a] Kinetic energy = decrease in PE
= qE*d
Alpha particle has larger KE because of larger charge.
KE = 2*1.6e-19*200*0.04
= 2.56*10^-18 J
b] speed v = sqrt(2a*d)
d is same for both, but acceleration a of proton is larger as a=qE/m and for alpha a=2qE/4m
so speed of proton is higher.
v = sqrt(2*1.6e-19*200/1.67e-27*0.04)
= 39153 m/s
A proton (m, = 1.67 X 10-27 kg, qo=e= 1.60 X 10-19 C) and an alpha...
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If a proton m = 1.67 x 10-27 kg and charge q = 1.60 x 10-19 C is released from rest at the + plate in the below electric field, what will be its speed at the-plate? 3 cm E = 5000 N/C
A proton of mass m_p = 1.67 \times 10^{-27}~kgm p =1.67×10 −27 kg and a charge of q_p = 1.60 \times 10^{-19} ~Cq p =1.60×10 −19 C is moving through vacuum at a constant velocity of 10,000 ~m/s10,000 m/s directly to the east when it enters a region of uniform electric field that points to the south with a magnitude of | \vec{E} | = ∣ E ⃗ ∣=2.60e+3~N/C N/C . The region of uniform...
A proton of mass m_p = 1.67 \times 10^{-27}~kgm p =1.67×10 −27 kg and a charge of q_p = 1.60 \times 10^{-19} ~Cq p =1.60×10 −19 C is moving through vacuum at a constant velocity of 10,000 ~m/s10,000 m/s directly to the east when it enters a region of uniform electric field that points to the south with a magnitude of | \vec{E} | = ∣ E ⃗ ∣=2.35e+3~N/C N/C . The region of uniform...
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In a vacuum, a proton (charge = +e, mass = 1.67 times 10^-27 kg) is moving parallel to a uniform electric field that is directed along the +x axis (see figure below). The proton starts with a velocity of +4.30 times 10^4 m/s and accelerates in the same direction as the electric field, which has a value of +5.70 times 10^3 N/C. Find the velocity of the proton when its displacement is +2.0 mm from the starting point. Number Units...
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