Question

Two point charges $+q$ and $-2q$ are placed at a distance $6\text{m}$ part on a horizontal plane. Find the locus of a point on this plane where the electric potential has zero value.

• A. Circle with radius $4\text{m}$ and centre $(-2,0)$.
• B. Circle with radius $4\text{m}$ and centre $(0,0)$.
• C. Circle with radius $2\text{m}$ and centre $(-2,0)$.
• D. Circle with radius $4\text{m}$ and centre $(0,-2)$.

Answer 1

The correct option is A Circle with radius $4\text{m}$ and centre $(-2,0)$.

Let a point $P$ be at a distance $r_1$ from point charge $+q$, and $r_2$ from point charge $-2q$ as shown in figure.


Let us assume that the net electric potential $\left(V_{net}\right)$ at $P$ due to charge $+q$ and $-2q$ is zero.
$V_1=$ Electric potential at $P$ due to charge $+q$.
$\Rightarrow V_1=\frac{kq}{r_1}\dots \left(i\right)$
$V_2=$ Electric potential at $P$ due to charge $-2q$.
$\Rightarrow V_2=\frac{-2kq}{r_2}\dots \left(ii\right)$

Net electric potential at $P$ is
$V_{net}=V_1+V_2=0$
Substitute values from equation ($i$) and ($ii$)
$\Rightarrow V_{net}=\frac{kq}{r_1}+\frac{(-2kq)}{r_2}=0$
$\Rightarrow kq\{\frac{1}{r_1}+\left(\frac{-2}{r_2}\right)\}=0$
$\Rightarrow \frac{1}{r_1}-\frac{2}{r_2}=0$
$\Rightarrow r_2=2r_1\dots \left(iii\right)$

It is given that charge $+q$ is located at the origin and $-2q$ is at $x=6\text{m}$

$\Rightarrow r_1^2=x^2+y^2$
and
$r_2^2={(6-x)}^2+y^2$

Substitute $r_1$ and $r_2$ in equation ($iii$)

$r_2^2=4r_1^2$
$\Rightarrow {(6-x)}^2+y^2=4\{x^2+y^2\}$
$\Rightarrow 36-12x+x^2+y^2=4x^2+4y^2$
$\Rightarrow 3x^2+3y^2+12x=36$
$\Rightarrow x^2+y^2+4x=12$
$\Rightarrow x^2+y^2+4x+4=12+4$
$\Rightarrow x^2+4x+4+y^2=16$
${(x+2)}^2+y^2={16}^2$

Comparing this equation with the equation of a circle having radius $r$ and centre at $(h,k)$ .

${(x-h)}^2+{(y-k)}^2=r^2$
$\Rightarrow r=4,(h,k)=(-2,0)$

So, the locus of point $P$ is a circle with radius $4\text{m}$ and centre $(-2,0)$

Hence, option (a) is correct.

$\begin{array}{c}\text{Why this question ?}\\ \text{Concept: In this question we learn}\\ \text{how to calculate the locus of}\\ \text{point having zero potential.}\end{array}$