TS EAMCET 18-July-2022 Shift 1 Solved Paper

The given situation is shown below.

When electron reaches at point B , then change in its potential energy is equal to gain in kinetic energy. i.e. ∆U=K
⇒ PE at point B− PE at point A=‌

1

2

mev2
⇒‌‌‌

1

4πε0

‌

q1⋅q

r22

−‌

1

4πε0

⋅‌

q1q

r12

=‌

1

2

mev2
⇒‌‌‌

1

4πε0

[‌

eq

r22

−‌

eq

r12

]=‌

1

2

mev2
⇒‌‌9×109[‌

eq

22

−‌

eq

42

]=‌

1

2

mev2
⇒‌‌9×109(‌

3eq

16

)=‌

1

2

mev2
⇒‌‌v=√‌

3eq×9×109

8me

⇒‌‌v=√‌

3×1.6×10−19×2×10−8×9×109

8×9×10−31

=3.46×106
Which is nearest to 4×106m∕ s .