JEE Main 2022 · 29 Jun · Shift 2 · Physics Q33

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PhysicsSingle correct+4 / -1Question 33 of 90

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:

x=d

x=\dfrac d2

x=\dfrac d{\sqrt2}

x=\dfrac d{2\sqrt2}

Answer and worked solution

Official answer: D

Force. $F\propto\dfrac{x}{\left(x^2+d^2/4\right)^{3/2}}$ ; setting $dF/dx=0$ gives $x^2=\frac{d^2}8$ , i.e. $x=\dfrac d{2\sqrt2}$ .

(D)

Topics:Coulomb's law and superposition Electric field

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JEE Main 2022 · 29 Jun · Shift 2, Physics question 33