/K₁r₁ – 1/K₂r₂)]

The capacitance is given by C = Q/V = 4πε₀/[(1/K₂R – 1/K₁R) + (1/K₁r₁ – 1/K₂r₂)]

[You can obtain the above expression from the expression for the capacitance of a spherical capacitor with a material of dielectric constant K, given in the post dated 14^th July 2008:

C = 4πε₀K r_a r_b /(r_b – r_a)

The radii r_a and r_b are to be replaced by r₁ and r₂. Further, C will be the series combined value of two spherical capacitors C₁ and C₂ where C₁ is made by spherical shells of radii r₁ and R, with material of dielectric constant K₁ in between and C₂ is made by spherical shells of radii R and r₂, with material of dielectric constant K₂ in between. C is given by the reciprocal relation, 1/C = 1/C₁ + 1/C₂.

Since you have to borrow the expression for the simple spherical capacitor, this method will not be as rigorous as the one we used above]