Two point charges -Q and +Q / sqrt{3} are pla | vedclass

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(C) Let a point $P(x, y)$ be on the equipotential circle where the potential $V = 0$.The potential at point $P$ due to charges $-Q$ at $(0,0)$ and $+Q

Vedclass · Accepted Answer

$1.73, 3$

Vedclass · Answer

(C) Let a point $P(x, y)$ be on the equipotential circle where the potential $V = 0$.The potential at point $P$ due to charges $-Q$ at $(0,0)$ and $+Q / \sqrt{3}$ at $(2,0)$ is given by:$V_P = \frac{k(-Q)}{r_1} + \frac{k(Q / \sqrt{3})}{r_2} = 0$where $r_1 = \sqrt{x^2 + y^2}$ and $r_2 = \sqrt{(x - 2)^2 + y^2}$.Rearranging the equation:$\frac{kQ}{r_1} = \frac{kQ}{\sqrt{3} r_2} \implies \frac{1}{\sqrt{x^2 + y^2}} = \frac{1}{\sqrt{3} \sqrt{(x - 2)^2 + y^2}}$Squaring both sides:$3((x - 2)^2 + y^2) = x^2 + y^2$$3(x^2 - 4x + 4 + y^2) = x^2 + y^2$$3x^2 - 12x + 12 + 3y^2 = x^2 + y^2$$2x^2 - 12x + 2y^2 + 12 = 0$Dividing by $2$:$x^2 - 6x + y^2 + 6 = 0$Completing the square for $x$:$(x^2 - 6x + 9) + y^2 = -6 + 9$$(x - 3)^2 + y^2 = 3 = (\sqrt{3})^2$Comparing this with the standard circle equation $(x - b)^2 + y^2 = R^2$,we get:$b = 3$ and $R = \sqrt{3} \approx 1.73$.Thus,$R = 1.73$ and $b = 3$.

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