There is a uniform electric field of strength | vedclass

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(C) The body moves along a parabolic path because it experiences a constant force in the $y$-direction while moving with constant velocity in the $x$-

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$10\sqrt{2}$

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(C) The body moves along a parabolic path because it experiences a constant force in the $y$-direction while moving with constant velocity in the $x$-direction.Given: $E = 10^3 \, V/m$,$m = 1 \, g = 10^{-3} \, kg$,$q = 10^{-6} \, C$,$u_x = 10 \, m/s$,$u_y = 0$,$t = 10 \, s$.For vertical motion ($y$-axis): The force $F_y = qE$ causes an acceleration $a_y = \frac{qE}{m} = \frac{10^{-6} 	imes 10^3}{10^{-3}} = 1 \, m/s^2$.Using $v_y = u_y + a_y t$,we get $v_y = 0 + (1)(10) = 10 \, m/s$.For horizontal motion ($x$-axis): Since there is no force in the $x$-direction,the velocity remains constant: $v_x = u_x = 10 \, m/s$.The resultant speed $v$ after $10 \, s$ is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2} \, m/s$.

There is a uniform electric field of strength $10^3 \, V/m$ along the $y$-axis. $A$ body of mass $1 \, g$ and charge $10^{-6} \, C$ is projected into the field from the origin along the positive $x$-axis with a velocity of $10 \, m/s$. Its speed in $m/s$ after $10 \, s$ is (Neglect gravitation).

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