two equal point charges A and B are R distance apart A third point charge placed on the prependicular - Physics - Electric Charges And Fields

Question

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

Jyoti Pant · Accepted Answer

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From the figure,Total force on third charge
q=2×FsinθWhere,F=14π∈0q2r2 (Suppose the third
charge shown on the top of triangle is 'r' distance apart from the
second charge shown on the right )So,
f=2×14π∈0q2r2×dr
(sinθ=dr)=2×14π∈0×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

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

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