Question

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

1) d= R/2*sq.root(2)

2) d=R/sq.root(2)

3) d=R*sq.root(2)

4) d=R*2*sq.root(2)

Answer 1

answer : option (1)

explanationTwo equal point charges A and B are R distance apart. A third point charge placed on the perpendicular bisector at a distance 'd' from the centre.

Let charge on each particle is q and charge on third particle is Q.

so, net force acting on Q, F = 2fcos$\theta$ , where f = 4kqQ/(R² + 4y²)

and $cos\theta$ = 2y/√(R² + 4y²)

so, net force , F = 8kqQy/(R² + 4y²)^(3/2)

now, differentiating F with respect to y,

dF/dy = 8kqQ/(R² + 4y²)³ [(R² + 4y²)^(3/2) - y × 3/2 × 8y√(R² + 4y²)]

= 8kqQ/(R² + 4y²)^5/2 [ R² + 4y² - 12y² ]

= 8kqQ/(R² + 4y²)^5/2 [R² - 8y²]

dF/dy = 0, at y = ±R/2√2

but only y = R/2√2 , d²F/dy² < 0 and force will be maximum.

so, at y = R/2√2 electrostatic force will be maximum.

Answer 2

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