Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsPotential & potential energyEasy2 minPYQ_2020

PhysicsEasysingle choice

A particle $A$ of charge $1 μC$ is held fixed at a point $P$ in free space. Another particle $B$ of same charge and mass $4 mg$ 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 $\frac{1}{4 {πε}_{0}} = 9 \times 10^{9} {Nm}^{2} C^{- 2}$ )

Options:

Answer:

Solution:

Loss in $PE =$ Gain in $KE$

$K \times 10^{- 6} \times 10^{- 6} [\frac{1}{10^{- 3}} - \frac{1}{9 \times 10^{- 3}}] = \frac{1}{2} {mv}^{2}$

$9 \times 10^{9} \times \frac{10^{- 6} \times 10^{- 6}}{10^{- 3}} \times \frac{8}{9} = \frac{1}{2} \times 4 \times 10^{- 6} \times V^{2}$

$V = 2 \times 10^{3} m / s$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Potential & potential energy

⏱ 2mℹ️ Source: PYQ_2020

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