18.58 Applications of Electrostatics Points:4 An electron has an initial velocit

Question

18.58 Applications of Electrostatics Points:4 An electron has an initial velocity of 5.00x106 m/s in a uniform 2.00x105 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity, what is the direction of the electric field? OOpposite direction to its initial velocity O Same direction as its initial velocity O 90° to the right of its initial velocity (along the horizontal plane). O90o to the left of its initial velocity (along the horizontal plane Submit Answer Tries 0/5 How far does the electron travel before coming to rest? Submit Answer Tries 0/5 How long does it take the electron to come to rest? Submit Answer Tries 0/5 What is the electron's velocity when it returns to its starting point? Magnitude of the velocity Direction of the velocity is opposite to its initial velocity. the same direction as its initial velocity 900 to the right of its initial velocity (along the horizontal plane). 90° to the left of its initial velocity (along the horizontal plane) Subrmit Answer Tries 0/5

Explanation / Answer

A) Same direction as its initial velocity

B) Change in Kinetic Energy = Work Done

Therefore,

0.5*9.11*10^-31*(5*10^6)^2 = (2*10^5)*1.6*10^-19*s

s = 3.56*10^-4 m

C) As we know

v = u + at

Therefore,

t = v/a = (5*10^6)/((2*10^5*1.6*10^-19)/(9.11*10^-31)) = 1.42*10^-10 sec

D) As it returns to the starting point

Net Displacement = 0

As we know, v^2 - u^2 = 2as

Therefore,

As Displacement , s = 0

Therefore, v = - u

That is, Its velocity = - 5*10^6 m/sec

-ve sign means direction of velocity in the direction opposite to the inital velocity

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