TS EAMCET 18-July-2022 Shift 2 Solved Paper

Given, σ=8.85×10−6C∕m2
Kinetic energy of electron, K=8×10−17J
We know that, electric field due to metal plate is given by
E=‌

σ

ε0

∴ Force on electron in this electric field,
F=−eE=−‌

eσ

ε0

∴ Acceleration on electron, a=‌

F

m

=−‌

eσ

mε0

Let u be the initial velocity of electron, then
˙v2=u2+2as
but v=0
∴‌‌0=u2+2(‌

−eσ

mε0

)s
⇒‌‌s=‌

u2ε0m

2eσ

‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
but ‌‌K=‌

1

2

mu2
⇒‌‌u2=‌

2K

m

‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
∴ From Eqs. (i) and (ii), we get
s‌=‌

2Kε0m

2meσ

=‌

Kε0

eσ

s‌=‌

8×10−17×8.85×10−12

1.6×10−19×8.85×10−6

‌=5×10−4m=0.5‌mm