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GATE 2016 EE – Question 48

Electromagnetic Fields · Coulomb's Law, Electric Field Intensity, Electric Flux Density, Gauss's Law, Divergence · 2 marks · Multiple choice

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. $x^2 + y^2 = 2$
  4. $(x + 2)^2 + y^2 = 16$

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Correct answer: (D) $(x + 2)^2 + y^2 = 16$

Explanation

The potential is zero where $\frac{q}{r_1} = \frac{2q}{r_2}$, so $r_2 = 2r_1$. Squaring the distances: $(x - 6)^2 + y^2 = 4(x^2 + y^2)$. Expanding gives $x^2 - 12x + 36 + y^2 = 4x^2 + 4y^2$, so $3x^2 + 12x + 3y^2 = 36$ and $x^2 + 4x + y^2 = 12$. Completing the square, $(x + 2)^2 + y^2 = 16$.