JEE PYQ: Electrostatics Question 16
Question 16 - 2020 (02 Sep Shift 1)
A charged particle (mass $m$ and charge $q$) moves along $X$ axis with velocity $V_0$. When it passes through the origin it enters a region having uniform electric field $\vec{E} = -E\hat{j}$ which extends upto $x = d$. Equation of path of electron in the region $x > d$ is:
(1) $y = \frac{qEd}{mV_0^2}(x - d)$
(2) $y = \frac{qEd}{mV_0^2}\left(\frac{d}{2} - x\right)$
(3) $y = \frac{qEd}{mV_0^2}x$
(4) $y = \frac{qEd^2}{mV_0^2}x$
Show Answer
Answer: (2)
Solution
Inside field ($0 < x < d$): $y_0 = \frac{1}{2}\frac{qE}{m}\left(\frac{d}{V_0}\right)^2 = \frac{qEd^2}{2mV_0^2}$. Slope at exit: $\tan\alpha = \frac{qEd}{mV_0^2}$. For $x > d$: straight line $y = -\frac{qEd}{mV_0^2}x + c$. Using boundary condition at $x = d$: $y = \frac{qEd}{mV_0^2}\left(\frac{d}{2} - x\right)$.
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