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/(4π ₀) = 9 × 10⁹N m² C⁻² ]

Electrostatics: JEE Main 2019 Physics question

JEE Main 2019 · 10 Apr, Evening shift · Q7007

JEE Main 201910 Apr, Evening shiftPhysicsMediumElectrostatics167× since 2002

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[ {Take\,{1 \over {4\pi { \in _0}}} = 9 \times {{10}^9}N{m^2}{C^{ - 2}}} \right]

JEE Main 2019 Electrostatics Question with Solution (10 Apr Evening shift, Q7007) | JEE Arc

JEE Main 2019 · 10 Apr, Evening shift · Q7007

JEE Main 201910 Apr, Evening shiftPhysicsMediumElectrostatics167× since 2002

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[ {Take\,{1 \over {4\pi { \in _0}}} = 9 \times {{10}^9}N{m^2}{C^{ - 2}}} \right]