Q.

A point charge moving with velocity v=(3m/s)i^ is released in a uniform electric field E=(ai^+bj^). The subsequent trajectory of the particle will be:

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a

A parabola

b

A rectangular hyperbola

c

A straight line

d

An ellipse

answer is A.

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Detailed Solution

x= 3t +12qamt2

y= 12qbmt2

The above two equations come in the form of parabola.

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