A proton is accelerated from rest through a potential difference of 3kV. It then

Question

A proton is accelerated from rest through a potential difference of 3kV. It then enters a region where its velocity is perpendicular to an electric field. The field is created by two parallel plates separated by a distance (d) in cm with a potential difference of 250V across them. The magnitude of the magnetic field ? to E needed to allow the proton to pass undeflected between plates is 3.3*10^-3T.
a. Find the separation d
b. If the E field turned off, find the radius of the curvature R if the proton trajectory.

beyond lost!!! This book I have isn't helping me!! I really need a walk through or a step-by-step of this question

Explanation / Answer

A proton is accelerated from rest through a potential difference of V = 3.0*10^3 V     sped of the proton is 1/2 mv^2 = q V                                       1/2 *167*10^-27 kg v^2 = 1.6*10^-19 V * 3.0*10^3 V                                                  v = 7.74*10^5 m/s         v = E / B magnetic field is B = 3.3*10^-3 T        Electric field E = 3.3*10^-3 T * 7.74*10^5 m/s                                 = 2.55*10^3 N/C    The field is created by two parallel plates separated by a distance with a potential difference of 250V across them.                    V = E * d                    d = 250V / 2.55*10^3 N/C                         = 9.78 cm b )   radius of the curvature R if the proton trajectory is r = m v / qB                           = 1.67*10^-27 kg * 7.74*10^5 m/s / 1.6*10^-19 C * 3.3*10^-3 T                           = 2.44m

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