two equal point charges A and B are R distance apart. A third point charge placed on the prependicular bisector at a distance 'd' from the center will experience maximum electrostatic force when

1. d = R/2 X 1.414

2 d = R/ 1.414

3. d = R 1.414

4. d = 2 X 1.414 R


From the figure,Total force on third charge q=2×FsinθWhere,F=14π0q2r2   (Suppose the third charge shown on the top of triangle is 'r' distance apart from the second charge shown on the right.)So, f=2×14π0q2r2×dr    (sinθ=dr)=2×14π0×q2[d2+(R/2)2]3/2So, f(d)=Kdd2+R24) 3 For maximum force,f'(d)=0 (d2+R24) 3×1-d×32(d2+R24)1/2×2d=0or( d2+R24)-3d2=0or d=R22=R2×3.14

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