A charged particle (mass m and charge q) move | vedclass

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(A) $1$. Inside the region $0 \le x \le d$,the particle experiences a constant force $F_{y} = -qE$ in the negative $y$-direction. The acceleration is

Vedclass · Accepted Answer

$y = \frac{qEd}{mV_{0}^{2}} \left( \frac{d}{2} - x \right)$

Vedclass · Answer

(A) $1$. Inside the region $0 \le x \le d$,the particle experiences a constant force $F_{y} = -qE$ in the negative $y$-direction. The acceleration is $a_{y} = -\frac{qE}{m}$.$2$. The time taken to reach $x = d$ is $t_{0} = \frac{d}{V_{0}}$.$3$. At $x = d$,the vertical displacement is $y_{0} = \frac{1}{2} a_{y} t_{0}^{2} = -\frac{1}{2} \frac{qE}{m} \left( \frac{d}{V_{0}} \right)^{2} = -\frac{qEd^{2}}{2mV_{0}^{2}}$.$4$. The velocity components at $x = d$ are $v_{x} = V_{0}$ and $v_{y} = a_{y} t_{0} = -\frac{qE}{m} \cdot \frac{d}{V_{0}} = -\frac{qEd}{mV_{0}}$.$5$. For $x > d$,there is no electric field,so the particle moves in a straight line with constant velocity. The slope of this line is $m_{slope} = \frac{v_{y}}{v_{x}} = \frac{-qEd/mV_{0}}{V_{0}} = -\frac{qEd}{mV_{0}^{2}}$.$6$. The equation of the line passing through $(d, y_{0})$ with slope $m_{slope}$ is $y - y_{0} = m_{slope} (x - d)$.$7$. Substituting the values: $y - \left( -\frac{qEd^{2}}{2mV_{0}^{2}} \right) = -\frac{qEd}{mV_{0}^{2}} (x - d)$.$8$. $y = -\frac{qEd}{mV_{0}^{2}} x + \frac{qEd^{2}}{mV_{0}^{2}} - \frac{qEd^{2}}{2mV_{0}^{2}} = -\frac{qEd}{mV_{0}^{2}} x + \frac{qEd^{2}}{2mV_{0}^{2}}$.$9$. Factoring out $\frac{qEd}{mV_{0}^{2}}$,we get $y = \frac{qEd}{mV_{0}^{2}} \left( \frac{d}{2} - x \right)$.

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