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Two electric charges q and -2q are placed at (0,0) and (6,0) on the x-y plane. The equation of the zero equipotential curve in the x-y plane is
1. x = - 2
2. y = 2
3. x2 + y2 = 2
4. (x + 2)2 + y2 = 16

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Correct Answer - Option 4 : (x + 2)2 + y2 = 16

From the given question, potential for q and -2q

\({V_q} = \frac{q}{{4\pi\epsilon \sqrt {{x^2} + {y^2}} }}\)

\({V_{ - 2q}} = \frac{{ - 2q}}{{4\pi\epsilon \left( {\sqrt {{{\left( {x - 6} \right)}^2} + {y^2}} } \right)}}\)

For equipotential region in x-y plane,

\({V_{total}} = 0 = \frac{q}{{4\pi\epsilon \left( {\sqrt {{x^2} + {y^2}} } \right)}} + \frac{{ - 2q}}{{4\pi\epsilon \left( {\sqrt {{{\left( {x - 6} \right)}^2} + {y^2}} } \right)}}\)

\(\begin{array}{l} \sqrt {{{\left( {x - 6} \right)}^2} + {y^2}} = 2\left( {\sqrt {{x^2} + {y^2}} } \right)\\ 3{x^2}\; + \;3{y^2}\; + 12x\; = \;36\\ {x^2}\; + \;{y^2}\; + \;4x\; = \;12\\ {\left( {x\; + \;2} \right)^2}\; + \;{y^2}\; = \;16 \end{array}\)

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