1. Electric Charges and Fields
hard

A uniform electric field of $10\,N / C$ is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy $0.5\,eV$. The length of each plate is $10\,cm$. The angle $(\theta)$ of deviation of the path of electron as it comes out of the field is  $.........$(in degree).

A

$44$

B

$43$

C

$42$

D

$45$

(JEE MAIN-2023)

Solution

$0.5\,e =\frac{1}{2}\,mv _{ x }^2 \Rightarrow v _{ x }=\sqrt{\frac{ e }{ m }}$

Along $x L =v_x t=\sqrt{\frac{ e }{ m }} t$

Along y $v _{ y }=\frac{e e }{ m } t$

$\operatorname{dividing} \frac{ v _{ y }}{ L }= E \sqrt{\frac{ e }{ m }}= Ev _{ x }$

$\Rightarrow \operatorname{Tan} \theta=\frac{ v _{ y }}{ v _{ x }}= E \times L =10 \times 0.1=1$

$\theta=45^{\circ}$

Standard 12
Physics

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