A proton of mass mp= 1.67×10−27 kg and a charge of qp= 1.60×10−19 C is moving...
A proton of mass mp= 1.67×10−27 kg and a charge of qp= 1.60×10−19 C is moving through vacuum at a constant velocity of 10,000 m/s directly to the east when it enters a region of uniform electric field that points to the south with a magnitude of E =3.62e+3 N/C . The region of uniform electric field is 5 mm wide in the east-west direction. How far (in meters) will the proton have been deflected towards the south by the time it exits the region of uniform electric field. You may neglect the effects of friction and gravity, and assume that the electric field is zero outside the specified region. Answer is to be in units of meters.
Homework Answers
acceleration of proton towards south, a = q*E/m
= 1.6*10-19*3.62*103/(1.67*10-27)
= 3.46*1011 m/s2
time taken to cross the electric field region, t = x/vx
= 5*10-3/10000
= 5*10-7 s
deflection in south direction, y = voy*t + (1/2)*a*t^2
= 0 + (1/2)*3.46*1011*(5*10-7)2
= 0.0432 m or 4.32*10-2 m
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