There is a uniform electric field of strength $10^{3} \, V/m$ along y-axis. A body of mass 1g and charge 10 –6 C is projected into the field from origin along the positive x-axis with a velocity 10m/s. Its speed in m/s after 10s is (Neglect gravitation)
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Body moves along the parabolic path.
For vertical motion : By using v = u + at
⇒ ⇒ $v_y = 0 + \frac{QE}{m} . t = \frac{10^{-6} \times 10^{3}}{10^{-3}} \times 10 = 10 \, m/sec$
For horizontal motion – It’s horizontal velocity remains the same i.e. after 10 sec, horizontal velocity of body v x = 10 m/sec.
Velocity after 10 sec $v = \sqrt{v_x^2 + v_y^2}$ $= 10 \sqrt{2} \, m/sec$
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