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:

x= 3t +12qamt2y= 12qbmt2The above two equations come in the form of parabola.

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:

x= 3t +12qamt2y= 12qbmt2The above two equations come in the form of parabola.

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A point charge moving with velocity $\vec{v} = (3 m / s) \hat{i}$ is released in a uniform electric field $\vec{E} = (a \hat{i} + b \hat{j})$ . The subsequent trajectory of the particle will be:

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A parabola

A rectangular hyperbola

A straight line

An ellipse

answer is A.

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x= 3t + $\frac{1}{2}$ $\frac{q a}{m}$ $t^{2}$

y= $\frac{1}{2}$ $\frac{q b}{m}$ $t^{2}$

The above two equations come in the form of parabola.

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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 point charge moving with velocity $\vec{v} = (3 m / s) \hat{i}$ is released in a uniform electric field $\vec{E} = (a \hat{i} + b \hat{j})$ . The subsequent trajectory of the particle will be: