PLEASE DON\'T COPY AND PASTE FROM ANOTHER ANSWER Protons are projected with an i

Question

PLEASE DON'T COPY AND PASTE FROM ANOTHER ANSWER

Protons are projected with an initial speed vi = 9.93 km/s from a field-free region through a plane and into a region where a uniform electric field E = ?720 N/C is present above the plane as shown in in the figure below. The initial velocity vector of the protons makes an angle ? with the plane. The protons are to hit a target that lies at a horizontal distance of R = 1.18 mm from the point where the protons cross the plane and enter the electric field. We wish to find the angle ? at which the protons must pass through the plane to strike the target.

Explanation / Answer

a)

a = F/m = qE/m = 1.6e-19*720/1.673e-27 = 6.8858e10 m/s2

R = v^2 sin(2 theta)/a

==> 1.18e-3 = (9.93e3)*(9.93e3)/6.8858e10 sin(2*theta)

==> sin(2*theta) = 0.82402

==> 2 * theta = 55.4893

==> theta = 27.7 degrees

==> theta = 62.3 degrees

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