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(C) The electric field is given by $E = 2x^2 - 4$. For the charge to cross the origin from infinity with minimum velocity,it must just reach the point

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The kinetic energy at $x = \sqrt{2} \, m$ must be zero.

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(C) The electric field is given by $E = 2x^2 - 4$. For the charge to cross the origin from infinity with minimum velocity,it must just reach the point where the electric field becomes zero,as the field opposes the motion.Setting $E = 0$:$2x^2 - 4 = 0$$x^2 = 2$$x = \sqrt{2} \, m$ (considering the region of interest).At $x = \sqrt{2} \, m$,the electric field changes sign. Since the charge is released from infinity (where potential is zero) and moves towards the origin,it experiences a retarding force until it reaches $x = \sqrt{2} \, m$. For the minimum velocity condition,the kinetic energy must be zero at the point where the electric field is zero,which is $x = \sqrt{2} \, m$.

In a region,the electric field varies as $E = 2x^2 - 4$,where $x$ is the distance in $m$ from the origin along the $x$-axis. $A$ positive charge of $1 \,\mu C$ is released with minimum velocity from infinity to cross the origin. Then:

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