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 B – Some Interesting Multiple Choice Practice Questions on Atomic and Nuclear Physics

“Non-violence leads to the highest ethics, which is the goal of all evolution. Until we stop harming all other living beings, we are still savages.”

– Thomas A. Edison

The sections ‘atomic physics and quantum effects’ and ‘nuclear physics’ in the AP Physics B syllabus will appear to be interesting to most of the AP Physics B aspirants. Today we will discuss a few practice questions (multiple choice) in these sections:

(1) The de Broglie wave length of a particle with kinetic energy E is λ. If its kinetic energy is increased  to 4E, its de Broglie wave length will be

(a) λ/4

(b) λ/2

(c) λ

(d) 2λ

(e) 4λ

Since the kinetic energy is directly proportional to the square of momentum, the momentum of the particle is doubled when its kinetic energy is quadrupled.

[Note that kinetic energy E = p²/2m where p is the momentum and m is the mass].

The de Broglie wave length λ is given by

             λ = h/p where h is Planck’s constant and p is the momentum.

Therefore, when p is doubled, λ is halved [Option (b)].

(2) Particles A and B have masses m and 4m respectively but they carry the same charge. When they are accelerated by the same voltage, their de Broglie wave lengths are in the ratio

(a) 1 : 1

(b) 2 : 1

(c) 4 : 1

(d) 1 : 4

(e) 1 : 8

Let V represent the common accelerating voltage and q represent the common charge of the particles. If p₁ and p₂ are the momenta of the particles A and B, on equating their kinetic energies, we have

             p₁²/2m = p₂²/(2×4m)

[Remember that the kinetic energy of a particle of charge q accelerated by a voltage V is qV. The kinetic energies of A and B are equal since they have the same charge and they are accelerated by the same voltage]

The above equation gives

             p₁/ p₂ = ½

Since the de Broglie wave length λ is given by

             λ = h/p where h is Planck’s constant and p is the momentum, the ratio of the de Broglie wave lengths of A and B is given by

              λ₁/ λ₂ = p₂/ p₁ = 2, as given in option (b).

(3) A metallic surface is found to emit photo-electrons when monochromatic light rays of frequencies n₁ and n₂ (n₂ > n₁) are incident on it. If the maximum values of kinetic energy of the photo-electrons emitted in the two cases are in the ratio 1 : 3, the threshold frequency of the metallic surface is

(a) (3n₁ – n₂)/ 2

(b) (2n₁ – n₂)/ 2

(c) (3n₁ – n₂)/ 3

(d) (n₂ – n₁)/ 3

(e) (n₂ – n₁)/ 2

If the threshold frequency is n₀ and the maximum values of kinetic energy in the two cases are E₁ and E₂ respectively, we have

             hn₁ = hn₀ + E₁ and

             hn₂ = hn₀ + E₂

Therefore, E₁/E₂ = (n₁ – n₀)/(n₂ – n₀)

Since the ratio is 1 : 3 we have

             (n₁ – n₀)/(n₂ – n₀) = 1/3

Or, 3n₁ – 3n₀ = n₂ – n₀

This gives n₀ = (3n₁ – n₂)/ 2, as given in option (a).

(4) The energy that must be added to an electron to reduce its de Broglie wave length from 2 nm to 1 nm is

(a) half the initial energy

(b) equal to the initial energy

(c) twice the initial energy

(d) thrice the initial energy

(e) four times the initial energy

The de Broglie wave length λ is given by

             λ = h/p where h is Planck’s constant and p is the momentum.

Since the de Broglie wave length of the electron is to be reduced to half the initial value (from 2 nm to 1 nm), the momentum of the electron is to be doubled. But when the momentum is doubled, its kinetic energy becomes four times the initial value. Therefore the energy that must be added is three times the initial energy [Option (d)].

(5) An alpha particle of mass m and speed v  proceeds directly towards a heavy nucleus of charge Ze. The distance of closest approach of the alpha particle is directly proportional to

(a) 1/Ze²

(b) 1/Ze

(c) v

(d) m

(e) 1/v²

The alpha particle has to move towards the nucleus with difficulty, doing work against the electrostatic repulsive force. When the alpha particle reaches the distance of closest approach, the entire kinetic energy gets converted into electrostatic potential energy. Therefore we have

             ½ mv² = (1/4πε₀)(2Ze²/r) where r is the distance of closest approach.

[Remember that the charge on the alpha particle is 2e].

This gives r = Ze²/πε₀mv²

This shows that r is directly proportional to 1/v².

AP Physics B & C - Multiple Choice Practice Questions on One Dimensional Kinematics

"Whenever you are confronted with an opponent, conquer him with love."

– Mahatma Gandhi

Let us discuss a few interesting multiple choice practice questions on one dimensional motion. Here are some questions beneficial for AP Physics B as well as AP Physics C aspirants:

(1) The adjoining figure shows the velocity time graph of an object. Total displacement suffered by the object during the interval when it has non-zero acceleration and retardation is

(a) 80 m

(b) 70 m

(c) 60 m

(d) 40 m

(e) 30 m

The object has non-zero acceleration and retardation during the time intervals from 5 sec to 15 sec and from 20 sec to 30 sec. The total area under the velocity time graph during these intervals gives the required displacement.

Displacement from 5 sec to 15 sec = 30 m

Displacement from 20 sec to 30 sec = 40 m

Therefore, total displacement = 70 m

(2) Successive positions (x) of an object (moving from left to right) at equal time intervals are shown in the following figure:

Which one among the following position-time graphs best represents the motion of the object? 

 

Change of position is slowest in the beginning and in the end. The motion is best represented by graph (c).

[Note that graph (d) is not the answer since the change of position in the beginning and in the end is shown as fastest in it]

(3) Which one among the following velocity-time graphs best represents the motion of the object mentioned in question no.(2)? 

The graph (c) is the answer.

The following questions are meant for AP Physics C aspirants:

(4) A particle projected vertically upwards attains the maximum height h in time t. While returning from the highest point it takes an additional time t₁ to fall to the height h.2. If air resistance is negligible, how is t₁ related to t?

(a) t₁ = t/2

(b) t₁ = √(t/2)

(c) t₁ = t/√2

(d) t₁ = t/√3

(e) t₁ = t/3

For the upward motion we have

             0 – u² = – 2gh …………(i)

[We have used the equation of motion, v² – u² = 2as. The sign of the gravitational acceleration g is negative since it is opposite to the direction of the velocity of projection u]

Using the equation of motion, v = u + at we have

             0 = u – gt from which u = gt

Substituting this value of u in Eq (i), we have

             g²t² = 2gh

Therefore, t =√(2h/g) ……..(ii)

For the fall through h/2 from the highest point we have

             h/2 = 0×t₁ + ½ gt₁²

[We have used the equation of motion, s = ut + ½ gt²]

Therefore, t₁ =√(h/g)

Comparing this with the value of t given in Eq (ii) we obtain t₁ = t/√2.

(5) A ball projected vertically up has the same vertical displacement h at times t₁ second and t₂ second. If air resistance is negligible, the maximum height reached by the ball is

(a) gt₁t₂/2

(b) g(t₁² + t₂²)/2

(c) g(t₁ + t₂)²/4

(d) g(t₁ + t₂)²/8

(e) 2g(t₁ + t₂)²

The time taken by the ball to move from height h to the top of its trajectory and back to the height h is t₂ – t₁. Therefore, the time taken to move from height h to the top of the trajectory is (t₂ – t₁)/2.

The total time taken for the upward journey (from ground to the top most point) is evidently t₁ + t₂ – t₁)/2 = (t₁ + t₂)/2.

Time taken for the return journey (from the top most point to the ground) also is equal to (t₁ + t₂)/2. Therefore, the maximum height H (using the equation, s = ut + ½ gt²) is given by

             H = 0 + ½ g [(t₁ + t₂)/2]²

Or, H = g(t₁ + t₂)²/8

             Questions on kinematics were discussed earlier on this site. You can access them either by clicking on the label ‘kinematics’ below this post or by trying a search for                  ‘kinematics’ using the search box provided on this page.

AP Physics B - Multiple Choice Practice Questions on Atomic Physics and Quantum Effects

"Men often become what they believe themselves to be. If I believe I cannot do something, it makes me incapable of doing it. But when I believe I can, then I acquire the ability to do it even if I didn't have it in the beginning

– Mahatma Gandhi

 

AP Physics 2012 exams are just a few days away. Your final preparations for the exam must be in full swing and there is no time to waste. Today I give you a few multiple choice practice questions on atomic physics and quantum effects. Questions in this section posted earlier on this site (with solution) can be accessed by clicking on the label ‘atomic physics and quantum effects’ given below this post. Or, you may try a search for ‘atomic physics and quantum effects’ using the search box provided on this page.

(1) The adjoining figure shows the graphical relation between the frequency of incident radiation and the magnitude of stopping potential in the case of metals A and B. Note that the straight line graphs have the same slope. Which one among the following statements is correct in this case?

(a) A and B have the same work function

(b) A and B have the same threshold wave length

(c) Maximum kinetic energy of photoelectron in the case of metals A and B is directly proportional to the increment in frequency of incident radiation over the threshold frequency.

(d) Metal B is a better photosensitive material than metal A.

(e) For a given change in the frequency of incident radiation, the changes in stopping potentials are different for metals A and B

Since we have straight line graphs for both metals, maximum kinetic energy of photoelectron is directly proportional to the increment in frequency of incident radiation over the threshold frequency. Therefore option (c) is correct.

(2) The de Broglie wave length of a particle can be reduced to half its initial value by changing its kinetic energy to

(a) half the initial value

(b) twice the initial value

(c) three times the initial value

(d) four times the initial value

(e) a quarter of the initial value

Kinetic energy E of a body of mass m is given by

             E = p²/2m where p is the momentum

[This follows from E = ½ mv² = m²v²/2m. Here v is the velocity and mv = p]

The above relation shows that the kinetic energy becomes four times when the momentum is doubled.

The de Broglie wave length λ is given by

             λ = h/p where h is Planck’s constant.

Therefore, the de Broglie wave length of a particle can be reduced to half its initial value by changing its momentum to twice the initial value. Evidently the kinetic energy of the particle the becomes four times the initial value [Option (d)].

[Suppose the above question is modifie as follows:

A particle has de Broglie wave length λ when its kinetic energy is E. What additional kinetic energy is to be aded to it in order to reduce the de Broglie wave length to λ/2?

(a) E

(b) 2E

(c) 3E

(d) 4E

(e) E/4

The answer is 3E since you are asked to find the additional kinetic energy].

(3) Uranium (atomic number 92) has an isotope of mass number 235. It can undergo successive disintegrations to get transformed into lead (₈₂Pb²⁰⁷). How many α-particles and β-particles are emitted during this transformation?

(a) α = 7, β = 4

(b) α = 4, β = 3

(c) α = 7, β = 0

(d) α = 7, β = 7

(e) α = 4, β = 7

Beta particle emission does not affect the mass number. In order to reduce the mass number by 28 (from 235 to 207), the number of α-particles to be emitte must be 7. Since each α-particles carries two fundamental units of positive charge, the atomic number of the end product gets reduced by 14. But the final product (₈₂Pb²⁰⁷) has its atomic number reduced by 10 only. The extra 4 units must be obtained by the emission of four β-particles. The correct option therefore ia (a).   

[Note that when a β-particles (electron) is emitted from the nucleus, the nuclear charge increases by one unit. This happens as a result of the transformation of a neutron in the nucleus into a proton].

(4) Fundamental forces in nature are gravitational force, electromagnetic force, nuclear force and weak force. If these forces act over very short distances of the order of nuclear dimensions, how do you arrange them in decreasing order (starting with the strongest?

(a) Gravitational force, electromagnetic force, nuclear force, weak force

(b) Electromagnetic force, gravitational force, nuclear force, weak force

(c) Electromagnetic force, nuclear force, weak force, gravitational force

(d) Gravitational force, nuclear force, electromagnetic force, weak force

(e) Nuclear force, electromagnetic force, weak force, gravitational force

The correct option is (d).

[Don’t get carried away by the term ‘weak force’. The weakest force is gravitational force where as the strongest is nuclear force].

(5) Two protons are separated by a distance of 50 Ǻ. If the electromagnetic force between them is F₁ and the nuclear force between them is F₂, which one among the following is the most reliable statement?

(a) F₁ >> F₂

(b) F₂ >> F₁

(c) F₁ > F₂

(d) F₂ > F₁

(e) F₂ = F₁

This question is similar to question No. (4) in the sense that it is meant for checking your knowledge of nuclear physics. The correct option is (a). Nuclear  force is a very short range force. At a separation of 50 Ǻ which is very large compared to the size of a nucleus, nuclear force (strong interaction) between two protons is negligible compared to the electrostatic force. Therefore the correct option is (a).