A charged particle (mass m and charge q) moves along X-axis with velocity V0. When it passes through the origin it enters a region having uniform electric field E = -Ej which extends upto x = d. Equation of path of electron in the region x > d is (a) y = (qEd)/(mV₀²)(x - d) (b) y = (qEd)/(mV₀²)(d/2 - x) (c) y = (qEd)/(mV₀²)x (d) y = (qEd²)/(mV₀²)x

Answer: (b). Inside the field ( 0 d the path is straight: y = yd - (qEd)/(mV₀²)(x - d) y = (qEd)/(mV₀²)(-d/2 - x + d) = (qEd)/(mV₀²)(d/2 - x) So the answer is (b).

A charged particle (mass m and charge q) moves along X-axis with… | JEE Main 2020 (2 Sep, Shift 1)

Q34JEE Main 2020 · 2 SEP · SHIFT 1Medium

Motion of Charges in an Electric Field

A charged particle (mass m and charge q)

moves along X-axis with velocity V0. 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

Show answer and solution

Answer: (b)

Inside the field ( $0 < x < d$ ): $a_y = -\frac{qE}{m}$ , time $t = \frac{d}{V_0}$ .
At $x = d$ : $y_d = -\frac{qEd^2}{2mV_0^2}$ and slope $\frac{dy}{dx} = \frac{v_y}{V_0} = -\frac{qEd}{mV_0^2}$ .
For $x > d$ the path is straight: $y = y_d - \frac{qEd}{mV_0^2}(x - d)$
$y = \frac{qEd}{mV_0^2}\left(-\frac{d}{2} - x + d\right) = \frac{qEd}{mV_0^2}\left(\frac{d}{2} - x\right)$
So the answer is (b).

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Motion of Charges in an Electric Field question from JEE Main 2020 (2 Sep, Shift 1)

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