AP Physics Resources

Friday, September 13, 2013

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

(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

(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

(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

(d) 16/3 μF

(e) 8/3 μF

Since the capacitors C₁, C₂, C₃and C₄are 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 C₄in 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

(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.

Friday, August 9, 2013

AP Physics B - Multiple Choice Practice Questions on Geometric Optics

“A person who won't read has no advantage over one who can't read.”

– Mark Twain

Questions on geometric optics were discussed on many occasions on this site. Try a search for ‘geometric optics’ using the search box provided on the side bar this page or click on the label ‘geometric optics’ below this post, to access all posts related to geometric optics.

Today we shall discuss a few more multiple choice practice questions on geometric optics:

(1) Focal power or optical power (or simply, power) of a lens is the reciprocal of its focal length. The focal length has to be expressed in metre for obtaining the power in its popular unit, dioptre.

Two thin converging lenses of focal lengths 40 cm and 60 cm are kept in contact so as to have a common principal axis. The power of this combination is approximately

(a) 1

(b) 2

(d) 4

(e) 5

If the focal length of the combination of the lenses is F metre, the power of the combination is 1/F.

When two thin lenses of focal lengths f₁ and f₂ are kept in contact so as to have a common principal axis, the focal length F of the combination is given by the reciprocal relation,

1/F = 1/f₁ + 1/f₂

[In fact, this is a relation connecting the powers of the individual lenses to the power of the combination]

Therefore we have 1/F = 1/0.4 + 1/0.6 = (0.6 + 0.4)/(0.4×0.6) = 1/0.24 ≈ 4

(2) A bright object O is placed on the principal axis of a thin converging lens. The lens produces a real magnified image of O at the position I (Fig.) at a distance of 44 cm from the lens. When a convex mirror of focal length 12 cm is interposed between the image and the lens as shown in the figure, a real image of the same size as the object is obtained side by side with the object. What is the distance between the lens and the mirror?

(a) 12 cm

(b) 16 cm

(d) 22 cm

(e) 24 cm

Since the final image is obtained side by side with the object, the rays of light must fall normally on the mirror. This means that the initial image I must be formed at the centre of curvature of the mirror. The radius of curvature of the mirror is twice its focal length and is therefore equal to 24 cm. The initial image I must therefore be at a distance of 24 cm from the mirror. Since the distance between the lens and the initial image I is 44 cm, the distance between the lens and the mirror must be (44 – 24) cm = 20 cm.

(3) An object is placed in front of a convex mirror of focal length f . The distance of the object is greater than f but less than 2f. The image of the object is

(a) in front of the mirror at distance less than f

(b) in front of the mirror at distance greater than f but less than 2f

(d) behind the mirror at distance less than f

(e) behind the mirror at distance greater than f but less than 2f

You may try drawing a ray diagram to locate the image, as shown in the adjoining figure. A ray of light proceeding towards the centre of curvature C of the mirror gets reflected from the mirror and retraces its path. Another ray proceeding parallel to the principal axis gets reflected from the mirror and proceeds as shown, as though it diverges away from the focus F of the mirror. The image is virtual and is located behind the mirror at distance less than f.

The correct option is (d).

[Note that the image will be located behind the mirror at distance less than f for all object distances].

(4) A student tries to burn a sheet of paper by focusing sun light on it using a converging lens of focal length f. The radius of the sun is R and its distance from the earth is D. The diameter of the image of the sun obtained on the sheet of paper is

(a) 2Rf/D

(b) 2RD/f

(d) RD/2f

(e) Rf/2D

Since the sun is far away from the lens, a real diminished image of the sun is formed at the focus of the lens.

[The student has to keep the sheet of paper at the focal plane of the lens for burning it].

The formation of the image is shown in the adjoining figure in which the radius of the image is shown as r.

From the similar triangles, we have

R/r = D/f

Or, r = Rf/D

The diameter of the image = 2r = 2Rf/D

Thursday, July 11, 2013

AP Physics C – Magnetic Fields – Answer to Free Response Practice Question on Magnetic Force on Moving Charges

A free response practice question on magnetic force on moving charges was posted on 8^th July 2013. As promised I give below a model answer for your benefit, along with the question:

The adjoining figure represents a cathode ray tube in a cathode ray oscilloscope. The voltage applied (with respect to the cathode) to the final accelerating anode is V and the distance between the anode and the fluorescent screen of the cathode ray tube is L. The mass of the electron is m and the magnitude of its charge is e. The cathode ray tube is so oriented that in the undeflected condition the electron beam is horizontal and is along the magnetic meridian, with the electrons traveling from south to north. The horizontal component of the earth’s magnetic field at the place is B_h. Now, answer the following questions:

(a) Calculate the velocity with which the electrons strike the fluorescent screen, assuming that the space between the accelerating anode and the fluorescent screen is devoid of electric fields and the electrons start from the cathode with negligible speed.

(b) The cathode ray tube is rotated through 90º about a verti8cal axis so that the electrons travel from east to west. In what direction will the spot of light (produced by the impact of the electron beam) on the fluorescent screen be deflected? Select the correct option from the following, by putting a tick mark (√) against one of the following:

Upwards………

Downwards………

Leftwards………

Rightwards……..

Justify your answer.

(c) If the cathode ray tube is rotated through 90º in the opposite direction, will the direction of deflection of the spot of light on the screen change? Justify your answer.

(d) Derive an expression for the specific charge (charge to mass ratio) of the electron in terms of the given data and the shift s of the electron spot on the screen, on rotating the cathode ray tube through 90º. Assume that the shift s is small compared to the distance between the accelerating anode and the fluorescent screen.

Answer:

(a) The electrons acquire kinetic energy under the action of the accelerating potential V applied on the accelerating anode. If the final velocity acquired by the electrons is v, we have

½ mv² = eV

Therefore v = √(2eV/m)

This is the velocity with which the electrons strike the fluorescent screen since their velocity after leaving the aperture of the accelerating anode remains unchanged in the field free space between the anode and the screen.

(b) Upwards.

Earth’s magnetic field is directed from south to north. Electrons traveling from east to west make a conventional current flowing from west to east (since the electrons are negatively charged).

[The current produced by electrons traveling from east to west has the same direction as the current produced by positive charges traveling from west to east]

Therefore, in order to apply Fleming’s left hand rule for fining the direction of the magnetic force on the electron beam, we have to hold the forefinger along the south to north direction and the middle finger along the west to east direction. The thumb will then point upwards.

(c) If the cathode ray tube is rotated through 90º in the opposite direction, the electron beam will be deflected downwards since the electrons travel from west to east.

(d) When the electrons travel parallel to the horizontal component (B_h)of earth’s magnetic field, there is no magnetic force on them. This follows from the expression for magnetic force F:

F =evB_hsinθ where θ is the angle between the magnetic field and the direction of the conventional current. Since θ = 180º, F = 0.

When the cathode ray tube is rotated through 90º, the electrons travel at right angles to the horizontal component (B_h) of earth’s magnetic field. In this case the magnetic force is maximum (F_max) and is given by

F_max = evB_h.

Since the deflection (s) produced is small (Fig.), we may assume that this force acts vertically throughout the path of the electrons.

The above vertical force produces a vertical acceleration and the vertical displacement s suffered by the electrons is given by the equation of uniformly accelerated motion,

s = ut + ½ at² ……….. (i)

where u is the initial vertical velocity (at the instant of leaving the accelerating anode), which is zero, t is the time taken by the electron to travel from the accelerating anode to the fluorescent screen and a is the vertical acceleration which is given by

a = F_max/m = evB_h/m

The time taken by the electron to travel the distance L between the anode and screen is given by

t = L/v

Substituting for u, a and t in Eq. (i) we have

s = 0 + ½ (evB_h/m) (L/v )² = ½ (e/m) (B_hL²/v)

The speed v of the electrons, as shown in part (a) is √(2eV/m)

Substituting for v we have

s = ½ (e/m) (B_hL²/√(2eV/m)

This gives e/m = 8s²V/ B_h²L⁴

Monday, July 8, 2013

AP Physics C – Magnetic Fields – Free Response Practice Question on Magnetic Force on Moving Charges

“One of the deep secrets of life is that all that is really worth the doing is what we do for others.”

– Lewis Carroll

The cathode ray oscilloscope is one of the most versatile instruments which is used as a test instrument in a variety of forms in almost all fields of scientific activity. Most of you may be familiar with this impressive instrument which is essentially a highly evacuated glass bulb containing an electron gun and a fluorescent screen, with provisions for deflecting the electron beam, usually by applying voltages across deflecting plates. I give below a free response question related to the cathode ray tube: