In a certain region,a uniform electric field | vedclass

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(C) The particle has an initial velocity $v_0$ in the $yz$-plane at an angle $	heta$ with the $y$-axis. The components of velocity are $v_y = v_0 \co

Vedclass · Accepted Answer

$\frac{mv_0 \cos 	heta}{qE}$

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(C) The particle has an initial velocity $v_0$ in the $yz$-plane at an angle $	heta$ with the $y$-axis. The components of velocity are $v_y = v_0 \cos 	heta$ and $v_z = v_0 \sin 	heta$.The electric field $E$ is along the negative $y$-axis (opposite to the $y$-axis direction if we consider the standard setup where $E$ acts to oppose the motion). The force due to the electric field is $F_E = qE$,which acts to decelerate the component of velocity along the $y$-axis.The speed of the particle is $v = \sqrt{v_y(t)^2 + v_z(t)^2}$. The speed is minimum when the velocity component along the direction of the electric field becomes zero.Using the equation of motion $v_y(t) = v_y(0) - \frac{qE}{m}t$,we set $v_y(t) = 0$ to find the time $t$:$0 = v_0 \cos 	heta - \frac{qE}{m}t$$t = \frac{mv_0 \cos 	heta}{qE}$.

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