1) An electron of mass = 9.11 x 10^-31kg and a charge of q = 1.60 × 10^19 C is r
ID: 1573594 • Letter: 1
Question
1) An electron of mass = 9.11 x 10^-31kg and a charge of q = 1.60 × 10^19 C is released, from rest, in a region of uniform electric field that points to the right with a magnitude of 285 N/C . How long (in seconds) does it take for the electron to reach a speed of E = 7320 m/s? Assume the experiment is performed in a vacuum and that you can ignore the effects of gravity and friction. Give you answer in seconds using the exponential form (i.e., use 1.23e-4 to represent 123.4 microseconds rather than 0.0001234).
2)A proton of mass = 1.67 × 10^27 kg and a charge of q = 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 = 3000 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.
Question 23 An electron of mass mp-9.11 × 10-31 kg and a charge of qp-1.60 × 10-19 C is released, from rest, in a region of uniform electric field that points to the right with a magnitude of |E-285 N/C. How long (in seconds) does it take for the electron to reach a speed of 7320 m/s? Assume the experiment is performed in a vacuum and that you can ignore the effects of gravity and friction. Give you answer in seconds using the exponential form (i.e, use 1.23e-4 to represent 123.4 microseconds rather than 0.0001234). Enter answer here seconds O of 5 attempts used Question 24 A proton of mass m,-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| =3000 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. Enter answer here metersExplanation / Answer
23)
electric force acting on the electron, Fe = q*E
m*a = q*E
a = q*E/m
= 1.6*10^-19*285/(9.11*10^-31)
= 5.00*10^13 m/s^2
time taken, t = (vf -vi)/a
= (7320 - 0)/(5*10^13)
= 1.46*10^-10 s
=1.46e-10 s
24)
acceleration of proton towards sounth, a = q*E/m
= 1.6*10^-19*3000/(1.67*10^-27)
= 2.87*10^11 m/s^2
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)*2.87*10^11*(5*10^-7)^2
= 0.0359 m or 3.59e-2 m
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.