A uniform electric field exists in the region between two oppositely charged par

Question

A uniform electric field exists in the region between two oppositely charged parallel plates 1.69 cm apart. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate in a time interval 1.45×10^- .

Use 1.60×10^-19 for the magnitude of the charge on an electron and 1.67×10^-27 for the mass of a proton.

answer should be in N/C

B: Find the speed of the proton at the moment it strikes the negatively charged plate.

M/s


I tried doing this but kept getting wrong answers im not sure if I have calculation errors: I NEED CALCULATIONS PLEASE

take the equation of motion
s = u t + (1 / 2) a t2
as u = 0 and
s = 1.70 cm
= 0.017 m
t = 1.44×10-6
acceleration a = ....... m / s2
from the newtons second law of motion
F = m a
= ........ N
the electric field will be
E = F / q
= ...... N / C
for the velocity we use the equation ofmotion
v = u + a t
= a t
= ........ m / s

Explanation / Answer

v(ave) = delta x/delta t = 0.0161 m/1.57 x10^-7s = 0.01026 x 10^7m/s v(ave) = (1/2)(vi+vf) = 0.01026 x 10^7m/s since vi=0 vf = 0.02052 x 10^7m/s = 2.052 x 10^5 m/s or about 200,000 m/s Now to find the electric field E = F/q = ma/q =(m/q)a =(m/q)(delta v/delta t) = (1.67×10-27 kg/ 1.60×10-19 C)(2.052 x 10^5 m/s)/(1.57×10-6 s) E=(1.04375 x 10-8kg/C)(1.307 x 10^11m/s/s) E = 0.799 x 10^3 = 8 x 10^3 = 8000 V/m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.