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Electrostatics question

2022 · 29 Jun · Shift 2 · Q46
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Electrostatics question

2022 · 29 Jun · Shift 2 · Q46

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Two point charges Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is :
  1. A
    x = d
  2. B
    x=d2x = {d \over 2}x=2d​
  3. C
    x=d2x = {d \over {\sqrt 2 }}x=2​d​
  4. D
    x=d22x = {d \over {2\sqrt 2 }}x=22​d​
View written solutionFree

Correct answer: D

  1. Set up the geometry

Place the two equal charges QQQ at coordinates (−d2,0)\left(-\dfrac d2,0\right)(−2d​,0) and (d2,0)\left(\dfrac d2,0\right)(2d​,0).

The third charge qqq is placed on the perpendicular bisector at

(0,x).(0,x).(0,x).

The distance of qqq from each charge QQQ is

r=x2+(d2)2.r=\sqrt{x^2+\left(\frac d2\right)^2}.r=x2+(2d​)2​.
  1. Force due to one charge

Magnitude of force on qqq due to one charge QQQ:

F1=kQqr2.F_1=\frac{kQq}{r^2}.F1​=r2kQq​.

By symmetry, the horizontal components cancel, and the vertical components add.

If θ\thetaθ is the angle made by the line joining a charge to qqq with the perpendicular bisector, then

cos⁡θ=xr.\cos\theta=\frac{x}{r}.cosθ=rx​.

So vertical component of force due to one charge is

F1cos⁡θ=kQqr2⋅xr=kQq xr3.F_1\cos\theta=\frac{kQq}{r^2}\cdot \frac{x}{r}=\frac{kQq\,x}{r^3}.F1​cosθ=r2kQq​⋅rx​=r3kQqx​.

Hence net force on qqq is

F=2⋅kQq xr3=2kQq x(x2+d24)3/2.F=2\cdot \frac{kQq\,x}{r^3} =\frac{2kQq\,x}{\left(x^2+\frac{d^2}{4}\right)^{3/2}}.F=2⋅r3kQqx​=(x2+4d2​)3/22kQqx​.
  1. Maximize the force

We need to maximize

f(x)=x(x2+d24)3/2.f(x)=\frac{x}{\left(x^2+\frac{d^2}{4}\right)^{3/2}}.f(x)=(x2+4d2​)3/2x​.

Differentiate:

f′(x)=(x2+d24)−3/2−3x2(x2+d24)5/2.f'(x)=\left(x^2+\frac{d^2}{4}\right)^{-3/2} -\frac{3x^2}{\left(x^2+\frac{d^2}{4}\right)^{5/2}}.f′(x)=(x2+4d2​)−3/2−(x2+4d2​)5/23x2​.

Taking common factor:

f′(x)=(x2+d24)−3x2(x2+d24)5/2=d24−2x2(x2+d24)5/2.f'(x)=\frac{\left(x^2+\frac{d^2}{4}\right)-3x^2}{\left(x^2+\frac{d^2}{4}\right)^{5/2}} =\frac{\frac{d^2}{4}-2x^2}{\left(x^2+\frac{d^2}{4}\right)^{5/2}}.f′(x)=(x2+4d2​)5/2(x2+4d2​)−3x2​=(x2+4d2​)5/24d2​−2x2​.

For maximum force,

f′(x)=0⇒d24−2x2=0.f'(x)=0 \Rightarrow \frac{d^2}{4}-2x^2=0.f′(x)=0⇒4d2​−2x2=0.

Thus,

2x2=d24⇒x2=d28⇒x=d22.2x^2=\frac{d^2}{4} \Rightarrow x^2=\frac{d^2}{8} \Rightarrow x=\frac{d}{2\sqrt2}.2x2=4d2​⇒x2=8d2​⇒x=22​d​.
  1. Match with options
x=d22x=\frac{d}{2\sqrt2}x=22​d​

which is Option D.

  1. Comparison with stored answer

Stored correct answer: D

This matches the derived result.

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