Two charges Q_1 = q and Q_2 = mq are placed at the points P_1(a, b) and P_2(ma, mb), respectively, in the XY plane, where a, b 0 and m 0, 1. If V_1 is the potential at a point in the XY plane due to charge Q_1 and V_2 is the potential at that point due to charge Q_2. Correct statement(s) for the points at which |V_1| = |V_2| is/are:
C. For m = -2, the locus of these points is a circle of radius 2 a^2 + b^2 centered at (2a, 2b)
The potential at a point (x, y) due to charge Q_1 is V_1 = 1 4 _0 q (x-a)^2 + (y-b)^2 . The potential at (x, y) due to charge Q_2 is V_2 = 1 4 _0 mq (x-ma)^2 + (y-mb)^2 . Given |V_1| = |V_2|, we have: 1 (x-a)^2 + (y-b)^2 = m^2 (x-ma)^2 + (y-mb)^2 Cross-multiplying and expanding both sides: (x-ma)^2 + (y-mb)^2 = m^2[(x-a)^2 + (y-b)^2] x^2 + m^2a^2 - 2max + y^2 + m^2b^2 - 2mby = m^2(x^2 + a^2 - 2ax + y^2 + b^2 - 2by) x^2 + y^2 - 2m(ax+by) + m^2(a^2+b^2) = m^2(x^2+y^2) - 2m^2(ax+by) + m^2(a^2+b^2) Canceling m^2(a^2+b^2) from both sides and rearranging: (m^2-1)(x^2+y^2) - 2m(m-1)(ax+by) = 0 Since m 1, dividing by m-1 gives the general equation of the locus: (m+1)(x^2+y^2) - 2m(ax+by) = 0 For m = -1: 0 - 2(-1)(ax+by) = 0 ax+by = 0. This represents a straight line. For m = 2: 3(x^2+y^2) - 4(ax+by) = 0 x^2+y^2 - 4 3 ax - 4 3 by = 0. This represents a circle with center ( 2 3 a, 2 3 b ) and radius ( 2 3 a )^2 + ( 2 3 b )^2 = 2 3 a^2+b^2 . For m = -2: (-1)(x^2+y^2) - 2(-2)(ax+by) = 0 x^2+y^2 - 4ax - 4by = 0. This represents a circle with center (2a, 2b) and radius (2a)^2 + (2b)^2 = 2 a^2+b^2 . For m = -3: (-2)(x^2+y^2) - 2(-3)(ax+by) = 0 x^2+y^2 - 3ax - 3by = 0. This represents a circle, not a straight line. Answer: For m = -1, locus of these points is ax + by = 0.; For m = 2, the locus of these points is a circle of radius 2 3 a^2 + b^2 centered at ( 2 3 a, 2 3 b ); For m = -2, the locus of these points is a circle of radius 2 a^2 + b^2 centered at (2a, 2b)
Related: Physics — Electrostatics · All PYQ Banks