Protons are projected with an initial speed v_1 = 9.72 km/s from a field-free re

Question

Protons are projected with an initial speed v_1 = 9.72 km/s from a field-free region through a plane and into a region where a uniform electric field Evector = -720 N/C is present above the plane as shown in the figure. The initial velocity vector of the protons makes an angle theta with the plane. The protons are to hit a target that lies at a horizontal distance of R = 1.16 mm from the point where the protons cross the plane and enter the electric field. We wish to find the angle theta at which the protons must pass through the plane to strike the target. Find the two possible values of the angle theta. 18.9 degree or 71.1 degree 28.9 degree or 61.1 degree 38.9 degree or 51.1 degree 36.9 degree or 53.1 degree Cannot be determined

Explanation / Answer

E = -720j N/C
vi = 9.72 km/s
a = F/m = qE/m = 7.2*10^10 m/s/s
R = 1.16*10^-3 m
theta = ?
R = v^2*sin(2*theta)/g
theta = 31.06 deg or 58.934 deg

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