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Electrostatics question

2019 · 10 Apr · Shift 2 · Q65
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Electrostatics question

2019 · 10 Apr · Shift 2 · Q65

JEE MainPhysicsElectrostaticsMCQ+4 / −1
In free space, a particle A of charge 1 μ\muμ C is held fixed at a point P. Another particle B of the same charge and mass 4 μ\muμ 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 14π∈0=9×109Nm2C−2]\left[ {Take\,{1 \over {4\pi { \in _0}}} = 9 \times {{10}^9}N{m^2}{C^{ - 2}}} \right][Take4π∈0​1​=9×109Nm2C−2]
  1. A
    1.0 m/s
  2. B
    6.32 ×\times× 104 m/s
  3. C
    2.0 ×\times× 103 m/s
  4. D
    1.5 ×\times× 102 m/s
View written solutionFree

Correct answer: B

  1. Given data
  • Charge on particle AAA: q1=1 μC=10−6 Cq_1 = 1\,\mu C = 10^{-6}\,Cq1​=1μC=10−6C
  • Charge on particle BBB: q2=1 μC=10−6 Cq_2 = 1\,\mu C = 10^{-6}\,Cq2​=1μC=10−6C
  • Mass of particle BBB: m=4 μg=4×10−9 kgm = 4\,\mu g = 4 \times 10^{-9}\,kgm=4μg=4×10−9kg
  • Initial distance: r1=1 mm=10−3 mr_1 = 1\,mm = 10^{-3}\,mr1​=1mm=10−3m
  • Final distance: r2=9 mm=9×10−3 mr_2 = 9\,mm = 9 \times 10^{-3}\,mr2​=9mm=9×10−3m
  • Coulomb constant: k=14πε0=9×109 Nm2C−2k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\,Nm^2C^{-2}k=4πε0​1​=9×109Nm2C−2

Since both charges are positive, particle BBB is repelled and moves away from AAA.

  1. Apply conservation of energy

Initially, particle BBB is at rest, so initial kinetic energy is zero.

Electrostatic potential energy: U=kq1q2rU = \frac{k q_1 q_2}{r}U=rkq1​q2​​

So, kq1q2r1=kq1q2r2+12mv2\frac{k q_1 q_2}{r_1} = \frac{k q_1 q_2}{r_2} + \frac{1}{2}mv^2r1​kq1​q2​​=r2​kq1​q2​​+21​mv2

Therefore, 12mv2=kq1q2(1r1−1r2)\frac{1}{2}mv^2 = k q_1 q_2\left(\frac{1}{r_1} - \frac{1}{r_2}\right)21​mv2=kq1​q2​(r1​1​−r2​1​)

  1. Substitute values

First, kq1q2=9×109×10−6×10−6=9×10−3k q_1 q_2 = 9 \times 10^9 \times 10^{-6} \times 10^{-6} = 9 \times 10^{-3}kq1​q2​=9×109×10−6×10−6=9×10−3

Now, 1r1−1r2=110−3−19×10−3=1000−10009\frac{1}{r_1} - \frac{1}{r_2} = \frac{1}{10^{-3}} - \frac{1}{9\times10^{-3}} = 1000 - \frac{1000}{9}r1​1​−r2​1​=10−31​−9×10−31​=1000−91000​

=9000−10009=80009= \frac{9000 - 1000}{9} = \frac{8000}{9}=99000−1000​=98000​

Hence, 12mv2=9×10−3×80009=8\frac{1}{2}mv^2 = 9\times10^{-3} \times \frac{8000}{9} = 821​mv2=9×10−3×98000​=8

So, 12(4×10−9)v2=8\frac{1}{2}(4\times10^{-9})v^2 = 821​(4×10−9)v2=8

2×10−9v2=82\times10^{-9} v^2 = 82×10−9v2=8

v2=82×10−9=4×109v^2 = \frac{8}{2\times10^{-9}} = 4\times10^9v2=2×10−98​=4×109

v=4×109=2×104.5v = \sqrt{4\times10^9} = 2\times10^{4.5}v=4×109​=2×104.5

v=2×109=2×31622.8≈6.32×104 m/sv = 2\times \sqrt{10^9} = 2\times 31622.8 \approx 6.32\times10^4\,m/sv=2×109​=2×31622.8≈6.32×104m/s

  1. Match with options

This corresponds to:

Option B: 6.32×104 m/s6.32 \times 10^4\,m/s6.32×104m/s

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