Question

A proton of mass m_p = 1.67 \times 10^{-27}~kgm ​p ​​ =1.67×10 ​−27 ​​ kg and a charge of q_p = 1.60 \times 10^{-19} ~Cq ​p ​​ =1.60×10 ​−19 ​​ C is moving through vacuum at a constant velocity of 10,000 ~m/s10,000 m/s directly to the east when it enters a region of uniform electric field that points to the south with a magnitude of | \vec{E} | = ∣ ​E ​⃗ ​​ ∣=2.35e+3~N/C N/C . The region of uniform electric field is 5 mm wide in the east-west direction. How far (in meters) will the proton have been deflected towards the south by the time it exits the region of uniform electric field. You may neglect the effects of friction and gravity, and assume that the electric field is zero outside the specified region.

Answer 1

Initial velocity towards south u=0,

​​​​​​acceleration is only south direction. No acceleration in East direction.

In m d=28.15 ×10^(-3) m