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(D) The electric field $E$ acts upwards. Since the charge $q$ is negative,the force $F = qE$ acts downwards. The acceleration of the particle is $a =

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$0.2$

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(D) The electric field $E$ acts upwards. Since the charge $q$ is negative,the force $F = qE$ acts downwards. The acceleration of the particle is $a = \frac{|q|E}{m} = \frac{1 	imes 10^{-6} 	imes 10^3}{2 	imes 10^{-3}} = 0.5 \ ms^{-2}$ downwards.For the particle not to hit the upper plate,its maximum vertical displacement $h_{\max}$ must be less than or equal to the plate separation $d = 2 \ cm = 0.02 \ m$.The vertical component of velocity is $u_y = u \sin 45^{\circ} = \frac{u}{\sqrt{2}}$.At maximum height,the vertical velocity becomes zero. Using $v_y^2 = u_y^2 - 2ah_{\max}$,we get $0 = (\frac{u}{\sqrt{2}})^2 - 2ah_{\max}$,so $h_{\max} = \frac{u^2}{4a}$.Setting $h_{\max} = 0.02 \ m$ and $a = 0.5 \ ms^{-2}$:$0.02 = \frac{u^2}{4 	imes 0.5} = \frac{u^2}{2}$$u^2 = 0.04 \implies u = 0.2 \ ms^{-1}$.

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