An electron is fired at 4.0 times 10^6 m/s horizontally between the parallel pla

Question

An electron is fired at 4.0 times 10^6 m/s horizontally between the parallel plates, as shown, starting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 times 10^2 N/C. The separation of the plates is 2.0 cm. a) Find the acceleration of the electron between the plates. b) Find the horizontal distance travelled by the electron when it hits the plate. c) Find the velocity of the electron as it strikes the plate.

Explanation / Answer

Given that

speed u=4*10^6 m/s

electric field E=4*10^3 N/c

distance b/w the plates d=2 cm

basing on the concept of the electrostatices

now we find the acceleration b/w the plates

acceleration a=qE/m=1.6*10^-19*4*10^3/9.1*10^-31=0.7*10^15 =7*10^14 m/s

now we find the horizantal distance travelled by electrons hit the plates

horizantal distance X=u[2y/a]^1/2

=4*10^6[2*2*10^-2/7*10^14]^1/2

=3*10^-2=3 cm

now we find the velocity f the electron strike the plate

v^2-(4*10^6)^2=2*7*10^14*2*10^-2

v^2=16*10^12+28*10^12

v^2=44*10^12

speed after hits =>V=6.6*10^6 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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