JEE Challenger
More from Electrostatic Potential and Capacitance
PhyMedAdv 2026 • Paper 2

Locus of Points of Equal Electric Potential from Two Point Charges

Two charges Q1=qQ_1 = q and Q2=mqQ_2 = mq are placed at the points P1(a,b)P_1(a, b) and P2(ma,mb)P_2 (ma, mb), respectively, in the XYXY plane, where a,b≠0a, b \neq 0 and m≠0,1m \neq 0, 1. If V1V_1 is the potential at a point in the XYXY plane due to charge Q1Q_1 and V2V_2 is the potential at that point due to charge Q2Q_2. Correct statement(s) for the points at which ∣V1∣=∣V2∣|V_1| = |V_2| is/are:

Options

A

For m=−1m = -1, locus of these points is ax+by=0ax + by = 0.

Correct
B

For m=2m = 2, the locus of these points is a circle of radius 23a2+b2\frac{2}{3}\sqrt{a^2 + b^2} centered at (23a,23b)\left(\frac{2}{3}a, \frac{2}{3}b\right)

Correct
C

For m=−2m = -2, the locus of these points is a circle of radius 2a2+b22\sqrt{a^2 + b^2} centered at (2a,2b)(2a, 2b)

Correct
D

For m=−3m = -3, locus of these points is 3bx+3ay=03bx + 3ay = 0.

Topics & Concepts

Electrostatic Potential and CapacitanceElectric Potential

Step-by-Step Solution

To find the locus of points (x,y)(x, y) in the XYXY plane where the magnitudes of the electric potentials due to charges Q1Q_1 and Q2Q_2 are equal, i.e., ∣V1∣=∣V2∣|V_1| = |V_2|, we begin by writing the expressions for V1V_1 and V2V_2:

V1=14πε0Q1(x−a)2+(y−b)2=14πε0q(x−a)2+(y−b)2V_1 = \frac{1}{4\pi\varepsilon_0} \frac{Q_1}{\sqrt{(x-a)^2 + (y-b)^2}} = \frac{1}{4\pi\varepsilon_0} \frac{q}{\sqrt{(x-a)^2 + (y-b)^2}}

V2=14πε0Q2(x−ma)2+(y−mb)2=14πε0mq(x−ma)2+(y−mb)2V_2 = \frac{1}{4\pi\varepsilon_0} \frac{Q_2}{\sqrt{(x-ma)^2 + (y-mb)^2}} = \frac{1}{4\pi\varepsilon_0} \frac{mq}{\sqrt{(x-ma)^2 + (y-mb)^2}}

Given ∣V1∣=∣V2∣|V_1| = |V_2|, we have:

∣q∣(x−a)2+(y−b)2=∣mq∣(x−ma)2+(y−mb)2\frac{|q|}{\sqrt{(x-a)^2 + (y-b)^2}} = \frac{|mq|}{\sqrt{(x-ma)^2 + (y-mb)^2}}

Squaring both sides and simplifying yields:

(x−ma)2+(y−mb)2=m2[(x−a)2+(y−b)2](x - ma)^2 + (y - mb)^2 = m^2 \left[ (x - a)^2 + (y - b)^2 \right]

Expanding both sides:

x2−2max+m2a2+y2−2mby+m2b2=m2(x2−2ax+a2+y2−2by+b2)x^2 - 2max + m^2a^2 + y^2 - 2mby + m^2b^2 = m^2(x^2 - 2ax + a^2 + y^2 - 2by + b^2)

x2+y2−2m(ax+by)+m2(a2+b2)=m2(x2+y2)−2m2(ax+by)+m2(a2+b2)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 m2(a2+b2)m^2(a^2 + b^2) from both sides:

(m2−1)(x2+y2)−2m(m−1)(ax+by)=0(m^2 - 1)(x^2 + y^2) - 2m(m - 1)(ax + by) = 0

Since m≠1m \neq 1, we can divide the entire equation by (m−1)(m - 1):

(m+1)(x2+y2)−2m(ax+by)=0— (1)(m + 1)(x^2 + y^2) - 2m(ax + by) = 0 \quad \text{--- (1)}

Analysis of the Options:

  1. For m=−1m = -1: Substituting m=−1m = -1 into Equation (1): (−1+1)(x2+y2)−2(−1)(ax+by)=0(-1 + 1)(x^2 + y^2) - 2(-1)(ax + by) = 0 2(ax+by)=0  ⟹  ax+by=02(ax + by) = 0 \implies ax + by = 0 This represents a straight line passing through the origin. Thus, Option A is correct.

  2. For m=2m = 2: Substituting m=2m = 2 into Equation (1): (2+1)(x2+y2)−2(2)(ax+by)=0(2 + 1)(x^2 + y^2) - 2(2)(ax + by) = 0 3(x2+y2)−4ax−4by=03(x^2 + y^2) - 4ax - 4by = 0 x2+y2−43ax−43by=0x^2 + y^2 - \frac{4}{3}ax - \frac{4}{3}by = 0 This is the equation of a circle of the form x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0, where: Center=(−g,−f)=(23a,23b)\text{Center} = (-g, -f) = \left(\frac{2}{3}a, \frac{2}{3}b\right) Radius=g2+f2−c=(23a)2+(23b)2−0=23a2+b2\text{Radius} = \sqrt{g^2 + f^2 - c} = \sqrt{\left(\frac{2}{3}a\right)^2 + \left(\frac{2}{3}b\right)^2 - 0} = \frac{2}{3}\sqrt{a^2 + b^2} Thus, Option B is correct.

  3. For m=−2m = -2: Substituting m=−2m = -2 into Equation (1): (−2+1)(x2+y2)−2(−2)(ax+by)=0(-2 + 1)(x^2 + y^2) - 2(-2)(ax + by) = 0 −(x2+y2)+4ax+4by=0-(x^2 + y^2) + 4ax + 4by = 0 x2+y2−4ax−4by=0x^2 + y^2 - 4ax - 4by = 0 This is the equation of a circle with: Center=(2a,2b)\text{Center} = (2a, 2b) Radius=(2a)2+(2b)2=2a2+b2\text{Radius} = \sqrt{(2a)^2 + (2b)^2} = 2\sqrt{a^2 + b^2} Thus, Option C is correct.

  4. For m=−3m = -3: Substituting m=−3m = -3 into Equation (1): (−3+1)(x2+y2)−2(−3)(ax+by)=0(-3 + 1)(x^2 + y^2) - 2(-3)(ax + by) = 0 −2(x2+y2)+6ax+6by=0  ⟹  x2+y2−3ax−3by=0-2(x^2 + y^2) + 6ax + 6by = 0 \implies x^2 + y^2 - 3ax - 3by = 0 This represents a circle, not the straight line 3bx+3ay=03bx + 3ay = 0. Thus, Option D is incorrect.

Correct Options: A, B, C