Electrostatics

Concept:The condition $|V_1| = |V_2|$ gives a relation between distances from the two charges, leading to a geometric locus (line or circle).Explanation:Let $P(x,y)$ be any point in the $XY$ plane.Potential due to $Q_1 = q$ at $P_1(a,b)$: $V_1 = \frac{kq}{\sqrt{(x-a)^2+(y-b)^2}}$.Potential due to $Q_2 = mq$ at $P_2(ma,mb)$: $V_2 = \frac{k m q}{\sqrt{(x-ma)^2+(y-mb)^2}}$.Set $|V_1| = |V_2|$: $\frac{1}{\sqrt{(x-a)^2+(y-b)^2}} = \frac{|m|}{\sqrt{(x-ma)^2+(y-mb)^2}}$.Squaring gives: $(x-ma)^2+(y-mb)^2 = m^2\big[(x-a)^2+(y-b)^2\big]$.Now test each value of $m$ from the options.For $m = -1$:$(x+a)^2+(y+b)^2 = (-1)^2[(x-a)^2+(y-b)^2]$.Expand and simplify: $4ax+4by = 0 \Rightarrow ax+by=0$. (A line) → Option A is correct.For $m = 2$:$(x-2a)^2+(y-2b)^2 = 4[(x-a)^2+(y-b)^2]$.Simplify to $3x^2-4ax+3y^2-4by=0$.Complete squares: $(x-\frac{2}{3}a)^2+(y-\frac{2}{3}b)^2 = \frac{4}{9}(a^2+b^2)$.Centre $(\frac{2}{3}a,\frac{2}{3}b)$, radius $\frac{2}{3}\sqrt{a^2+b^2}$ → Option B is correct.For $m = -2$:$(x+2a)^2+(y+2b)^2 = 4[(x-a)^2+(y-b)^2]$.Simplifies to $x^2-4ax+y^2-4by=0$.Complete squares: $(x-2a)^2+(y-2b)^2 = 4(a^2+b^2)$.Centre $(2a,2b)$, radius $2\sqrt{a^2+b^2}$ → Option C is correct.For $m = -3$:$(x+3a)^2+(y+3b)^2 = 9[(x-a)^2+(y-b)^2]$.Simplifies to $x^2-3ax+y^2-3by=0$, which is a circle, not the line $3bx-3ay=0$ → Option D is incorrect.Answer:Options A, B, and C are correct.