TS EAMCET 5-Aug-2021 Shift 2 Question Paper

Given,
Mass of particle be m,
Charge be q
Electric field be E
Initial velocity, ux=v.
∵‌F=qE=ma
⇒‌a=‌

qE

m

where, F is force and a is acceleration. Using equation of motion (s=ut+‌

1

2

at2) in the direction of electric field,
we get y=uyt+‌

1

2

ayt2
y=‌

1

2

(‌

qE

m

)t2
[∵uy=0]
⇒‌t2‌=‌

2ym

qE

‌‌ Since, ‌‌‌x‌=uxt
⇒‌x2‌=ux2t2⇒t2=x2∕ux2
Substituting in Eq. (i), we get
‌

2ym

qE

=‌

x2

ux2

=‌

x2

v2

⇒‌‌y=‌

x2qE

2mv2

⇒αx2=‌

x2qE

2mv2

⇒‌‌α=‌

qE

2mv2