Animated Solution for Physics - Electrostatics: 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:
Select Answer:
* Multiple Correct
Visualized Solution
CoordinateSetup
Let P(x,y) be the point where ∣V1∣=∣V2∣.
ElectricPotential
V1=r1Kq
V2=r2K(mq)
EquatingPotentials
∣V1∣=∣V2∣⇒r1K∣q∣=r2K∣mq∣
⇒r11=r2∣m∣
DistanceFormula
r22=m2r12
⇒(x−ma)2+(y−mb)2=m2[(x−a)2+(y−b)2]
ExpandingSquares
x2−2max+m2a2+y2−2mby+m2b2=m2(x2−2ax+a2+y2−2by+b2)
SimplifyingEquation
x2(1−m2)+y2(1−m2)−2max(1−m)−2mby(1−m)=0
GeneralLocusEquation
Divide by (1−m) since m=1:
(1+m)(x2+y2)=2m(ax+by)
CheckingOptionA
For m=−1:
0⋅(x2+y2)=−2(ax+by)
⇒ax+by=0
CheckingOptionB
For m=2:
3(x2+y2)=4(ax+by)
⇒x2+y2−34ax−34by=0
CirclePropertiesform=2
Center: (32a,32b)
Radius: (32a)2+(32b)2=32a2+b2
CheckingOptionC
For m=−2:
−(x2+y2)=−4(ax+by)
⇒x2+y2−4ax−4by=0
Center: (2a,2b), Radius: 2a2+b2
CheckingOptionD
For m=−3:
−2(x2+y2)=−6(ax+by)
⇒x2+y2−3ax−3by=0
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The Sigma Insight: Electric Potential and Potential Difference
Solution Diagram
Analyzing the Setup
Imagine an XY plane where we have two point charges. The first charge, Q1=q, is placed at point P1(a,b). The second charge, Q2=mq, is placed at point P2(ma,mb). We are tasked with finding the locus of a point P(x,y) such that the magnitude of the electric potential due to Q1 is equal to the magnitude of the electric potential due to Q2.
The electric potential V at a distance r from a point charge Q is given by the formula V=rKQ. Therefore, the potentials at point P due to Q1 and Q2 are:
V1=r1Kq
V2=r2K(mq)
where r1 and r2 are the distances from P to P1 and P2, respectively.
The Master Equation
The problem states that ∣V1∣=∣V2∣. Equating the magnitudes, we get:
r1K∣q∣=r2K∣mq∣
Canceling out the common terms K and ∣q∣, we are left with:
r11=r2∣m∣
To eliminate the square roots that will arise from the distance formula, we square both sides:
r22=m2r12
Now, we substitute the expressions for r12 and r22 using the distance formula:
(x−ma)2+(y−mb)2=m2[(x−a)2+(y−b)2]
Expanding both sides carefully, we get:
x2−2max+m2a2+y2−2mby+m2b2=m2(x2−2ax+a2+y2−2by+b2)
Notice that the terms m2a2 and m2b2 appear on both sides and cancel out. Grouping the remaining terms, we obtain:
x2(1−m2)+y2(1−m2)−2max(1−m)−2mby(1−m)=0
Since the problem specifies that $m
eq 1$, we can safely divide the entire equation by (1−m). This simplifies our expression to the master equation:
(1+m)(x2+y2)=2m(ax+by)
Evaluating the Options
Now, we can test each option by substituting the given value of m into our master equation.
Checking Option A (m=−1):
Substituting m=−1 into the master equation yields:
0⋅(x2+y2)=−2(ax+by)
ax+by=0
This is the equation of a straight line passing through the origin. Thus, Option A is correct.
Checking Option B (m=2):
Substituting m=2 gives:
3(x2+y2)=4(ax+by)
Dividing by 3, we get the standard form of a circle:
x2+y2−34ax−34by=0
The center of this circle is (32a,32b), and its radius is (32a)2+(32b)2=32a2+b2. This perfectly matches Option B.
Checking Option C (m=−2):
Substituting m=−2 gives:
−1(x2+y2)=−4(ax+by)
x2+y2−4ax−4by=0
The center of this circle is (2a,2b), and its radius is (2a)2+(2b)2=2a2+b2. This matches Option C.
Checking Option D (m=−3):
Substituting m=−3 gives:
−2(x2+y2)=−6(ax+by)
x2+y2−3ax−3by=0
This represents a circle, not the straight line 3bx+3ay=0 as claimed in Option D. Therefore, Option D is incorrect.
