Two Kerala Engineering Entrance 2007 Questions from Electrostatics

When you calculate the electrostatic potential energy of a pair of charged particles, it is enough to consider the energy of one particle in the electric field of the other. If you have three charged particles (as for example A, B & C), you will have three pairs (AB, AC & BC) and the total energy will be the algebraic sum of the energies due to these three pairs. But, if you have four charged particles (A, B, C & D), you will have six pairs (AB, AC, AD, BC, CD & BD).

Consider the following MCQ which appeared in KEAM (Engineering) 2007 question paper:

The electrostatic potential energy between proton and electron separated by a distance 1 Ǻ is

(a) 13.6 eV (b) 27.2 eV (c) 14.4 eV (d) 1.44 eV (e) 28.8 eV

This is a simple question since we have just one pair of charges. The electrostatic potential energy (EPE) of the system is (1/4πε₀)(q₁q₂/r) where q₁ and q₂ are the charges of proton and electron.

Therefore, EPE = (1/4πε₀)(1.6×10^–19)²/10^–10 joule

= 9×10⁹×1.6×10^–19/10^–10 electron volt

= 14.4 eV.

[When you find the EPE of more than two charged particles, you will have to be careful to consider the sign of the charges since the total EPE is the algebraic sum of the EPE’s of all the pairs of charges].

The following simple MCQ also is from the KEAM (Engineering) 2007 question paper:

The work done in moving an alpha particle between two points having potential difference 25 volt is

(a) 8×10^–18 J (b) 8×10^–19 J (c) 8×10^–20 J (d) 8×10^–10 J (e) 4×10^–18 J

This question is intended to test your basic understanding of the work done in moving a charge in an electric field. An electric field will exist only if there is a potential gradient and the wok done in moving a charge is the product of the charge and the potential difference. The alpha particle has two protons in it, each carrying positive charge of 1.6×10^–19 coulomb.

Therefore work done = (2×1.6×10^–19)×25 joule = 8×10^–18 J

Imagination is more important than knowledge.

– Albert Einstein   

Two Questions (MCQ) on Electric Potential

(1) An infinite number of point charges each equal to +Q coulomb are arranged at random around a point P such that the distances of the charges from the point P are 1m, 2m, 4m, 8m, 16m,…….etc... The electric potential at P is

(a) zero (b) infinite (c) negligibly small

(d) Q/2πε₀ (e) Q/4πε₀

Note that the electric potential is a scalar quantity. Therefore, the direction of the location of the charge does not matter and the potentials simply add up. The resultant potential (V) at P is given by

V = (Q/4πε₀) × [(1/1) + (1/2) + (1/4) + (1/8) + ………]

The infinite series within the square bracket yields a value equal to 2 so that V = Q/2πε_0.

(2) A thin spherical conducting shell of radius R has a charge +Q. Another point charge –q is placed at the centre of the shell. The electrostatic potential at a point P distant R/2 from the centre of the shell is

(a) (Q/4πε₀R) – (2q/4πε₀R) (b) (q/4πε₀R) – (2Q/4πε₀R)

(c) – (2q/4πε₀R) (d) (Q/4πε₀R) (e) zero
The electrostatic potential at any point within the shell due to the charge Q on the shell is constant and is equal to Q/4πε₀R.

The potential at distance R/2 due to the charge –q placed at the centre of the shell is – q/4πε₀(R/2) = – 2q/4πε₀R.

Therefore, the net potential (V) at the point P distant R/2 from the centre of the shell is given by

V = (Q/4πε₀R) – (2q/4πε₀R), given in option (a).

[Note that positive charges will produce positive potential where as negative charges will produce negative potential].

You can find more posts on electrostatics by clicking on the label electrostatics below this post or on the left side of this page.

You will find similar multiple choice questions with solution at physicsplus also.

Two KEAM (Engineering) 2007 Multiple Choice Questions on Gravitation

The following questions which appeared in KEAM 2007 question paper are typical and of the type repeatedly asked in various entrance examinations:

(1) The change in potential energy when a body of mass ‘m’ is raised to a height nR from earth’s surface is (R = radius of the earth)

(a) mgR(n/n–1) (b) mgR (c) mgR(n/n+1)

(d) mgR(n²/n² +1) (e) mgR/n

This question appears often in entrance question papers. There may be slight change in the wording (as for example, “what is the work done in lifting a body of mass m, from the earth’s surface, through a height nR?”).

Gravitational potential energy of a mass ‘m’ at a height ‘h’ is given by

U = – GMm/(R+h) where M is the mass of the earth and G is the Gravitational constant.

Since h = nR, the change in potential energy is

– GMm/(R+nR) – (–GMm/R) = (GMm/R)[1 – 1/(1+n)] = (GMm/R)[n/(1+n)]

Since g = GM/R², the change in potential energy becomes mgRn/(n+1), given in option (c).

(2) The escape velocity of a body on the surface of the earth is 11.2 km/s. If the mass of the earth is doubled and its radius is halved, the escape velocity becomes

(a) 5.6 km/s (b) 11.2 km/s (c) 22.4 km/s

(d) 44.8 km/s (e) 67.2 km/s

The escape velocity (v_e) of a body on the surface of the earth is given by

v_e = √(2GM/R)

Therefore we have √(2GM/R) = 11.2 km/s

If the mass (M) of the earth is doubled and its radius (R) is halved, the escape velocity becomes √(2×4GM/R) = 2×√(2GM/R) = 22.4 km/s.

Two Multiple Choice Questions on Work and Energy

The following MCQ which appeared in KEAM (Engineering) 2007 question paper is worth noting:

A particle is released from a height S. At a certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively

(a) S/4, 3gS/2 (b) S/4, [√(3gS)]/2 (c) S/2, [√(3gS)]/2

(d) S/4, √(3gS/2) (e) S/3, √(3gS/2)

Suppose the given conditions occur at the instant when the particle is at height ‘h’.

Potential energy at the height h = mgh.

Since the initial energy is completely potential and is equal to mgS, the kinetic energy at the height ‘h’ is equal to the loss of potential energy and is equal to mg(S–h).

Since the K.E. is three times the P.E., we have

mg(S–h) = 3mgh, from which h = S/4.

But, kinetic energy = ½ mv².

Therefore, ½ mv² = mg(S–h) = mg[S– (S/4)] = 3 mgS/4 so that

v =√(3gS/2)

Now, consider the following MCQ which is simple. But simple qiestions may cheat you some times. Here is the question:

A uniform chain of mass M and length L is placed on a table with 20% of its length overhanging from the edge of the table. The minimum work that has to be done to place the hanging portion on the table is

(a) MgL/5 (b) MgL/10 (c) MgL/20

(d) MgL/25 (e) MgL/50

The hanging portion has mass M/5 and therefore weight Mg/5. The length of the hanging portion is L/5 and hence the centre of gravity of the hanging portion is at a depth L/10 from the surface of the table. The entire weight of the hanging portion can be imagined to act through its centre of gravity. When you pull the hanging portion up to keep it on the table, you are moving the point of application of the force (equal to the magnitude of the the weight Mg/5) through a distance L/10 along the direction of the force.

Therefore, the work done is (Mg/5)×(L/10) = MgL/50