Question

A uniform electric field has a magnitude of 2.3x103 N/C. A proton initially at rest is accelerated in this field. What is the kinetic energy of the proton after it travels a distance of 2 cm in the field? Charge of a proton is +1.6x10-19 C.

Answer 1

v²=u²+2as (equation of motion for constant acceleration)

where v is final velocity, u is initial velocity , a is acceleration and s is displacement.

Here , a=(eE)/m_p

e=charge of proton=1.602*10^-19

mp=mass of proton=1.67*10^-27

E=electric field=2.3*10³ N/C

u=0 ,s=0.02m

So v=(2as)^.5 =93943.54273 m/s

K.E=0.5m_pv²=7.3692×10⁻¹⁸    Joules

So option (d) is correct.