Electrostatic Potential and Capacitance · Question 5

Two charges Q1=q and Q2=mq are placed at the points P1(a,b) and P2(ma,mb), respectively, in the XY plane, where a,b 0 and m 0,1. If V1 is the potential at a point in the XY plane due to charge Q1 and V2 is the potential at that point due to charge Q2. Correct statement(s) for the points at which |V1|=|V2| is/are:

1. For m=-1, locus of these points is ax+by=0.
2. For m=2, the locus of these points is a circle of radius 23 a2+b2 centered at ( 23a, 23b )
3. For m=-2, the locus of these points is a circle of radius 2 a2+b2 centered at (2a,2b)
4. For m=-3, locus of these points is 3bx+3ay=0.

The difference in the capacitance (in F) between the metal plates at t=0 and that at t=500 s is (8-n) 0, where 0 is the permittivity of free space. The value of n is:

A positive point charge of 10-8 C is kept at a distance of 20 cm from the center of a neutral conducting sphere of radius 10 cm. The sphere is then grounded and the charge on the sphere is measured. The grounding is then removed and subsequently the point charge is moved by a distance of 10 cm further away from the center of the sphere along the radial direction. Taking 14 0=9 109 Nm2/C2 (where 0 is the permittivity of free space), which of the following statements is/are correct:

1. Before the grounding, the electrostatic potential of the sphere is 450 V.
2. Charge flowing from the sphere to the ground because of grounding is 5 10-9 C.
3. After the grounding is removed, the charge on the sphere is -5 10-9 C.
4. The final electrostatic potential of the sphere is 300 V.

Four identical thin, square metal sheets, S1, S2, S3 and S4, each of side a are kept parallel to each other with equal distance d ( a) between them, as shown in the figure. Let C0= 0 a2/d, where 0 is the permittivity of free space. [figure] Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

1. P → 3; Q → 2; R → 4; S → 5
2. P → 2; Q → 3; R → 2; S → 1
3. P → 3; Q → 2; R → 4; S → 1
4. P → 3; Q → 2; R → 2; S → 5