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(D) The electron is projected in a uniform electric field directed from $A$ to $B$. Since the electron is negatively charged,it experiences a force $F

Vedclass · Accepted Answer

$1600$

Vedclass · Answer

(D) The electron is projected in a uniform electric field directed from $A$ to $B$. Since the electron is negatively charged,it experiences a force $F = q_e E$ in the direction opposite to the electric field,i.e.,towards plate $A$. However,the problem states the electron is projected from $A$ and we want to ensure it does not hit $B$. This implies the electric field must be directed from $B$ to $A$ or the electron is positively charged. Given the context of the trajectory shown,the electron experiences a downward acceleration $a = \frac{q_e E}{m_e}$.For the electron not to hit plate $B$,its maximum vertical displacement $h_{\max}$ must be less than or equal to the separation distance $d = 4 \ cm = 0.04 \ m$.The vertical component of initial velocity is $u_y = v \sin 30^{\circ} = \frac{v}{2}$.The acceleration is $a = \frac{q_e E}{m_e} = \frac{1.6 	imes 10^{-19} 	imes 45.5}{9.1 	imes 10^{-31}} = \frac{72.8 	imes 10^{-19}}{9.1 	imes 10^{-31}} = 8 	imes 10^{12} \ m \ s^{-2}$.Using the kinematic equation $v_y^2 = u_y^2 - 2ah$,at maximum height $v_y = 0$:$0 = (\frac{v}{2})^2 - 2ah_{\max} \implies h_{\max} = \frac{v^2}{8a}$.Setting $h_{\max} = 0.04 \ m$:$0.04 = \frac{v^2}{8 	imes 8 	imes 10^{12}}$$v^2 = 0.04 	imes 64 	imes 10^{12} = 2.56 	imes 10^{12}$$v = \sqrt{2.56 	imes 10^{12}} = 1.6 	imes 10^6 \ m \ s^{-1} = 1600 \ km \ s^{-1}$.

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