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 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 B & C Multiple Choice Practice Questions on Electrostatics

“I never did a day's work in my life. It was all fun.”

– Thomas A. Edison

Questions in the section ‘electrostatics’ were discussed on many occasions earlier on this site. You may access them by clicking on the label ‘electrostatics’ below this post. Since the number of posts displayed per page is limited, you will have to use the ‘older posts’ button to access all the posts.

All posts in this section can equally well be accessed by trying a search for ‘electrostatics’ using the search box provided on this page.

Today we will discuss a few more multiple choice practice questions on electrostatics:

(1) Point charges +q and –3q are arranged at points A and B on the x-axis (Fig.). At which point (or points) is the electric field due to this system of charges zero?

(a) At two points on the x axis, one point to the right of the charge –3q  and the other to the left of the charge +q.

(b) Somewhere on the x axis to the left of the charge +q.

(c) Somewhere on the x axis to the right of the charge –3q.

(d) Somewhere on the x axis, in between the two charges.

(e) The electric field cannot be zero anywhere.

The electric field lines due to a positive charge proceed radially outwards (diverge) from the charge where as the electric field lines due to a negative charge proceed radially inwards (converge) to the charge. At points on the x-axis lying in between A and B the fields due to these charges are directed along the positive x-direction. Therefore, the fields add up and the resultant field cannot be zero. But at points on the x-axis to the left of the charge +q, the fields are in opposite directions (along the negative x-direction due to the charge +q and along the positive x-direction due to the charge –3q) so that they can cancel at a point where the magnitudes of the fields due to the charges are the same. So the correct option is (b).

[At points on the x-axis to the right of the charge –3q, the fields are in opposite directions (along the positive x-direction due to the charge +q and along the negative x-direction due to the charge –3q. But the two fields cannot cancel anywhere here since the magnitude of the field due to the charge –3q is greater than that due to  the charge +q (since the charge –3q is nearer as far as points on the right of B are concerned]

(2) A large flat plate is positively charged so that it has uniform surface charge density σ. If the electric field near the central region of the plate at a distance of 1 cm from the plate is 9 NC^–1, what will be the electric field at a distance of 3 cm from the plate?

(a) 81 NC^–1

(b) 27 NC^–1

(c) 9 NC^–1

(d) 3 NC^–1

(e) 1 NC^–1

The electric field due to the surface charge on the plate is normal to the surface of the plate. Since the plate is large, the electric field near the central region of the plate is uniform (with the electric field lines proceeding normal to the flat surface and therefore parallel). This means that the electric field at a point is independent of the distance of the point from the plate so that electric field at 3 cm from the plate is 9 NC^–1 itself.

[For AP Physics C aspirants: 

To find the electric field near the surface of the charged plate, we can apply Gauss theorem and accordingly imagine a Gaussian surface shaped as a rectangular parallelepiped of cross section area A (Fig.). The electric field due to the surface charge on the plate is normal to the surface of the plate. Therefore the electric field is directed normal to the end faces of the rectangular parallelepiped so that the total electric flux through the closed suface (rectangular parallelepiped) is 2EA where E is the magnitude of the electric field. (The flux through the side surfaces of the parallelepiped is zero since the electric field is parallel to these surfaces). The total charge enclosed by the closed surface is σA. Therefore, by Gauss theorem we have

             2EA = σA/ε₀ where ε₀ is the permittivity of free space.

Therefore E = σ/2ε₀, which is independent of distance]

(3) Charges +q, –q and –q are placed at the vertices of an equilateral triangle ABC as indicated in the adjoining figure. What is the direction of the net electric field at the central point P?

(a) Along AP

(b) Along PA

(c) Along CP

(d) Along BP

(e) There is no field at P

The direction of the electric fiel at P is the direction of the force acting on a test positive charge placed at P. The charge +q will exert a repulsive force on the test positive charge. This is irected along AP. The charges –q and –q at B and C will exert attractive forces along PB and PC respectively. These two forces being of equal magnitudes, their resultant will be directed along AP. Therefore the resultant electric field due to all the three charges will be along AP [Option (a)]. 

(4) Points 1, 2 and 3 lie on the axis of an electric dipole AB. Point 4 lie on the equatorial line (perpendicular bisector of AB) of the dipole. Out of the following choices which one correctly gives the directions of the electric fields at the points 1, 2, 3 and 4?        

When you place a positive test charge at point (1) the net force on it is repulsive since the charge +q of the dipole is nearer than the charge –q. The direction of the force (and the electric field) is therefore leftwards.

At point (2) the charge +q exerts a repulsive force on the positive test charge where as the charge –q exerts an attractive force on it. Thus both forces act rightwards an hence the electric field at point (2) acts rightwards.

At point (3) the net force on the positive test charge is attractive since the charge –q of the dipole is nearer than the charge +q. The electric field at point (3) is therefore leftwards.

At point (4) the positive test charge  is repelled by the charge +q and attracted by the charge –q. these forces are equal in magnitude and are inclined equally upwards and downwards respectively.  Their resultant acts rightwards and hence the electric field at point (4) is rightwards.

The above facts are correctly given in option (e).

If you have a mental picture of the distribution of the electric field lines due to an electric dipole (See the figure below), you will be able to answer the above question in no time].

 

(5) In the above question, in the case of the points (1), (2), (3) and (4), where do you find zero electric potential?

(a) Nowhere

(b) At point (1)

(c) At points (1), (2) and (3)

(d) At points (2) and (4)

(e) At point (2)

Points (2) and (4) are equidistant from the charges +q and –q. These equal and opposite charges produce positive and negative potentials of equal values so that they cancel each other, giving rise to zero potential at points (2) and (4). 

Now see these questions with solution.

AP Physics C - Answer to Free Response Practice Question on Electrostaics

"I am a friend of Plato, I am a friend of Aristotle, but truth is my greater friend".

–Sir Isaac Newton

In the post dated 11^th January 2012, a free-response question for practice was given to you. As promised, I give below a model answer. The question also is given for convenience:

A thin circular nonconducting disc of radius ‘a’ resting horizontally on the ground has positive charges sprayed on its top surface so that it has a uniform surface charge density σ. The centre of the disc is O and the axis of the disc is vertical (Fig.). There are no other electric charges in the vicinity for producing any appreciable interaction on the charges on the disc and the acceleration ue to gravity at the place is g. Now answer the following questions:

(a) Derive an expression for the electric potential at a point on the axis of the disc at distance x from its centre.

(b) What is the electric potential at the centre of the disc?

(c) A positively charged particle of mass m and charge q with specific charge (q/m) of 8ε₀g/σ is released from rest from a point P on the axis of the disc. The distance of point P from the centre of the disc is d. If the particle just reaches the centre O of the disc, determine the value of d.

(d) Explain qualitatively the nature of motion of the charged particle after its release from the point P stated in part (c) above.

[This question was asked in a different form in IIT entrance examination earlier].

(a) The disc can be imagine to be made of a large number of concentric rings of radii ranging from zero to a. To derive an expression for the potential at a point P (Fig.) on the axis of the disc at distance x from the centre, consider one such ring of radius r and width dr as shown in the figure. The total charge on the ring is 2πrdrσ since the area of the top surface of the ring is 2πrdr.

All charges on the ring are at the same distance √(r² + x²) from the point P so that the electric potential at P due to the charges on the ring is (1/4πε₀){2πrdrσ/√(r² + x²)} where ε₀ is the permittivity of free space.

The electric potential at P due to the charges on the entire disc [_(P)] is obtained by adding the contributions of all the rings forming the entire disc. Therefore we have

V_(P) = ₀∫^a [(1/4πε₀){2πrdrσ/√(r² + x²)}]dr

Or, V_(P) = (σ/2ε₀) ₀∫^a [r/√(r² + x²)] dr

Since the value of the integral is √(r² + x²) and the upper and lower limits (of r) are a and zero, we get

V_(P) = (σ/2ε₀)[√(a² + x²) – x]

(b) The electric potential at the centre of the disc [V_(O)] is obtained by putting x = 0 in the above general expression for the potential on the axis.

Therefore, V_(O) = σa/2ε₀

(c) When the particle falls from height d and just comes to rest (momentarily) at O, as is evident from the question, it gains electric potential energy at the cost of gravitational potential energy. Therefore, by the law of conservation of energy we have

mgd =q[V_(O) – V_(P)] where V_(P) here is the electric potential energy when the point P is at height d and is equal to (σ/2ε₀){√(a² + d²) – d}. Therefore we have

mgd =q[(σa/2ε₀) – (σ/2ε₀){√(a² + d²) – d}]

Or, mgd =(qσ/2ε₀) [a –{√(a² + d²) – d}]

Therefore, d =(q/m)(σ/2ε₀g) [a –{√(a² + d²) – d}]

Since the specific charge q/m =8ε₀g/σ, as given in the question, we get

d =(8ε₀g/σ)(σ/2ε₀g) [a –{√(a² + d²) – d}] = 4[a –{√(a² + d²) – d}]

Or, 4√(a² + d²) = 3d + 4a

Squaring, 16a² + 16d² = 9d² + 16a² + 24ad

Or, 7d² = 24ad from which d = 24a/7

(d) Initially the particle moves down with an acceleration because of the gravitational pull mg even though the electrostatic repulsive force due to the like charges on the disc opposes the motion. As the particle approaches the disc, the electric force increases where as the gravitational force remains constant. But the particle overshoots its final equilibrium position because of inertia. The downward motion of the particle gets decelerated because of the increasing electric repulsive force and it comes to rest momentarily at the central point O. The particle then moves upwards because of the electric repulsive force which is greater in magnitude here, compared to the gravitational pull. The particle oscillates about its final equilibrium position and finally comes to rest at the final equilibrium position which is at a height from the point O.