An electron is released from a distance of 4 m from a stationary point charge $20 nC$ . What will be the speed of the electron when it is 2 m away from the point charge?
[Charge of electron $= 1.6 \times 10^{- 19} C$ , mass of electron $= 9 \times 10^{- 31} kg, \frac{1}{4 π ϵ} = 9 \times 10^{9}$ S.I unit]

Question

An electron is released from a distance of 4 m from a stationary point charge

20 nC

. What will be the speed of the electron when it is 2 m away from the point charge?
[Charge of electron

= 1.6 \times 10^{- 19} C

, mass of electron

= 9 \times 10^{- 31} kg, \frac{1}{4 π ϵ} = 9 \times 10^{9}

S.I unit]

Accepted Answer

Correct answer: B. (b) By law of conservation of mechanical energy 0- 44 ϵ_0 20 10^-9 1.6 10^-194 = -14 ϵ_0 20 10^-9 1.6 10^-192+ 12 mV^2 ⇒ 9 10^9 32 10^-284+ 9 10^9 32 10^-282 = 12 9.1 10^-31 V^2 ⇒-7.2 10^-18+14.4 10^-18=4.55 10^-31 V^2 ⇒-7.2 10^-18=4.55 10^-31 V^2 ⇒V=3.977 10^6 ⇒V=4 10^6 m/s

Solution

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