JEE Main 2020 · PhysicsMultiple choiceDiagram basedMediumMulti-step

JEE Main 2 September 2020, Shift 1, Physics Q11

Question 11 of 75 in this shift, Physics question 11 of 25.

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 :

The figure, in words

X-Y axes with origin O; particle enters along +X with velocity V0; uniform field E pointing down (-Y) between x=0 and x=d (two vertical lines), no field beyond x=d.

(1) $y=\frac{qEd}{mV_0^{2}}\,x$
(2) $y=\frac{qEd^{2}}{mV_0^{2}}\,x$
(3) $y=\frac{qEd}{mV_0^{2}}\,(x-d)$
(4) $y=\frac{qEd}{mV_0^{2}}\left(\frac{d}{2}-x\right)$ Official answer

Official answer

Option 4

NTA final key (Sep 2020).

Chapter: Electric Charges and Fields
Topic: Electric Field
Idea tested: Charge Trajectory in Uniform Field

JEE Main 2 September 2020, Shift 1, Physics Q11

Same idea in other shifts