1. Electric Charges and Fields
medium

A particle of mass $\mathrm{m}$ and charge $\mathrm{q}$ is released from rest in a uniform electric field. If there is no other force on the particle, the dependence of its speed $v$ on the distance $x$ travelled by it is correctly given by (graphs are schematic and not drawn to scale)

A
B
C
D
(JEE MAIN-2020)

Solution

$\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}$

$\mathrm{v}^{2}=0+2\left(\frac{\mathrm{qE}}{\mathrm{m}}\right) \mathrm{x}$

$\mathrm{v}^{2}=\frac{2 \mathrm{q} \mathrm{E}}{\mathrm{m}} \mathrm{x}$

Standard 12
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.