12 A proton is shot vertically upward with a speed of 200 x

12. A proton is shot vertically upward with a speed of 2.00 x 105 m/s in a downward-directed electric field of 500 NIC. How high will it rise?

Solution

Downward accleration or decleration of proton, a = qE/m

So at the maximum height its speed will become zero can the decleration is constant, so we can use third equation of kinematics

0 = u² - 2as, u =initial speed

s = u²/2a = mu²/2qE

Charge on a proton q = 1.6×10¹⁹ Coulomb,

Mass of proton m=1.67 × 10^-27 kg

s = (1.67 × 10^-27 )x(2x10⁵)²/2x1.6×10¹⁹ x500 = 41.75 centimeter