Fill in the blanks Protons are projected with an initial spe

Fill in the blanks. Protons are projected with an initial speed v_i = 9.75 times 10^3 m/s into a region where a uniform electric field E = (-71 0j) N/C is present, as shown in the figure below. The protons are to hit a target that lies at a horizontal distance of 1.18 mm from the point where the protons are launched. (a) Neglecting the effects of gravity, find the two projection angles theta that will result in a hit. Round your answers to three significant figures. Take the charge of the proton to be q = 1.6 times 10^-19 C, and the mass of the proton to be m_p = 1.67 times 10^-27 kg. degree (smallest angle), degree (largest angle) (b) Neglecting the effects of gravity, find the total time of flight for each trajectory. Round your answers to three significant figures. ns (for the smallest angle), ns (for the largest angle) v_i = 9.75 times 10^3 m/s

Solution

in electric field, proton will feel electric force.
Fe =q E

hence acceleration in vertical direction will be

a = q E / m = (1.609 x 10^-19 x 710)/(1.67 x 10^-27)

a = 6.828 x 10^10 m/s^2

in this question, range of this projectile have to be 1.18 mm.

Applying Range, R = v^2 sin(2@) / a

1.18 x 10^-3 = (9.75 x 10^3)^2 sin(2@) / (6.828 x 10^10 )

sin2@ = 0.848

2@ = 58 deg Or 122 deg

@ = 29 deg Or 61deg

(B)
time of flight, T = 2 v sin@ / a

T1 = (2 x 9.75 x 10^3 x ain29) / (6.828 x 10^10)

T1 = 1.385 x 10^-7 sec

T2 = (2 x 9.75 x 10^3 x sin61) / (6.828 x 10^10)

T2 = 2.5 x 10^-7 sec