Categories > Electrostatics > electric-field-potential > An electron moving with the speed $5 \times {10^6}$ per sec is shooted parallel to the electric field of intensity $1 \times {10^3}N/C$. Field is responsible for the retardation of motion of electron. Now evaluate the distance travelled by the electron before coming to rest for an instant (mass of $e = 9 \times {10^{ - 31}}Kg.$ charge $= 1.6 \times {10^{ - 19}}C)$

Electrostatics

Question: An electron moving with the speed $5 \times {10^6}$ per sec is shooted parallel to the electric field of intensity $1 \times {10^3}N/C$. Field is responsible for the retardation of motion of electron. Now evaluate the distance travelled by the electron before coming to rest for an instant (mass of $e = 9 \times {10^{ - 31}}Kg.$ charge $= 1.6 \times {10^{ - 19}}C)$


1) 7 m
2) 0.7 mm
3) 7 cm
4) 0.7 cm
Solution: Explanation: Explanation
electric-field-potential

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