If two electric charges $q$ and $- 2 q$ are placed at distance $6 a$ apart, then locus of point in the plane of charges, where electric potential is zero is: (Take charge $q$ at origin and $- 2 q$ lies on positive x-axis)

x^{2} + y^{2} + 4 a x - 6 a^{2} = 0

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x^{2} + y^{2} + 4 a x - 12 a^{2} = 0

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x^{2} + y^{2} + 2 a x - 12 a^{2} = 0

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x^{2} + y^{2} - 4 a x + 12 a^{2} = 0

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Solution

The correct option is B $x^{2} + y^{2} + 4 a x - 12 a^{2} = 0$
$\frac{k q}{\sqrt{x^{2} + y^{2}}} + \frac{- 2 q}{\sqrt{{(x - 6 a)}^{2} + y^{2}}} = 0$ (sum of forces has to be zero)

$\Rightarrow$ ${(x - 6 a)}^{2} + y^{2} = 4 (x^{2} + y^{2})$

$\Rightarrow$ $x^{2} + y^{2} + 4 a x - 12 a^{2} = 0$

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If two electric charges q and −2q are placed at distance 6a apart, then locus of point in the plane of charges, where electric potential is zero is: (Take charge q at origin and −2q lies on positive x-axis)

The correct option is B x2+y2+4ax−12a2=0kq√x2+y2+−2q√(x−6a)2+y2=0 (sum of forces has to be zero)⇒ (x−6a)2+y2=4(x2+y2)⇒ x2+y2+4ax−12a2=0

If two electric charges $q$ and $- 2 q$ are placed at distance $6 a$ apart, then locus of point in the plane of charges, where electric potential is zero is: (Take charge $q$ at origin and $- 2 q$ lies on positive x-axis)

The correct option is B $x^{2} + y^{2} + 4 a x - 12 a^{2} = 0$
$\frac{k q}{\sqrt{x^{2} + y^{2}}} + \frac{- 2 q}{\sqrt{{(x - 6 a)}^{2} + y^{2}}} = 0$ (sum of forces has to be zero)

$\Rightarrow$ ${(x - 6 a)}^{2} + y^{2} = 4 (x^{2} + y^{2})$

$\Rightarrow$ $x^{2} + y^{2} + 4 a x - 12 a^{2} = 0$