Monday, October 09, 2006

Multiple Choice Questions on Electrostatics:

The following M.C.Q. appearered in the question paper of A.I.E.E.E.2004 (It may appear to be time consuming on the first reading, but it is simple):Two spherical conductors B and C having equal radii and carrying equal charges on them repel each other with a force F when kept apart at some distance. A third spherical conductor A having the same radius as that of B but uncharged is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is
(a) F/4 (b) 3F/4 (c) F/8 (d) 3F/8
If Q is the initial charge on B and C and ‘r’ is the distance between their centres, the initial repulsive force between B and C is F= (1/4πε0)(Q2/ r2). When the uncharged sphere A is brought in contact with B, they share the charge Q and each will have a charge Q/2 since they are identical spheres. When the sphere A is then brought in contact with C they will share the total charge Q + Q/2 = 3Q/2 equally and each will have a charge 3Q/4. On removing the sphere A we have sphere B with charge Q/2 and the sphere C with charge 3Q/4. The new repulsive force between B and C is evidently (1/4πε0)(Q/2)(3Q/4)/r2 = (1/4πε0)3Q2/8r2 = 3F/8 [Option (d)].
Take a special note of the following M.C.Q. which often finds a place in entrance test papers:
A charge ‘q’ is placed at the centre of the line joining two equal point charges, each equal to +Q. This system of three charges will be in equilibrium if 'q' is equal to
(a) +Q (b) +Q/2 (c) –Q/2 (d) +Q/4 (e) –Q/4

+Q____ q_____+Q

The force on the charge ‘q’ placed at the centre will be zero irrespective of the sign of ‘q’. For the system to be in equilibrium, the net force on the charge +Q should be zero. Therefore, (1/4πε0)[Q2 /d2 + Qq/(d/2)2] = 0 where ‘d’ is the separation between the charges Q&Q. From this we get q = -Q/4 [Option (e)].The following M.C.Q is quite simple because of the symmetry in the arrangement of charges. Similar questions can be seen in entrance test papers:Equal charges of value -Q each are arranged at the eight vertices of a non-conducting skeleton cube of side ‘a’. If a point charge +Q is placed at the centre of the cube, the electrostatic force exerted by the eight negative charges on the positive charge at the centre is
(a) Q2/3πε0a2 (b) 4Q2/3πε0a2 (c) 8Q2/3πε0a2 (d) 16Q2/3πε0a2 (e) zero

The net electrostatic force on the charge Q at the centre is zero since the force on Q due to each negative charge is balanced by the force due to the negative charge at the diagonally opposite corner of the cube.
The following M.C.Q. which appeared in the A.I.E.E.E.2004 question paper is worth noting:

Four charges equal to –Q are placed at the four corners of a square and a charge ‘q’ is at the centre.If the system is in equilibrium, the value of ‘q’ is
(a) –Q(1+2√2)/4 (b) Q(1+2√2)/4 (c) –Q(1+2√2)/2 (d) Q(1+2√2)/2
The system will be in equilibrium if the attractive force on the charge –Q at a corner due to the central charge ‘q’ is balanced by the net repulsive force on the same charge –Q due to the remaining three charges (-Q each) at the remaining three corners. Hence we have (if ‘a’ is the side of the square),
(1/4πε0) [Qq/(√2 a/2)2] = (1/4πε0) [Q2/(√2 a)2] + √2 × [(1/4πε0) ×(Q2/a2)].
The term on LHS is the attractive force between –Q and q which are separated by the distance half of √2 a. The first term on RHS is the repulsive force between –Q and the diagonally opposite charge –Q (which are separated by √2 a). The second term on RHS is the net repulsive force between –Q and the remaining two charges (-Q each).
Hence we obtain 4q = Q + Q(1+2√2) from which q = Q(1+2√2)/4 [Option (b)].

The following question was asked at the Kerala Medical Entrance test of 2006:Three identical charges each of 2μC are placed at the vertices of a triangle ABC. If AB+AC=12cm and AB.AC=32cm2, the potential energy of the charge at A is
(a) 1.53J (b) 5.31J (c) 3.15J (d) 1.35J (e) 3.51J

The electrostatic potential energy of the charge at A ia (1/4πε0)(Q.Q/AB + Q.Q/AC) since the charges are of the same value, Q=2μC. Since AB.AC=32, AB = 32/AC. Substituting this in the equation, AB + AC = 12, we obtain 32/AC + AC = 12. Rearranging, (AC)2 – 12AC + 32 = 0, which yields AC = 8 or 4. Since AB.AC = 32 cm2, if AC = 8 cm, AB = 4 cm and vice versa.
Therefore potential energy = (1/4πε0)[(4×10-12)/(4×10-2) +(4×10-12)/(8×10-2)] joule.
Since 1/4πε0 =9×109, the potential energy works out to 1.35 J, given in option (d).
Now consider the following M.C.Q.:
Two isolated copper spheres of radii 3cm and 6cm carry equal positive charges of 30 units each. If they are connected by a thin copper wire and then the wire is removed, what will be the charge on the smaller sphere?
(a) 60 units (b) 40 units (c) 30 units (d) 20 units (e) 10 units
The total charge (Q1+Q2) in the system is 60 units. Since the connecting wire is thin, its capacitance can be neglected. Even though the potentials of the spheres are different initially, their potentials will become the same when they are connected by the copper wire.
The capacitance of a spherical conductor being 4πε0 r, the charges on the spheres will be directly proportional to their radii, since Q = CV and V is the same. Therefore we have Q1/Q2 = 3/6 = 1/2 and Q1+Q2 = 60. So, Q1 = 60×1/3 = 20 units. The potential of the spheres will change after removing the connecting wire; but the charges on them will not change. So, the correct option is (d).
Let us discuss another M.C.Q.:
An infinite number of charges each equal to –q coulomb are placed on a straight line at x = 1m, 2m, 4m, 8m,16m,…….. What will be the potential at x = 0 due to these charges?
(a) infinite (b) 2q/4πε0 (c) –q/4πε0 (d) –2q/4πε0 (e) zero
The potential is certainly negative since a negative charge will produce negative potential only. You should supply the sign of the charge in the expression for the net potential. The potential at x = 0 is given by, V = (-q/4πε0)(1+ ½ + ¼ + 1/8 + 1/16 + ………). Since the infinite series yields a value equal to 2, the answer is –2q/4πε0.
Very simple questions are often likely to mislead even very bright students. Be careful. Here is a very simple question:Nine negative charges, each of magnitude Q are arranged symmetrically along the circumference of a circle of radius R. The electric field at the centre of the circle is
(a) (1/4πε0) Q/R2 (b) (1/4πε0) (9Q/R2) (c) -(1/4πε0) (9Q/R2)
(d) -(1/4πε0) (9Q/R2) cos(2π/9) (e) zero
When you place a positive test charge at the centre of the circle, it will experience zero net force since it is pulled equally by the nine charges arranged symmetrically all around. Therefore, the electric field at the centre is zero.
You will find all the posts on Electrostatics in this site by clicking on the label ‘electrostatics’ below this post.


  1. in question 4 the second term at RHS should be- 2x(kQQ/aa), why hv multiplied it with under root 2
    pls explain

  2. Forces are vectors and you can’t add them as scalars. Two forces F and F acting at right angles will produce a resultant force (√2)F