Two point charges -Q and +Q / √3 are placed in the x y -plane at the origin (0,0) and a point (2,0) , respectively, as shown in the figure. This…

Question

Two point charges -Q and +Q / √3 are placed in the x y -plane at the origin (0,0) and a point (2,0) , respectively, as shown in the figure. This results in an equipotential circle of radius R and potential V=0 in the x y -plane with its center at (b, 0) . All lengths are measured in meters. The value of b is ________ meter.

MARKS Content Team · Accepted Answer

Correct answer: 3. Let Ph,k be be a general point with potential to be zero. V_1+V_2=0 (as potential given =0) K−Q√(h^2+k^2)+KQ√3√(h−2^2+k^2)=0 1/(√3√(h−2^2+k^2))=1√(h^2+k^2) 3h−2^2+k^2=h^2+k^2 3h^2−4h+4+k^2=h^2+k^2 2h^2+2k^2−12h+12=0 h^2+k^2−6h+6=0 h^2−6h+6+k^2=0 h^2−6h+9+k^2=3 h−3^2+k^2=3 Standard equation of circle is given by x−x1^2+y−y1^2=R^2 ∴ x_1=3, y_1=0 and R=√3 Hence, R=√3 m and value of b=3 m

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