The following four questions which appeared in the All India Engineering/Architecture Entrance Examination 2009 (AIEEE 2009) will be beneficial to most of the entrance test takers. Those who prepare for AP Physics C exam and physics GRE may take special note of question no.2. Here are the questions with solution:
(1) A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electric force on Q is zero, then Q/q equals
(1) – 1/√2
(2) – 2√2
(3) –1 
(4) 1
The electric force FQQ (Fig.) on Q because of the charge Q at the opposite corner is repulsive and is given by
F1 = (1/4πε0)(Q2 /2a2) since the distance between the charges Q and Q is 2×a/√2 = a√2
The electric forces FQq and FQq (Fig.) on Q because of the charges q and q at the adjacent corners are perpendicular to each other. Each force has magnitude given by
FQq = (1/4πε0)(Qq/a2).
Their resultant F2 has magnitude √2 times the above value:
F2 = (1/4πε0)(√2 Qq/a2).
Since the net force on Q is zero, we have
F1 + F2 = 0
Therefore, (1/4πε0)(Q2 /2a2) + (1/4πε0)(√2 Qq/a2) = 0
This gives Q/q = – 2√2
[The charges Q and q must be of opposite sign so that the force between Q and q is attractive to ensure that F1 and F2 are opposite in directions and the net force on Q is zero as given in the question].
(2) Let P(r) = Qr/πR4 be the charge density distribution for a solid sphere of radius R and total charge Q. For a point ‘p’ inside the sphere at distance r1 from the centre of the sphere, the magnitude of electric field is
(1) Qr12 /3πε0 R4
(2) 0
(3) Q/4πε0 r12
(4) Qr12 /4πε0 R4
Charge dQ contained in the spherical shell of radius r and thickness dr is (Qr/πR4)(4πr2dr) = (4Q/R4) r3dr
Charge Q1 contained in the spherical volume of radius r1 is therefore given by
Q1 = 0 ∫r1 (4Q/R4) r3dr = (4Q/R4)(r14/4) = Qr14 /R4
If E is the electric field at distance r1 from the centre of the sphere, we have from Gauss theorem,
4πr12E = Qr14 /R4ε0 from which
E = Qr12 /4πε0 R4
[Remember that the charges distributed with spherical symmetry outside the point p will produce zero field at p and the field at p is indeed given correctly by the above expression].
(3) Two points P and Q are maintained at the potentials of 10 V and – 4 V, respectively. The work done in moving 100 electrons from P to Q is
(1) 2.24×10–16 J
(2) – 9.60×10–17 J
(3) 9.60×10–17 J
(4) – 2.24×10–16 J
This is a simple question. The potential difference ∆V between P and Q is 10 V – (– 4 V) = 14 V. Since the electrons are negatively charged, external work (positive work) has to be done to move them from the higher potential point P to the lower potential point Q.
The work done = q∆V = (100×1.6×10–19)×14 J = 2.24×10–16 J
(4) This question contains Statement-1 and statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement 1 : For a charged particle moving from point P to point Q, the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q.
Statement 2 : The net work done by a conservative force on an object moving along a closed loop is zero.
(1) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statment-1.
(2) Statment-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.
(3) Statement-1 is false, Statement-2 is true.
(4) Statement-1 is true, Statement-2 is false
Electrostatic field is conservative and the net work done by an electrostatic field on a charged particle is dependent only on the initial and final positions of the charged particle. (It is independent of the path connecting the initial and final positions).
The correct option is (1).
