In free space, a particle A of charge 1 μ C is held fixed at a point P. Another particle B of the same charge and mass 4 μ g is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is: [Take (1)/(4πε₀) = 9 × 10⁹ Nm²C⁻²] (a) 1.0 m/s (b) 6.32 × 104 m/s (c) 2.0 × 103 m/s (d) 1.5 × 102 m/s

Answer: (b). KE gained = PE lost: 1/2mv² = kq²(1/r₁ - 1/r₂) = 9×10⁹×10⁻¹²(1000 - 1000/9) = 8 J v = √2×84×10⁻⁹ = √(4×10⁹) = 6.32×10⁴ m/s So the answer is (b) 6.32×10⁴ m/s.

In free space, a particle A of charge 1 μ C is held fixed at a point… | JEE Main 2019 (10 Apr, Shift 2)

Q51JEE Main 2019 · 10 APR · SHIFT 2Medium

Electrostatic Potential Energy

In free space, a particle A of charge 1 $\mu$ C is held fixed at a point P. Another particle B of the same charge and mass 4 $\mu$ g is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is: $\left[\text{Take } \frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\text{ Nm}^2\text{C}^{-2}\right]$

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Answer: (b)

KE gained $=$ PE lost: $\frac{1}{2}mv^2 = kq^2\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$
$= 9\times10^9\times10^{-12}\left(1000 - \frac{1000}{9}\right) = 8$ J
$v = \sqrt{\frac{2\times8}{4\times10^{-9}}} = \sqrt{4\times10^9} = 6.32\times10^4$ m/s
So the answer is (b) $6.32\times10^4$ m/s.

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Electrostatic Potential Energy question from JEE Main 2019 (10 Apr, Shift 2)

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